The document provides a mathematical discussion on complex numbers, specifically proving certain identities and relationships involving equal modulus complex numbers and their imaginary components. It presents solutions to various problems, including proving equations involving trigonometric functions and identities. The document also includes detailed calculations and transformations related to these proofs.

1. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 1
COMPLEX NUMBERS
1. If Z1 , Z2 are non zero complex numbers of equal modulus and Z1 ≠ Z2 then prove that
1 2
1 2
Z Z
Z Z

 is purely imaginary.
SOLn
: Since Z1 and Z2 are two complex numbers with equal modulus (say r),
Let  1 1 1cos sinZ r i  
& 2 2 2(cos sin )Z r i  

 1 2 1 2 1 2
1 2 1 2 1 2 1 2
(cos cos ) (sin sin )
2cos .cos .2sin .cos
2 2 2 2
Z Z r i
r i
   
       
    
            
         
        
1 2
1 2 1 2 1 2
21 2
2 cos cos sin
2 2 2
2 cos . .........( )
2
i
r
r e i
 
     
 
 
 
 
         
       
      
 
  
 
Also    1 2 1 2 1 2cos cos sin sinZ Z r i         
1 2 1 2 1 2 1 2
1 2 1 2 1 2
1 2 1 2 1 2
2sin .sin .2cos .sin
2 2 2 2
1
2 sin cos sin
2 2 2
1
2 sin cos sin
2 2 2
r i
ir
i
ir i i
i
       
     
     
            
          
        
         
       
      
          
                

1 2
21 2
2 sin . ........( )
2
i
ir e ii
 
 
 
 
 


 
  
 
Dividing (i) by (ii) we get,
1 2 1 2 1 2
1 2
1
cot cot
2 2
z z
i
z z i
        
     
    
which is purely imaginary.
2. If
2 2
cos sin
3 3
r r
rx i 
   
    
   
prove that
(i) 1 2 3......... 1x x x    (ii) 1 2......... 1ox x x    
SOLn
: Now
2 2
cos sin
3 3
r r
rx i 
   
    
   
(i) Then 1 2 3........x x x 
2 2 3 3
2 2 2 2 2 2
cos sin cos sin cos sin
3 3 3 3 3 3
i i i     
               
                   
                   
2 3 2 3
2 2 2 2 2 2
cos sin
3 3 3 3 3 3
i     
              
                     
                 
2 2
2 2 2 2 2 2
cos 1 sin 1
3 3 3 3 3 3
i
                 
                    
                   

2. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 2
2 1 2 1
cos sin
2 23 31 1
3 3
2 2
cos 3 sin 3
3 3
cos2 sin 2
(1) (0)
1
i
i
i
i
 
 
 
   
   
      
    
   
   
         
 
 

[For G.P., Sum=
1
a
r
]
(ii) Also 0 1 2........x x x 
0 2 0 2
2 2 2 2 2 2
cos sin
3 3 3 3 3 3
     
              
                     
                 
2 2
2 2 2 2
cos 1 sin 1
3 3 3 3
i 
                
                    
                   
   cos 3 sin 3
cos3 sin3
( 1) (0)
1
i
i
i
 
 
   
 
  
 
3. If cos sin ,z i   prove that
2
1 tan
1 2
i
z

 

and
1
cot
1 2
z
i
z



SOLn
: (i) L.H.S.
2
1 z


2
2
1 cos sin
2
2cos 2sin .cos
2 2 2
1
cos cos sin
2 2 2
cos sin
2 2
cos
2
1 tan
2
. . .
i
i
i
i
i
R H S
 
  
  
 



 

   
   
   

 
 
 


 

(ii) L.H.S.
1
1
z
z



1 cos sin
1 cos sin
i
i
 
 
 

 

3. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 3
2
2
2cos 2sin cos
2 2 2
2sin 2sin cos
2 2 2
i
i
  
  
   
   
   
   
   
   
2cos cos sin
2 2 2
2sin sin cos
2 2 2
i
i
  
  
 
 
 
 
 
 
cos sin
2 2
cot
2
cos sin
2 2
i
i
 

 
 
 
 
 
  
 
cos sin
2 2
cot
2
cos sin
2 2
i
i
i
 

 
 
 
 
 
 
 
[Multiplying Numerator & Denominator by i]
cot
2
i

 . . .R H S
4. If   1 cos sin 1 cos2 sin2i i u iv        
Prove that (i)
2 2 2 2
16cos .cos
2
u v

  (ii)
3
tan
2
v
u


SOLn
: Now (1 cos sin )(1 cos2 sin2 )u iv i i        
 
 
2 2
2cos .2sin cos 2cos .2sin cos
2 2 2
2cos cos sin .2cos cos sin
2 2 2
4cos cos cos sin
2 2 2
i i
i i
i
  
  
  
  
  
  
 
   
 
 
   
 
    
       
    
3 3
4cos cos cos sin
2 2 2
u iv i
  

 
     
Comparing both sides, we get,
3
4cos .cos .cos .............( )
2 2
3
4cos .cos .sin ..............( )
2 2
u i
v ii
 

 



Squaring and adding (i) & (ii) we get,
2 2 2 2 2 23 3
16cos .cos cos sin
2 2 2
u v
  

 
   
 
2 2
16cos .cos
2


Dividing (ii) by (i) we get,

4. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 4
3
4cos cos .sin
2 2
3
4cos cos .cos
2 2
v
u
 

 


3
tan
2
v
u

 
5. If 1 i   , 1 i   and cot 1x   , prove that     ( )sin .cos
n n n n
x x ec         
SOLn
: Now 1 i   , 1 i   , cot 1x  
cot 1 1x i      
cot
cos
sin
cos sin
sin
cos (cos sin )
i
i
i
ec i



 

  
 
 


 
   cos cos sin ..........( )
n n
x ec n i n i      
Similarly cot 1 1x i     
cot i 
 cos cos sinec i    [As above]
   cos cos sin .......( )
n n
x ec n i n ii      
Subtracting (ii) from (i) we get,
     cos 2 sin
n n n
x x ec i n      
2 cos .sinn
i ec n 
 .cos .sinec n     [ 2 ]i  
6. Prove that 1
tan log
2
i i z
z
i z
  
  
 
SOLn
: Let 1
tan ..............( )z i

tanz  
Then
tan
tan
i z i
i z i


 

 
sin
cos
sin
cos
cos sin
cos sin
i
i
i
i




 
 






cos sin
cos sin
i
i
 
 



[Multiplying N & D by -i]
i
i
e
e



 2i
e 


5. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 5
log 2
1
log
2
log .............( )
2
i z
i
i z
i z
i i z
i i z
ii
i z



 
   
 
 
    
 
 
   
 
From (i) & (ii) we get
1
tan log
2
i i z
z
i z
  
  
 
7. If 5 3 3 5
sin6 .cos .sin .cos .sin .cos .sina b c         find the value of a, b, c.
Hence show that 4 2sin6
16cos 16cos 3
sin 2

 

  
SOLn
: Now  
6
cos6 sin6 cos sini i     
           
        
5 4 2 3 36 6 6 6
1 2 3
2 4 5 66 6
4 5
cos c cos sin c cos sin c cos sin
c cos sin c cos sin sin
i i i
i i i
      
    
   
  
6 5 4 2 3 3
cos 6 cos sin 15cos sin 20 cos sini i         
2 4 5 6
15cos sin 6 cos sin sini      
 6 4 2 2 4 6
cos 15cos sin 15cos sin sin        
 5 3 3 5
6cos sin 20cos sin 6cos sini        
Comparing imaginary part on both sides, we get,
5 3 3 5
sin6 6cos sin 20cos sin 6cos sin        
Comparing above equation with the given equation we get,
a = 6, b= -20, c = 6
Deduction:
5 3 3 5
sin6 6cos sin 20cos sin 6cos sin
sin 2 2sin cos
      
  
 

   
4 2 2 4
24 2 2 2
4 2 4 2 4
4 2
3cos 10cos sin 3sin
3cos 10cos 1 cos 3 1 cos
3cos 10cos 10cos 3 6cos 3cos
16cos 16cos 3
   
   
    
 
  
    
     
  
8. If 4 3
1 3 5 7sin cos cos cos3 cos5 cos7a a a a         ,prove that 1 3 5 79 25 49 0a a a a   
SOLn
: Let cos sin ,x i  
1
cos sini
x
   
Also cos sin ,n
x n i n  
1
cos sinn
n i n
x
   
Then
1
2cos ,x
x

 
  
 
1
2 sini x
x

 
  
 
,
1
2cosn
n
x n
x
  ,
1
2sinn
n
x n
x
 
Hence
4 3
4 3 1 1 1 1
sin .cos
2 2
x x
i x x
 
      
        
      
4 3
7 4
1 1 1
2
x x
i x x
   
     
   

6. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 6
3
7
3
2
2
6 2
2 6
7 3 5
5 3 7
7 5 3
7 5 3
1 1 1 1
2
1 1 1
128
1 1 3 1
3
128
1 3 1 3 1
3 3
128
1 1 1 1 1
3 3
128
x x x
x x x
x x
x x
x x x
x x x
x x x x
x x x x
x x x x
x x x x
     
        
     
  
    
  
  
      
  
 
         
       
              
       
 
 
1
2cos7 2cos5 6cos3 6cos
128
1
3cos 3cos3 cos5 cos7
64
   
   

 

   
   
Comparing this with the given equation we get,
1 3 5 7
3 3 1 1
, , ,
64 64 64 64
a a a a     
1 3 5 7
3 27 25 49
9 25 49 0
64
a a a a
  
     
9. Show that the  4
th
n power of
 
2
1 7
2
i
i


is ( 4)n
 where n is a positive integer.
SOLn
: Now
 
2 2
1 7 1 7
4 42
i i
i ii
 

 
  
  
   
1 7
3 4
1 7 3 4
3 4 3 4
3 28 21 4
9 16
25 25
1
25
i
i
i i
i i
i
i
i



 

 
  


 
   
Hence
 
 
4
4
2
1 7
1
2
n
ni
i
i
 
      
  
 
 
4
4
4
4
1
1
2
2 2
2 cos sin
4 4
n
n
n
n
i
i
i
 
 
  
   
  
 
  
 
 
   
2
2 cos sin
4 1 0 cos 1 ,sin 0
n
n nn
n i n
n n
 
 
 
        
   

( 4)n
 

7. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 7
10. Find the roots common to 6
0x i  and 4
1 0x   .
SOLn
: We have 6
x i
6
cos 2 sin 2
2 2
x k i k
 
 
   
       
   
[General polar form] [ cos sin ]
2 2
i i
 
 
   
1/6
cos 4 1 sin 4 1
2 2
x k i k
  
      
   cos 4 1 sin 4 1
12 12
x k i k
 
    
Putting k=0,1,2,3,4,5, we get the roots as,
5 5 9 9
cos sin , cos sin , cos sin ,
12 12 12 12 12 12
i i i
          
       
     
13 13 17 17 21 21
cos sin , cos sin , cos sin
12 12 12 12 12 12
i i i
          
       
     
5 5 3 3
. . cos sin , cos sin , cos sin ...........( )
12 12 12 12 4 4
i e i i i I
          
          
     
Also 4
1x  
   4
cos 2 sin 2x k i k        [General polar form] [ 1 cos sin ]i   
    1/4
[cos 2 1 sin 2 1 ]x k i k     
   cos 2 1 sin 2 1
4 4
x k i k
 
    
Putting k=0,1,2,3, we get the roots as,
3 3 5 5 7 7
cos sin , cos sin , cos sin , cos sin
4 4 4 4 4 4 4 4
i i i i
              
          
       
3 3
. . cos sin , cos sin ..................( )
4 4 4 4
i e i i II
      
      
   
From (I) and (II) we get the common roots as
3 3
cos sin
4 4
i
  
  
 
11. If , , ,    are the roots of 4 3 2
1 0x x x x     , find their values and show that
    1 1 1 1 5            2 3 4
[ 1 1 1 1 5]or        
SOLn
: Now 4 3 2
1 0x x x x    
  4 3 2
1 1 0x x x x x       [Multiplying both sides by (x-1)]
5
1 0x  
5
1x 
5
cos2 sin2x k i k   
2 2
cos sin
5 5
k k
i
 
  where k=0,1,2,3,4.
When k=0, Root cos0 sin0 1i  
When k=1, Root
2 2
cos sin
5 5
i
 
   (say)
When k=2, Root
4 4
cos sin
5 5
i
 
   (say) 2


8. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 8
When k=3, Root
6 6
cos sin
5 5
i
 
   (say) 3

When k=4, Root
8 8
cos sin
5 5
i
 
   (say) 4

Since , , ,    are the roots of 4 3 2
1 0x x x x     , we have
     4 3 2
1x x x x x x x x           
Putting x=1, we get
    1 1 1 1 1 1 1 1 1 5            
Note : 2 3 4
, ,       
Hence     2 3 4
1 1 1 1 5       
12. Prove that 5 2 2 3
1 ( 1) 2 cos 1 2 cos 1 0
5 5
x x x x x x
   
         
  
SOLn
: Consider 5
1 0...........( )x I 
5
1x 
5
cos2 sin2x k i k     1 cos0 sin0i 
 
1/5
cos2 sin 2x k i k   
2 2
cos sin
5 5
k k
x i
    
     
   
When k=0, cos0 sin0 1x i  
k=1,
2 2 3 3 3 3
cos sin cos sin cos sin
5 5 5 5 5 5
x i i i
     
 
       
               
       
k=2,
4 4
cos sin cos sin cos sin
5 5 5 5 5 5
x i i i
     
 
       
               
       
k=3,
6 6
cos sin cos sin cos sin
5 5 5 5 5 5
x i i i
     
 
       
               
       
k=4,
8 8 3 3 3 3
cos sin cos sin cos sin
5 5 5 5 5 5
x i i i
     
 
       
               
       
Also 5 3 3
1 ( 1) cos sin cos sin
5 5 5 5
x x x i x i
         
              
      
3 3
cos sin cos sin
5 5 5 5
x i x i
         
           
      
 
 
 
2 2 2 2
2 2
1 cos sin cos sin
5 5 5 5
3 3 3 3
cos sin cos sin
5 5 5 5
3 3
1 cos sin cos sin
5 5 5 5
3
1 2 cos 1 2 cos 1
5 5
x x i x i
x i x i
x x x
x x x x x
   
   
   
 
  
       
  
  
     
  
          
               
             
 
      
 
...............( )II

 
 
From(I) and (II) we have

9. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 9
 5 2 2 3
1 1 2 cos 1 2 cos 1 0
5 5
x x x x x x
   
         
  
13. If cos ,
4
ec ix u iv
 
   
 
then prove that (i) 2 2
2sec 2 ,u v h x  (ii)    
22 2 2 2
2u v u v  
SOLn
: (i) Now cos
4
u iv ec ix
 
   
 
cos
4
u iv ec ix
 
    
 
Then   2 2
u v u iv u iv   
cos cos
4 4
2
2sin sin
4 4
2
cos2 cos
2
2
cosh 2 0
2sec 2
ec ix ec ix
ix ix
ix
x
h x
 
 

   
     
   

   
    
   





(ii) Now cos
4
u iv ec ix
 
   
 
 
 
2 2
1
sin
4
1
sin cos cos sin
4 4
2
cosh sinh
2 cosh sinh
cosh sinh
2 cosh sinh
cosh 2
ix
ix ix
x i x
x i x
x x
x i x
x

 

 
 
 









Comparing both sides we get,
2 cosh 2 sinh
,
cosh 2 cosh 2
x x
u v
x x
 
Then  
22 2
22 2
2 2
2cosh 2sinh
cosh 2 cosh 2
x x
u v
x x
 
   
 

10. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 10
22 2
2
2
2
2
cosh sinh
4
cosh 2
cosh 2
4
cosh 2
4
.......................(I)
cosh 2
x x
x
x
x
x
 
  
 
 
   

Also  
2 2
2 2
2 2
2cosh 2sinh
2 2
cosh 2 cosh 2
x x
u v
x x
 
   
 
2 2
2
cosh sinh
4
cosh 2
x x
x
 
  
 
2
4
....................(II)
cosh 2x

From (I) and (II) we have,    
22 2 2 2
2u v u v  
14. If  x iy c cot u iv   then show that

 
sin sinh2 cosh2 cos2
yx c
u v v u
Soln
: Now  x iy c.cot u iv   ,
 x iy c.cot u iv   
Adding two equations we get,
   2 cot u iv cot u ivx c     
 
 
 
 
  
  
  
cos cos
sin sin
u iv u iv
c
u iv u iv
       
   
     
  
  
sin cos cos .sin
sin .sin
u iv u iv u iv u iv
c
u iv u iv
 
   
  
 
 
sin
2sin .sin
c u iv u iv
x
u iv u iv
 x =
sin2
cos2 cos2
c u
iv u

sin2 cosh2 cos2
x c
u v u


……..(i)
Similarly subtracting we get,
2iy = c [cot (u+ iv) – cot (u-iv)]
 iy =
 

sin 2
cos2 cos2
c iv
iv u
[as above]
 iy =
sinh2
cosh2 cos2
ic v
v u



sinh2 cosh2 cos2
y c
v v u



……(ii)
From (i) & (ii) we get,
sin2 sinh2 cosh2 cos2
x y c
u v v u

 

Alternately,

11. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 11
x+ iy =
 
 


cos
sin
c u iv
u iv
=
   
   
  
 
.2cos sin
2sin sin
c u iv u iv
u iv u iv
=
 

sin2 sin2
cos2 cos2
c u iv
iv u
=
 

sin2 sinh2
cosh2 cos2
c u i v
v u
Comparing both sides we get,


sin2
cosh2 cos2
c u
x
v u

sin2
x
u
=
cosh2 cos2
c
v u
……….(i)
and



sinh2
cosh2 cos2
c v
y
v u

sinh2
y
v

=
cosh2 cos2
c
v u
……….(ii)
From (i) and (ii) we get,
sin2 sinh2 cosh2 cos2
x y c
u v v u

 

15. If log (tan x) = y then prove that (i)  
1
sinh (tan cot )
2
n n
ny x x
(ii)    cosh n 1 y cosh n 1 y 2 cosh ny. cosec2x   
Soln
: Now log (tan x) = y
 tan & coty y
e x e x
 
(i) sinh ny =
2
ny ny
e e

=   
1
( ) ( )
2
y n y n
e e
=   
1
(tan ) (cot )
2
n n
x x
=   
1
tan cot
2
n n
x x
(ii)    
            
 
    
1 1 1 1
cosh n 1 y cosh n 1 y
2 2
n y n y n y n y
e e e e
   
  

. . . .
2
ny y ny y ny y ny y
e e e e e e e e
     
  

2
ny y y ny y y
e e e e e e
 
 

( )( )
2
ny ny y y
e e e e
 
   
   
  
2
2 2
ny ny y y
e e e e

12. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 12
 2 cosh ny. cosh y I 
Alternately,        cosh n 1 y cosh n 1 y cos i n 1 y cos i n 1 y      
2 cos i ny. cos iy
 2 cosh ny. cosh y I 
But


cosh
2
y y
e e
y
=
tan cot
2
x x
=
 
 
 
1 sin cos
2 cos sin
x x
x x
=
2 2
sin cos
2cos .sin
x x
x x

=
1
sin2x
= cosec 2x
Subs . in (I) we get,
cosh (n+ 1) y + cosh (n- 1) y = 2 cosh ny. cosec 2x
16. If  
1
tan z 1+ i
2
Prove that 
 11
tan 2 log5
2 4
i
z
Soln
: Let  z x iy I  
Now tan z =  
1
1
2
i
 tan(x+ iy) =  
1
1
2
i
& tan(x- iy) =
1
(1 )
2
i
Then tan 2x = tan [ (x+ iy) + (x- iy)]
=
   
   
  
  
tan tan
1 tan .tan
x iy x iy
x iy x iy
=
   
   
  
  
1 1
1 1
2 2
1 1
1 1 . 1
2 2
i i
i i
=
 
 
  
1 1
2
1 1
1 1 1 1
4 2
2x = 1
tan (2)
x =  11
tan 2
2

Also tan 2iy = tan[(x+ iy) – (x- iy)]
=
2
3
i
(try at home)

13. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 13
 i tanh 2y =
2
3
i
 2y =   
 
 
1 2
tanh
3
=
2
1
1 3log
22 1
3
 
 
 
 
 
=  
1
log 5
2
 y =  
1
log 5
4
Subs. in (I) we get, z =    
11
tan 2 log 5
2 4
i
17. Find the sum of the series 2 3
sin sin2 sin3 sin
..... .............
cos cos cos cosn
n
S
   
   
      
Soln
: Let
   
   
      2 3
cos cos2 cos3 cos
..... .............
cos cos cos cosn
n
C
 C +iS = 2 3
cos sin cos2 sin2 cos3 sin3
cos cos cos
i i i     
  
  
  ………
=
2 3
2 3
cos cos cos
i i i
e e e  
  
   .....
=
1 2 3
cos cos cos
i i i
e e e  
  
     
       
     
……...
= cos
1
cos
i
i
e
e





[
1
a
S
r
 

for a Geom. Series]
=
cos
i
i
e
e


 
= cos sin
cos cos sin
i
i
 
  

 
=
cos sin
sin
i
i
 



=
cot
1
i



= i cot 1 
Equating the imaginary parts, we get,
2 3
sin sin2 sin3
....... cot
cos cos cos
  

  
    
Equating the real parts, we get,
2 3
cos cos2 cos3
....... 1
cos cos cos
  
  
     

14. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 14
18. If u + iv =
1 1
log
1
i
i
ie
i ie


 
 
 
prove that
2
u

 and log(sec tan )v   
Soln
: Now
1
1
i
i
ie
ie




=
 
 
 
 
 
 
1 cos sin
1 cos sin
i i
i i
=
   
   
   
   
   
   
   

   
1 sin cos 1 sin cos
1 sin cos 1 sin cos
i i
i i
=
   
 
       
 
     
 
2 2
22
1 sin cos cos sin cos cos cos sin
1 sin ) (cos
i
=
 
    2 2
2cos
1 2sin sin cos
i
=
 


2 cos
2 1 sin
i
=
cos
1 sin
i 


1
log
1
i
i
ie
ie


 
 
 
=
cos
log log
1 sin
i


 
  
 
=
2
i 1 sin
log
cos


 
  
 
= log(sec tan )
2
i
   ………(i)
But u+ iv =
1
i
1
log
1
i
i
ie
ie


 
 
 
 u+ iv =
1
i
log(sec tan )
2
i
 
 
   
[Subs. from (i)]
=
1
log(sec tan )
2
i i
i

 
 
     

Comparing both sides we get,
, log(sec tan )
2
u v

   
19. Find the value of log [ sin(x+ iy) ]
Soln
: sin (x+ iy) = sin x.cos iy + cos x. sin iy
= sin x. cosh y + i cos x. sinh y
 [sin(x+ iy)] = log (sin x. cosh y + i cos x. sinh y)
= 2 2 2 2 1 cos .sinh
log( sin cosh cos sinh ) tan
sin .cosh
x y
x y x y i
x y
  
   
 
=             
2 2 2 2 11 tanh
log sin .cosh cos . cosh 1 tan
2 tan
y
x y x y i
x
=    
 2 2 11
log cosh cos tan cot .tanh
2
y x i x y
=    
  
 
11 1 cosh2 1 cos2
log tan cot .tanh
2 2 2
y x
i x y

15. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 15
=   
 
 
11 cosh2 cos2
log tan cot .tanh
2 2
y x
i x y
20. If
 
 
 



 

x iy
x iy
a ib
i
a ib
, find  and 
Soln
: Now i  =
 
 




x iy
x iy
a ib
a ib
   log i =            log logx iy a ib x iy a ib
=                    log log log logx a ib a ib iy a ib a ib ……….(I)
But     
     
 
2 2 1
log log tan
b
a ib a b i
a
    
     
 
2 2 1
log log tan
b
a ib a b i
a
Subs in (I) we get,
                
1 2 2
log 2 tan 2log
b
i x i iy a b
a
               
1 2 2 2 2 2 2
2 tan log( ) 2log ( ) log( )
b
i x y a b a b a b
a
i (say)
 cos sini
i e i
      
Hence  =cos , sin  where  1 2 2
2 tan log( )
b
x y a b
a
  
   
 
21. Prove that
   
   
 
      
           
1
2 2
2
2 tan
a b i a b ab
log i n
a b i a b a b
Soln
: Let a – b = x, a + b = y
Then
   
   
     
            
( ) ( )
a b i a b x iy
log log log x iy log y ix
a b i a b y ix
2 2 1 2 2 1
log 2 tan log 2 tan
y x
x y i p x y i q
x y
  
        
              
         
1 1
2 ( ) tan tan
y x
i p q i
x y
     
       
    
1
2 tan
1 .
y x
x y
i n i
y x
x y
 
 
 
   
 
  
(Putting n = p – q)
2 2
1
2 tan
1 1
y x
xy
i n 
  
  
   
  
    

16. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 16
2 2
1
2 tan
2
y x
i n
xy
 
  
   
  
   
  
 
    
    
     
2 2
1
2 tan
2
a b a b
i n
a b a b
1
2 2
2
2 tan
ab
i n
a b
   
      
22. Show that if (1 tan )
(1 tan ) i
i 
 
 has real values then one of them is
2
sec
(sec ) 

Soln
:   
   
 (1 tan ) log(1 tan )(1 tan )
(1 tan ) i i i
i e
 

   
    
  
2 1 tan
log 1 tan tan 1 tan
1
i i
e
     

logsec 1 tani i
e
          

logsec tan tan .logseci
e
For the given expression to be real we must have tan logsec 0   
tan logsec     ………(i)
Then value of expression    

log sec tan
e
  

2
logsec tan logsec
e [ Subs. from(i) ]
   

2
logsec 1 tan
e
  
     
2
logsec .sec 2 2
1 tan sece
   
     
2
sec (logsec )
sec sece
23. If     cos sin cos2 sin2 ..... cos sin 1i i n i n        
Then show that the general value of θ is
 
4
1
r
n n


Soln
: Now     cos sin cos2 sin2 .... cos sin 1i i n i n        
   
   
   1 1
2 2
cos 2 ...... sin 2 ...... 1
cos 1 2 ..... sin 1 2 ..... 1
cos sin 1
n n n n
n i n
n i n
i
 
     
 
 
      
            
     
   
 1
2sin 0
n n


  
  ……[By comparing imaginary parts]
 1
2 2
n n
r



   cos2 1&sin2 0r r  
 
4
1
r
n n

 
 
24. If Z1 , Z2 and Z3, Z4 are two pairs of conjugate complex numbers
Then show that (i) 31
4 2
,
zz
amp amp
z z
   
   
   
(ii) 31
4 2
mod mod 1
zz
z z
   
    
   
Soln
: Let 1 1
i
z re
 and 3 2
i
z r e

2 2
i
z r e
  and 4 2
i
z r e


17. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 17
Then
 1 1 1
4 22
i
i
iz r e r
z rr e
e


 


 
 1
4
z
zamp     &  1 1
4 2
mod z r
z r ……(i)
Also
 3 1 1
2 22
i
i
z ir e r
z rr e
e


 


 
 3
2
z
zamp     &  3 1
2 2
mod
z r
z r ………(ii)
Hence from (i) & (ii),
   31
4 2
zz
z zamp amp &    31
4 2
mod mod
zz
z z =1
25. If 1 2 1 2z z z z   Show that 2
1 2
z
amp
z
 
  
 
Soln
: Let  1 1 1 1cos sinz r i  
 2 2 2 2cos sinz r i  
   
   
1 2 1 1 2 2 1 1 2 2
1 2 1 1 2 2 1 1 2 2
cos cos sin sin
& cos cos sin sin
z z r r i r r
z z r r i r r
   
   
     
    
But 1 2 1 2z z z z   …..(given)
2 2
1 2 1 2z z z z   
   
2 2
1 1 2 2 1 1 2 2cos cos sin sinr r r r      
2 2 2 2 2 2
1 1 2 2 1 2 1 2 1 1 2 2cos cos 2 cos cos sin sinr r rr r r         
2 2 2 2
1 2 1 2 1 1 2 2 1 2 1 22 sin sin cos cos 2 cos cosrr r r rr        
2 2 2 2
1 1 2 2 1 2 1 2sin sin 2 sin sinr r rr     
∴ 1 2 1 2 1 2 1 24 cos cos 4 sin sin 0rr rr    
1 2 1 24 cos( ) 0rr     1 20, 0r r 
2 1 2

    
 1 2
z
zamp 
  
26. If     1 1 2 2 ... n nx iy x iy x iy x iy    
Show that (i)
1 1 1 11 2
1 2
tan tan ...........tan tann
n
yy y y
x x x x
         
        
      
(ii)     2 2 2 2 2 2 2 2
1 1 2 2 ..... n nx y x y x y x y    
Soln
: Let . i p
p p px iy r e
 
Where 2 2
p p pr x y 
&  1
tan p
p
y
p x 
 p=1,2,3…..n
Let . i
x iy r e
 
Where 2 2
r x y 
&  1
tan y
x 

Now ,     1 1 2 2 ...... n nx iy x iy x iy x iy     (given)
    1 2
1 2 ..........i i i n i
nre r e r e re   
 
   1 2 ....
1 2. ....... .n i
nr r r e r e
    
 

18. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 18
(i) Comparing amplitude we get,
1 2 ........ 0n    
i.e.        1 2
1 2
1 1 1 1
tan tan ....tan tann
n
yy y y
x x x x
   
  
(ii) Comparing modulii we get,
1 2 3. . .......r r r r r
∴
2 2 2 2
1 2. ....... nr r r r (Squaring both thet sides)
∴     2 2 2 2 2 2 2 2
1 1 2 2 .... n nx y x y x y x y    
27. Prove that 2cos
z z
zz

 
  
 
Soln
: Let  cos sin . i
z a i a e
   
 cos sin . i
z a i a e 
  
   
Then 2
i
i
i
z ae
e
aez



 
& 2iz
e
z


Hence
2 2
2 2
2 2cos2
2
i i
i iz z e e
e e
zz
 
 


  
     
 
28. If  
1/3
x iy a ib   then prove that  2 2
4
x y
a b
a b
  
Soln
: Let cosa r  , sinb r 
 3
cos3 sin3x iy r i    
Comparing both the sides,
3 3
cos3 , cos3x r y r  
Then
3 3
cos3 sin3
cos sin
yx r r
a b r r
 
   
 
2
2 sin3 cos cos3 sin
cos sin
sin4
cos sin
r
r    
 

 



 2
2 2sin cos cos2
cos sin
r   
   sin2 2sin cos  
 
   
 
2
2 2 2
2 2
2 2
4 cos 2
4 cos sin
4 cos sin
4
r
r
r r
a b

 
 

 
 
 
29. If
2 2 2
1a b c   and  1b ic a z   then prove that
1
1 1
a ib iz
c iz
 

 
Soln
: Now 1
b ic
az 

Then
 
 
1
1
11
1 1
b ic
a
b ic
a
iiz
iz i






 

19. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 19
   
   
1
1
1
.......( )
1
a ib c
a ib c
a ib c
i
a ib c
  

  
  

  
But
2 2 2
1a b c  
     
2 2 2
1
1 1
1
1
a b c
a ib a ib c c
a ib c
c a ib
   
     
 
 
 
   
   
11
.....( )
1 1
a ib ca ib c
ii
c a ib a ib c
   
  
    
[By equal ratio theorem]
From (i) & (ii) we get,
1
1 1
iz a ib
iz c
 

 
30. Find two complex numbers such that their difference is 10i and their product is 29
Soln
: Since thet difference between two complex numbers is imaginary and their product is real,
The two numbers must be conjugates
Let the numbers be 1z x iy  and 2z x iy 
Now 1 2 10z z i 
    10x iy x iy i    
2 10iy i 
5y 
Also 1 2 29z z 
( )( ) 29x iy x iy   
2 2
29x y  
 
22
5 29x  
2
24x 
2x  
Hence the two numbers are 2 5i & 2 5i
Or 2 5i  & 2 5i 
31. If  arg 2
4
z i

  and  
3
arg 2
4
z i

  , find z
Soln
: Let z x iy 
 2 2z i x i y    
 & 2 2z i x i y   
Now  arg 2
4
z i

 
1 2
tan
4
y
x
  
  
 
2
tan 1
4
y
x

  
2
2.........( )
y x
x y i
  
  
Also  
3
arg 2
4
z i

 

20. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 20
1 2 3
tan
4
y
x
  
  
 
2 3
tan 1
4
y
x

   
2
2.........( )
y x
x y ii
   
  
From (i) & (ii) we get,
2, 0x y 
Then 2 0 2z i  
32. If   2
0 1 21 .........
n n
nx p p x p x p x     then Show that (i)
/2
0 2 4.... 2 cos
4
n n
p p p

  
(ii)
/2
1 3 5.... 2 sin
4
n n
p p p

  
Soln
: Now  2 3 4 5
0 1 2 3 4 5 .... 1
n
p p x p x p x p x p x x       
Putting x i we get,
 0 1 2 3 4 5........ 1
n
p ip p ip p ip x      
   0 2 4 1 3 5
1
... ... 2
2 2
n
i
p p p i p p p
  
         
  
( 2) cos sin
4 4
n
n
i
  
  
 
/2
2 cos sin
4 4
n n n
i
  
  
 
(i) Comparing real parts we get,
/2
0 2 4 ..... 2 .cos
4
n n
p p p

   
(ii) comparing imaginary part we get,
/2
1 3 5 ..... 2 .sin
4
n n
p p p

   
33. If  1 3
n
n nx iy i   then prove that
1
1 1. . 4 3n
n n n nx y x y 
  
Soln
: Now
1 3
2
2 2
n
n n
i
x iy
  
     
   
2 cos sin
3 3
2 cos sin
3 3
n
n
n
i
n n
i
 
 
 
  
 
 
  
 
2 cos
3
n
n
n
x

  , 2 sin
3
n
n
n
y


Hence
 1
1
1
2 cos
3
n
n
n
x



 ,
 1
1
1
2 sin
3
n
n
n
y




Then
   1 1
1 1
1 1
. . 2 cos .2 sin 2 cos .2 sin
3 3 3 3
n n n n
n n n n
n nn n
x y x y
   
 
 
  

21. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 21
 
 2 1
2 1
2 1 2 2 1
1
2 sin sin cos cos sin sin
3 3
2 sin
3
3
2 . 2 . 3 4 . 3
2
n
n
n n n
nn
a b a b a b




  
 
        
 

  

34. If 1 2 3 0z z z   and 1 2 3z z z k   show that
1 2 3
1 1 1
0
z z z
  
Soln
: Now 1 2 3z z z k  
Let
   
   
   
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
cos sin cos sin
cos sin cos sin
cos sin cos sin
z r i k i
z r i k i
z r i k i
   
   
   
   
   
   
But 1 2 3 0z z z  
  1 2 3 1 2 3(cos cos cos ) sin sin sin 0k i            
1 2 3cos cos cos 0      ………(i)
1 2 3sin sin sin 0     ………(i)
Then      1 1 2 2 3 3
1 2 3
1 1 1 1 1 1
cos sin cos sin cos sini i i
z z z k k k
            
   1 2 3 1 2 3
1
cos cos cos sin sin sin
k
            
 
1
0 0i
k
    [ using (i)]
0
35. If sin sin ,cos cos 0     
Show that  ( )cos2 cos2 2cosi        
 ( )sin2 sin2 2sinii        
Soln
: Let cos sinx i  
cos siny i  
   cos cos sin sinx y i        
 0 0
0
i 

 
      
       
2
2 2
2 2
0
2 0
2
cos2 sin 2 cos2 sin 2 2 cos sin cos sin
cos2 cos2 sin 2 sin 2 2 cos sin
x y
x xy y
x y xy
i i i i
i i
       
       
  
   
   
       
          
   2 cos sini            
Comparing both thet sides we get,
 
 
cos2 cos2 2cos
sin 2 sin 2 2sin
    
    
   
   

22. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 22
36. If cos3 sin3 ,a i   cos3 sin3 ,b i   cos3 sin3c i  
Prove that  3 3 2cos
ab c
c ab
     
Soln
: Now
  
 
cos3 sin3 cos3 sin3
cos3 sin3
i iab
c i
   
 
 


   cos 3 3 3 sin 3 3 3i          
   
1/3
3 cos 3 3 3 sin 3 3 3
ab
i
c
             
   cos sin ......( )i i          
Hence    3 cos sin ......( )
c
i ii
ab
          
Then (i) + (ii) gives,
 3 3 2cos
ab c
c ab
     
37. If cos sina i   then show that   2
1 1 2cos cos sina a a i     
Soln
:    
22
1 1 cos sin cos sina a i i         
 
   
  
2
1 cos sin cos2 sin 2
1 cos2 sin 2 cos sin
2cos 2sin cos cos sin
2cos cos sin 1 cos sin
2cos 1 cos sin
i i
i i
i i
i i
i
   
   
    
    
  
    
    
   
   
  
38. Prove that
 2
1 cos sin
cot .
1 cos sin 2
i ai
e
i
  
 
 

 
Soln
: L.H.S.
1 cos sin
1 cos sin
i
i
 
 
 

 

23. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 23
2
2
2cos 2sin .cos
2 2 2
2sin 2sin .cos
2 2 2
2cos cos sin
2 2 2
2sin sin cos
2 2 2
cos sin
2 2
cot
2
cos sin
2 2 2 2
cot cos sin
2 2 2 2 2 2 2
cot
2
i
i
i
i
i
i
i
  
  
  
  
 

   
      




 
 
 
 
 
 
 
 
 
   
     
   
    
         
    

2
cos sin
2 2
cot .
2
. . .
i
i
e
R H S


 
 

 
 
 
    
      
    


39. If sin tani  prove that
1 tan
2cos sin tan
4 21 tan
2
i

 
 


 
    
 
Soln
: Now sin tani 
sin
sin
cos
cos 1
sin sin
i
i




 
 
 
cos sin 1 sin
cos sin 1 sin
i
i
  
  
 
 
 
[ Using componendo - dividendo]
   
2 2
2 2
cos sin 2sin cos
2 2 2 2cos sin . cos sin
cos sin 2sin cos
2 2 2 2
i i
   
   
   
 
   
 
2 2
1 cos sin
2 2
&sin 2sin cos
2 2
 
 

 
   
 
 
  
 
2
2
2
cos sin
2 2
cos sin
cos sin
2 2
cos sin
2 2cos sin
cos sin
2 2
i
i
 
 
 
 
 
 
 
 
   
 
 
 

  

1 tan
2
1 tan
2





[ Dividing N & D by cos
2

]

24. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 24
tan tan
4 2
1 tan .tan
4 2
 
 



1 tan
4
 
  

tan
4 2
  
  
 
40. Using De Moivre’s Theorem show that    2 2
2 1 cos8 4 2x x    where 2cosx 
Soln
: Now,  
4
cos4 sin 4 cos sini i     
Expanding R.H.S. by Binomial Theorem and Comparing real parts we get,
4 2
cos4 8cos 8cos 1    
4 2
8 8 1
2 2
x x   
     
   
 2cosx 
4
2
2 1
2
x
x  
4 2
2cos4 4 2x x   
 
22 4 2
4cos 4 4 2x x    ( Squaring both the sides )
   
24 2
2 1 cos8 4 2x x     2
2cos 4 1 cos8    
41. Show that
1
2 cot 1
1
1
a
ai b bi
e
bi


 
  
Soln
: Now
1
1 1
bi b i
bi b
 

 
[ Multiplying N & D by -i]
i
i
re
re


 Where
2
1r b  &
1 11
tan cot b
b
  
 
1
2
2 cot
i
i b
e
e




1
2 cot1
1
a
ai bbi
e
bi


 
 
 
Hence
1
2 cot 1
. 1
1
a
ai b bi
e
bi


  
 
 
42. If ,  are the root of the equation 2
3. 1 0x x   , prove that 2cos
6
n n n
  
Hence , deduce that 12 12
2  
Soln
: Now 2
3 1 0x x  
,  are its roots we have
      
 
2
3 3 4 1 1
,
2 1
 
    

3 1
2
 

Let
3
2
i



3
2 2
i
 

25. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 25
cos sin
6 6
i
 
 
cos sin
6 6
n n n
a i
 
 
Similarly cos sin
6 6
n n n
i
 
  
Hence 2cos
6
n n n
  
Putting n=12 we get
 12 12
2cos2 2 1 2     
43. If ,  are the roots of
2 2
sin sin2 1 0z z    . Prove that 2cos .cosn n n
n ec    
Soln
: Now
      
 
2
2
sin 2 sin 2 4 sin 1
,
2 sin
  
 

    

2 2 2
2
2
2sin cos 4sin cos 4sin
2sin
cos cos 1
sin
   

 

 

 

 cos cos sinec i    2 2
cos 1 sin     
 
 
cos cos sin
& cos cos sin
n n
n n
ec n i n
ec n i n
   
   
 
 

Adding we get, 2cos .cosn n n
ec n    
44. Find the continued product of  
1/
1
n

Soln
: Now    1 cos 2 sin 2k i k       
     
1/
1 cos 2 1 sin 2 1
n
k i k
n n
 
     
When 0, cos sink value i
n n
 
  
When
3 3
1, cos sink value i
n n
 
  
When
5 5
2, cos sink value i
n n
 
  
.
.
.
.
When    1, cos 2 1 sin 2 1k n value n i n
n n
 
     
 Continued product od all the values
   
     
3 5 3 5
cos ...... 2 1 sin ...... 2 1
cos 1 3 5 .......... 2 1 sin 1 3 5 .......... 2 1
n i n
n n n n n n n n
n i n
n n
       
 
   
               
   
               
Now 1+3+5+……. 2 1n  is an A.P. with 1, 2, 2 1a d l n   

26. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 26
 Its    2 1
2 2
n n
sum a n d a l      
 1 2 1
2
n
n  
2
n
Hence, Required Product of Values 2 2
cos . sin .n i n
n n
    
    
   
cos sin
( 1) (0)
( 1)
n
n
n i n
i
  
  
 
45. Find the cube root of  1 cos sini  
Soln
: Let
3
1 cos sinz i   
 
2
2sin 2sin cos
2 2 2
2sin sin cos
2 2 2
2sin cos sin
2 2 2 2 2
2sin cos sin
2 2 2 2 2
2sin cos 2 sin 2
2 2 2 2 2
2sin cos 4 1
2 2
i
i
i
i
k i k
k
  
  
    
    
    
 
 
 
 
  
 
    
       
    
    
       
    
    
         
    
    sin 4 1
2 2 2
i k
      
      
    
   
1/3
2sin cos 4 1 sin 4 1
2 2 2 6 6
z k i k
          
            
      
When K=0, Z1
1/3
2sin cos sin
2 6 6 6 6
i
          
         
      
When K=1, Z2
1/3
3 3
2sin cos sin
2 6 6 6 6
i
          
         
      
When K=3, Z3
1/3
7 7
2sin cos sin
2 6 6 6 6
i
          
         
      
46. Solve
4 3 2
1 0x x x x    
Soln
: Now   4 3 2
1 1 0x x x x x      [ Multiply by (x+1) on both the side]
   
   
5
5
1/5
1 0
1 cos 2 sin 2
cos 2 1 sin 2 1
x
x k i k
x k i k
   
 
  
      
      
   cos 2 1 sin 2 1
5 5
k i k
 
   
When k=0, Root cos sin
5 5
i
 
 
When k=1, Root
3 3
cos sin
5 5
i
 
 

27. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 27
When k=2, Root cos sin 1 (0) 1i i       
When k=3, Root
7 7
cos sin
5 5
i
 
 
When k=4, Root
9 9
cos sin
5 5
i
 
 
Discarded, 1x   ( as we have taken it in the equation)
Also
7 7 3 3 3 3
cos sin cos 2 sin 2 cos sin
5 5 5 5 5 5
i i i
     
 
   
         
   
&
9 9
cos sin cos 2 sin 2 cos sin
5 5 5 5 5 5
i i i
     
 
   
         
   
Hence Required Roots are
3 3
cos sin
5 5
i
 
 ,
3 3
cos sin
5 5
i
 

47. Given that 4 3 2
1 2 is one root of the equation x 3 8 7 5 0i x x x      . Find the other roots.
Soln
: Since 1 2 is one of the equationi
1 2 is is the other rooti 
 The equation with this root is
2
( ) ( ) 0x sum x product  
    2
2
2 4 3 2
. . x 1 2 1 2 1 2 1 2 0
. . x 2 5 0
x 2 5 must be factor of x 3 8 7 5
i e i i x i i
i e x
x x x x
       
  
      
For finding the other factor we have to divide
2
x 1 0Then x  
1 1 4 1 3
x=
2 2
the required roots are
1 3
1 2i,
2
Hence
  
 


48. Show that all the roots of    
7 7
1 1x x   are given by cot where k=1,2,3.
7
k
i
 
  
 
Soln
:
   
   
   
 
7 7
7
7
1 1
1
1
1
1
cos2 sin 2
1
1 2 2
cos sin
1 7 7
2 2
cos sin 11 1 7 7 Componendo-Dividendo
2 21 1 cos sin 1
7 7
x x
x
x
x
k k
x
x k k
x
k k
ix x
By
k kx x i
 
 
 
 
  
 
  
 
 
   
 

  

    
        
22
2 2
1 cos 2cos2cos 2sin .cos
27 7 7
2 2
2sin 2sin .cos 1 cos 2sin
7 7 7 2
k k k
i
x
k k k
i
   
   

 
   
   
    
  


28. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 28
2cos cos sin
7 7 7
2sin cos sin
7 7 7
cos sin
7 7
cot [Multiplying N & D by -1]
7
cos sin
7 7
=-icot
7
k k k
i
k k k
i
k k
i
k
i
k k
i
k
  
  
 

 

 
  
 
  
 
   
 
  
k=0, x= -icot0 not defined (Hence discard)
k=1, x= -icot
7
2
k=2, x= -icot
7
3
k=3, x= -icot
7
4 3 3
k=4, x= -icot cot cot
7 7 7
k=5
When
i i



  


 
    
 
5 2 2
, x= -icot cot cot
7 7 7
6
k=6, x= -icot cot cot
7 7 7
Hence the solution are given by icot where k=1,2,3
7
i i
i i
k
  

  


 
    
 
 
    
 

49. Show that the points representing the roots of the equation  
33
1z i z  on Argand’s diagram
are collinear.
Soln
:
3
cos 2 sin 2 i=cos sin
1 2 2 2 2
z
Now i k k i
z
   
 
       
                 

   
 
cos 4 1 sin 4 1 where k=0,1,2
1 6 6
cos sin where = 4 1
1 6
i
z
k i k
z
z
i e k
z

 

  
    

    

.i i
z e z e 
  
 1
1
i i
i
i
e z e
e
z
e
 


  
 

2
cos sin
cos sin 1
cos sin
2sin cos 2sin
2 2 2
i
i
i
i
 
 
 
  


 


   
   
   

29. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 29
 
cos sin
2sin cos sin
2 2 2
(cos sin )
[ Munltiplying N & D by-i]
2sin cos sin
2 2 2
= cos _ sin
2 22sin
2
1 1
cot where = 4 1
2 2 2 6
i
i
i i
i
i
i
k
 
  
 
  
 

 



 
 
 
 

 
 
 
  
 
 
  
For K=0,1,2 we get three values of Z.
All these values have the same real parts i.e.
1
2
Hence the points represented by the 3 numbers are collinear.
50. If 2 2
1 3 and n is an integer, prove that z 2 .z 0 is not a multiple of 3n n n n
z i     
Soln
: Now 1 3z i  
1 3
2
2 2
2 2
2 cos sin
3 3
i
i
 
 
    
 
 
  
 
2
3
2
3
2
2 2
cos sin
2 3 3
i
nn i
n
z
e
z n n
e i


 
 
 
 
 
 
 
 
   
2
3 2 2
cos sin
2 3 3
2 2
2cos
2 3
nn i
n
n n
n n
z n n
Similarly e i
z n
Hence
z

 

 
 
 
  
 
If n is not a multiple then,
Let 3 1 & n = 3k-2 where k is an integern k 
3 1,When n k 
Value of expression  
2
2cos 3 1
3
k

 
2
2cos 2
3
2
2cos
3
1
2cos 1
2
k



 
  
 
 
  
 
 
    
 
3 2,When n k 
Values of expression  
2
2cos 3 2
3
k

 

30. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 30
4
2cos 2
3
4
2cos
3
1
2 1
2
k



 
  
 
 
  
 
 
    
 
Subs in (i) we get,
2 2
2 2
2
1
2 2
2 0 if n is not a multiple of 3
n n
n n
n n n n
n n n n
z
z z
z z
z z z
  
   
   
51. Show that  5 1
cosh cosh5 5cosh3 10cosh
16
x x x x  
Soln
:
5
cosh
2
x x
e e
x

 
  
 
 
 
     
 
5
5 5 4 5 3 2 5 2 5 4 5
1 2 3 1
5 5 3 3
5 5 3 3
1
32
1
. . . .
32
1
10
32
1
5 10
16 2 2 2
1
cosh5 5cosh3 10cosh
16
x x
x x x x x x x x x x
x x x x x x
x x x x x x
e e
e c e e c e e c e e c e e e
e e e e e e
e e e e e e
x x x

    
  
  
 
     
       
        
        
      
  
52. Show that
3
1 tanh
cosh6 sinh6
1 tanh

 

 
  
 
Soln: L.H.S.
3
1 tanh
1 tanh


 
  
 
 
 
 
3
3
32
2 2
32
2 2
6
sinh
1
cosh
sinh
1
cosh
cosh sinh
cosh sinh
cosh sinh
cosh sinh
cosh coscosh sinh
sinh sinhcosh sinh
cos sin
i
i
i i i




 
 
 
 
  
  
 
 
 
  
 
 
 
  
 
 
  
  
    
         
 


31. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 31
cos 6 sin 6
cosh6 sinh
. . .
i i i
R H S
 
 
 
 

53. If  
2 2
1
2 2
cos . Prove that (i) 1
cosh sinh
x y
x iy i 
 

    
2 2
2 2
(ii) 1
cosh sinh
x y
 
 
Soln
: Now  cosx iy i   
cos cosh sin sinhi    
cos cosh & y=-sin sinhx     
 
2 2
2 2
2 2
2 2
( ) now cos = & sin =
cosh sinh
But cos sin 1
1 eq. of ellipse is is constant
cosh sinh
( ) Also cos = & sin =
cosh sinh
But cos sin 1
x y
i
x y
x y
ii
 
 
 

 
 
 
 

 
   

 
 
2 2
2 2
1 eq. of hyperbola is is constant
cosh sinh
x y

 
   
54. Show that  1 2
sin 2 log 1ix n i x x
   
Soln
: Let  1
sin ix u iv
 
 
 
sin
sin u cosh v + icos u sinh v = ix
Comparing both the we get
sin u coshv = 0 ......(i)
cos u sinh v = x......(ii)
From (i)
sin u = 0 cos 0
u=2n
Also, sinh v =
u iv ix
v

  




 
 
   
1 2
1 2
x cos cos2 1
v=sinh x=log 1
Hence sin =2n +ilog 1
u n
x x
ix x x




 
  
 

55. Prove that    1 2 1 1 2 1
2
( ) cosh 1 sinh (ii) cosh 1 tan
1
x
i x x x
x
     
     
 
Soln
: (i) Let  1 2
cosh 1 ......( )x i
 

32. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 32
 
 
2 2
2 2
2 2
2 2 2 2
1
1 2 1
cosh 1
cosh 1
cosh 1
sinh cosh sinh 1
sinh
sinh ......( )
From (i) & (ii) we have,
cos 1 sinh ( )
x
x
x
x
x
x ii
x x



  

 
 
  
  
  
     
 
 
 

(ii)
 
2
2
1
2
1 2 1
2
Now sinh = x
& cosh = 1
sinh
tanh =
cosh 1
= tanh .......( )
1
From (i) & (iii) we have
cosh 1 tanh
1
x
x
x
x
iii
x
x
x
x





 
 

 

 
 
 
 
   
 
56.  1
Prove that sec sin log cot
2
h

  
  
 
Soln
:  1
Let sec sec ......( )h x i

 1
sec sin
1
cosh cos
sin
cosh cos
hx
x ec
x ec




 
  
 
 
 
2
2
log cos cos 1
log cos cot
1 cos
log
sin
2cos
log
2sin .cos
1 cos
log ...........( )
sin
ec ec
ec
h
h h
ii
 
 



 


  
 
 
  
 
 
  
 
 
  
 
From (i) & (ii) we have,
 1
sec sin log cot
2
h

  
  
 
57.   1
Prove that tanh log
1
x
x
x



 5
Hence deduce that tanh log tanh log 7 1
3
 
   
 
Soln
:  Let tanh log .......( )x i

33. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 33
1
tan log
1 1 1
log log
2 1 2
1
1
2 1
[By componendo-dividendo]
2 1
1
.....( )
1
x
x
x
x
x
x
ii
x








 
 
  
 

 


 


 

From (i) & (ii) we get,
 
 
1
tanh log .......(I)
1
3
Putting and 7 resp. in (i) and then adding we get,
5
5
1
5 7 13tanh log tanh log 7
53 7 11
3
x
x
x
x x



 
  
       
2 6
8 8
1
 

58. If 1 1 1 2 2
sinh sinh sinh then prove that x=a 1 a 1a b x b b a  
    
Soln
: Now
1 1 1
sinh sinh sinha b x  
 
Let
1
sinh a=sinha  
 
 
1
1
sinh sinh
sinh x=sinh
Also + by data
b b
x
 
 
  


  
 

2 2
Then R.H.S. = a 1 1b b a  
 
 
2 2
sinh 1 sinh sinh 1 sinh
sinh .cosh sin .cosh
sinh
sinh
. . .
x
L H S
   
   
 
   
   
 
 
  



59. If        1 1 1 2 2 2
cosh cosh cosh prove that 2 1 2 1 1x iy x iy a a x a y a  
        
Soln
:
 
 
1
1
Let cosh
& cosh
x iy i
x iy i
 
 


  
  
 
 
cosh cosh .cos sin .sin
cosh cosh .cos sin .sin
x iy i i
x iy i i
     
     
     
     
Adding we get 2x = 2cosh .cos 
cosh .cosx   
Subtracting we get 2 2 sinh .siniy i  
sinh .siny   

34. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 34
   1 1 1
Also cosh cosh cosh [given]x iy x iy a  
   
    1
1
cosh
cosh 2
cosh 2
i i a
a
a
   




    
 
 
   
   
2 2 2
2 2
2
2 2 2 2
T.P.T. 2 1 2 1 1
2 2
. . 1 Dividing by a 1
1 1
2 cosh .cosh 2 sinh .sinh
L.H.S.
cosh 2 1 cosh 2 1
a x a y a
x y
i e
a a
   
 
    
     
 
 
2 2 2 2
2 2
2 2
2cosh .cosh 2sinh .sinh
2cosh 2cosh
cos sin
1
. . .R H S
   
 
 
 
 


60. If cosh secu  Prove that  sinh tani u 
 
 
tanh sin
log tan
4 2
ii u
iii u

 

  
   
  
Soln
:   2 2
Now sinh cosh 1i u u 
 2 2
2 2
sinh sec 1 cosh sec
sinh tan
sinh tan
u u
u
u
 


   
 
 

sinh
( ) tanh
cosh
u
ii u
u

tan
sec
sin
cos
cos
sin







 

( ) Now, tanh siniii u 
 tanhu u sin 
2
2
1 1 sin
log
2 1 sin
1 1 cos
log where 0
2 1 cos 2
2cos
1 2log
2 2sin
2


 




 
  
 
 
   
 
 
 
  
 
 

35. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 35
1
log cot
2 2
log tan
2 2
log tan
2 4 2

 
  
 
  
 
  
   
  
  
    
  
log tan
2 2
   
   
  
61. If cosh sec ,x   Prove that (i) log sec tanx   
 1
(ii) tan
2
(iii) tanh tan
2 2
x
e
x



 


Soln
:
 1
( ) Now cosh sec
cosh sec
i x
x



 
 
 
2
log sec sec 1
log sec tanx

 
  
 
 ( ) now sec tan ( )
1
sec tan
x
x
ii e from i
e
 
 

 
 

2
2 2
2
1 sin
cos cos
1 sin
cos
1 sin
where
cos 2
2sin
2
2sin 2cos
2 2
2tan
2
a

 


 



 

 



  


 
 
 
1
1
1
tan
2
2tan
2
Hence 2tan
2
x
x
x
e
e
e


 


 
 
 
 
   
 
2 2
( ) Now cosh sec
1
cosh
cos
1 tan 1 tan
2 2 [By componendo-Dividendo]
2 2
Hence tanh tanh
2 2
iii x
x
x
x





 
 
 


36. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 36
62.
    1 1 2
Find the sum of the series sin sin 2 sin 2 ........ to n terms
1.2 1.2.3
n n n n n
n   
  
  
Soln
: Let
    1 1 2
s sin sin 2 sin3 ........... terms
2! 3!
n n n n n
n n  
  
   
    
 
 
 
 
1 1 2
C 1 cos cos2 cos3 ........... 1 terms
2! 3!
1
C S 1 cos sin cos2 sin 2 .........
2!
n n n n n
n n
n n
i n i i
  
   
  
     

        
 
    
 
   
 
2 3
2
1 1 2
1 1 ............
2! 3!
1 By Binomial Expression of 1
1 cos sin
2cos .2sin cos
2 2 2
2cos cos sin
2 2 2
2cos cos
2 2
i i i i
n ni
n
n
n n
n
n n n n n
ne e ne e
e x
i
i
i
n
   

 
  
  
 
  
      
   
 
  
 
  
 
   
    
   
 
  
 
sin [By De Moivre's Theorem]
2
n
i
 
 
 
Equating imaginary parts we get,
2cos sin
2 2
n
n
S
    
    
   
63. Prove that  
2
coscos2
1 cos ................ cos sin
2!
xx
x e x
    
Soln
:
   
2
2
2
x cos2
Let C 1+x cos ..........
2!
x sin 2
S xsin ...............
2!
C+iS 1 cos sin cos2 sin 2 ..............
2!
x
x i i




   
  
  
      
2
2
2
1 ...........
2!
1 ............ Where z=xe
2!
i
i i
i
z
xe
x
xe e
z
z
e
e

 

   
      


 
 
 
cos sin
sincos
cos
2
cos
.
Comparing real parts we get
cos2
C=1 cos ................ cos sin
2!
x i
i xx
x
x
e
e e
e
x
x e x
 



 




   

37. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 37
64. If  
 
 
1
2 2
2tan /
Prove that
log
p x iy b ay
a ib m
x a b


  

Soln
: Now  
p x iy
a ib m 
 
   
 2 2 1
2 2
1
1
2 2
log log
log tan log
Comparing both the sides
log log ...............( )
& tan log ...............( )
Dividing (ii) by (i)
tan
log
p a ib x iy m
b
p a b i x iy m
a
p a b x m i
b
p y m ii
a
b
ya
xa b



   
  
      
  
 
 
 
 
 
 
  

 
1
2 2
2tan
log
b
y a
x a b
  
 
  

65. Prove that
4 1
log
4 1
i
n
i
m



Soln
: Now 1cos 2 sin 2 General Polar Form cos 0 ,sin 1
2 2 2 2
i k i k
   
 
     
            
     
2 2
log log 0 1 2
2
i i k


 
     
 
   0 4 1 4 1
2 2
i k i k
 
    
 
 
 
 
log
Then log
log
4 1 4 12
4 14 1
2
i
i
i
i
i n n
mi m



 
 

66. Prove that  sin log 1i
i
   
Soln
: Now  log log 2 2
i iii i i i
i e e e e
  
     
   
 
 
1 2
log log
2
Hence sin log sin 1
2
i
i e
i





 
   
 
    
67. If , prove that 2
2
ie
e i n  


  
Soln
: Now
ie
e i

Taking log(general of both sides) we get,
logi i 

38. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 38
2
2
i i n

  
  
    
  
Hence 2
2
n
 


 
68. If    
2 2
1 2 2
2 2
sin log prove that 1 where A B
sin cos
x y
x iy A iB e
 

      
Soln
: Now   2 1
log log A tan
B
A iB B i
A
  
     
 
 
1/22 1 2 2 2
1
log tan A (given)
where = tan
B
e i B e
A
B
i
A
 
  


 
       
 
 
   
 

   
 
 
1
1
But sin log
sin
x+iy= sin
= sin .cosh cos .sinh
x iy A iB
x iy i
i
i
 
 
   


  
   
 

69. If   1
1 , prove that tan log2
4 2
x iy x y
i
 

 
  
Soln
:  Now 1
x iy
a i i

  
     
 2 2 1 2 2 1
log log 1
1
log tan log 1 1 tan
1
i x iy i
i x iy i
 

 

 
    
    
          
    
 
1
log 2
2 4
log 2 log 2
2 4 4 2
x iy i
x y x y
i

 
 
   
 
   
      
   
Comparing imaginary parts of both the side we get,
1
tan log2
4 2
x y 

  
  
 
70. If
...
2 2
A+iB, prove that (i) , (ii) A
2
i BA B
i B e
A



   
Soln
: Now
...
A+iBi
i


A iB
i A iB
  
...
A+iBi
i

 
 
 

 
 
 
2 2 1
2 2 1
log
log tan
2
1
log tan
2 2 2
A iB i A iB
B
A iB i A B i
A
B A B
i A B i
A

 


   
 
      
 
  
      
 

39. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 39
 
 
 
1
2 2
2 2
2 2
( ) Comparing imaginary parts we get tan
2
tan
2
1
( ) Comparing imaginary parts we get log
2 2
log
B
A B
i
A
A B
A
B
ii A B
A B B
A B e 






 
  
 
 

 
   
  
71.  
 4 1
2
If cos sin , show that 4 1
2
i m
i
i i n e


  
 
   
Soln
: Now
   4 1 4 1
log 2 2
i i m m
i i i
i e e e
  
   
 
  
   
 
 
 
 
4 1
2
4 1
2
4 1 4 1
log 2 2
4 1
2
Then,
But i cos sin
Hence = 4 1
2
i i
m
i
m
e m i m
i i i
i n e
i i
i e e
e
i e
n e


 


 


 
 
  
    
  
 
 
 
 
 

  

72. Find the principle value of  
i
x iy & show that it is purely real if  2 21
log
2
x y is multiple of

Soln
: Now    log
;
ii x iy
x iy e 
    
 
 
   
2 2 1
1 2 2
1 2 2
1
log tan
1
tan log
2
1
tan log
2
tan
2 2 2 21 1
cos log sin log
2 2
y
x y i
x
y
x y
x
y
i x y
x
y
x
e
e
e e
e x y i x y




  
    
  
 
  
 
   
   
   
 
 
 


 
    
           
   
   
2 2
2 2
1
If is entirely real then sin log 0
2
1
log . . multiple of
2
i
x iy x y
x y n i e 
 
    
  
73. If  
..
cos sin
x
x
x a i 

  , prove that the general value of x is given by  cos sinr i 
Where
   2 sin log .cos 2 cos log .sin
log &
n a n a
r
a a
     

 
 
Soln
: If  
..
cos sin
x
x
x a i 

  i
ae

.i
ae i x i
x ae x ae


  
    

Taking log (general) of both sides we get,

40. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 40
      
     
 
 
log log log
cos sin log log 2 cos sin .
log cos sin log sin cos log 2
Comparing both side we get,
log cos sin log .......( )
log sin cos 2
i i
i
ae x a e
a i r i a i n x r i r e
a r ia r a i n
a r a i
a r n
 

      
       
  
   
 
          
      
 
 

   
   
   
   
.........( )
Then ( ) cos ( ) sin gives,
alogr log cos 2 sin
log cos 2 sin
log
Also ( ) cos ( ) sin gives,
2 cos log sin
2 cos log sin
ii
i ii
a n
a n
r
a
ii i
a n a
n a
a
 
   
   
 
    
   

  
  
 
 
  
  
 
 

41. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 41
Homework Problems
Part I: DeMoivre’s Theorem
1. cos2 sin2 , cos2 sin2 , cos2 sin2p i q i r i          
Show that (i) 2cos( )
p q
q p
   
(ii) 2 sin( )
p q
i
q p
    (iii) 2cos( )
pq r
r pq
      [M99]
2.
1 cos sin
cos sin
1 cos sin
n
i
n i n
i
 
 
 
  
  
  
3.
1 sin cos
cos sin
1 sin cos 2 2
n
i n n
n i n
i
   
 
 
      
        
      
[M04]
Hint: sin cos cos( ) sin( )
2 2
i i
 
       
4. Prove that [(cos cos ) (sin sin )] [(cos cos ) (sin sin )]n n
i i             
1
2 sin .cos[ ]
2 2
n n
n
          
    
   
5. If cos sin , cos sinx i y i       , Prove that tan
2
x y
i
x y
   
  
  
6. If cos sin , cos sin , cos sina i b i c i           then show that
( )( )( )
8cos cos cos
2 2 2
a b b c c a
abc
               
      
     
[6M06]
Hint: (cos cos ) (sin sin )a b i i       
2cos .cos 2 sin .cos
2 2 2 2
2cos cos sin
2 2 2
i
i
       
     
          
        
       
         
       
      
Also cos( ) sin( )abc i          
7. If ( , ) (cos sin )r r i    and in the Argand’s diagram if (1, ), (1, ), c (1, )a b     where
0a b c   then prove that 0.ab bc ca  
Hint: cos sin , cos sin , cos sina i b i c i          
8. If 1 2 3, ,z z z are three complex numbers with modulus ' 'r each and 1 2 3 0z z z   .
Prove that (i)
1 2 3
1 1 1
0
z z z
   (ii) 2 2 2
1 2 2 0z z z  
9. If sin sin sin cos cos cos 0a b c a b c          
Prove that (i) 3 3 3
cos3 cos3 cos3 3 cos( )a b c abc         
(ii) 3 3 3
sin3 sin3 sin3 3 sin( ]a b c abc          [M81,D84, D90, D93]

42. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 42
10. Using De Moivre’s Theorem, prove the following.
(i) 3 2 2 3
cos3 cos 3cos sin ,sin3 3cos sin sin          [D81]
(ii) 4 2 2 4
cos4 cos 6cos sin sin       [M84]
(iii) 6 4 3 2 5 7
sin7 7cos sin 35cos sin 21cos sin sin           [D84]
(iv) 7 5 3 3 5 7
sin8 8cos sin 56cos .sin 56cos .sin 8cos sin           
11. If 6 4 2 2 4 6
cos6 cos cos sin cos sin sina b c d        
Find the values of , , , .a b c d Ans : 1, 15, 15, 1a b c d     
12. Prove that 2 4 4sin7
7 56sin 112sin 64sin
sin

  

   
13. Prove that
3 5 7
2 4 6
7tan 35tan 21tan tan
tan7
1 21tan 35tan 7tan
   

  
  

  
. Hence deduce that
6 4 2
7tan 35tan 21tan 1 0
14 14 14
  
    . [M99]
14. Prove that
4 3 2
16 cos A – 8 cos A – 12 co
1 cos9
1
s 4cos A 1
A
co
A
sA

  

.
Hint : Now
2
2
9 9 9
2cos cos 2cos sin
1 9 sin5 sin 42 2 2 2
1 2cos cos 2cos sin
2 2 2 2
A A A A
cos A A A
A A A AcosA sinA
 
   

Now,  
5
cos 5A i sin 5A cos A i sin A   Find
sin5A
sinA
Similarly  
4
cos 4A i sin 4A cos A i sin A   Find
sin 4A
sinA
15. (i) Prove that 5 1
sin (sin 5 5 sin 3 10 sin )
16
      [M91, 5M06]
(ii) Expand 8
cos  as a series of cosines of multiples of 𝜃.
Ans: 1/128 (cos 8 8 cos 6 28 cos 4 56 cos 2 70)      
(iii) Expand 7
sin  as a series of sines of multiples of 𝜃.
Ans: 1/ 64 (sin 7 7 sin 5 21 sin 3 35 cos )      
16. (i) Express 6 6
cos sin  in terms ofcos 6 , cos 4 ,cos2   . [ M87,6D,07]
(ii) Show that 8 8 1
cos sin (cos 8 28cos 4 +35)
64
      [M82,M97,8D05]
17. Show that 5 3 7
1/ 2 ( 8 2 6 2 4 6 2 )cos sin sin sin sin sin          [ M02]
18. If 2 4
0 2 4 6cos sin A A cos 2 A cos 4 A cos 6        Prove that A0 + 9A2 +25A4 + 57A6 = 0.
19. If 2 cos x 1/ x, 2 cos y 1/ y     . Show that 2cos ( m n )
m n
n m
x y
y x
   [D93,D96 ]
20. If x+1/x = 2 cos α, y+1/y = 2 cos β, z+1/z = 2 cos 𝛾
Show that xyz + 𝑥𝑦𝑧 +
1
𝑥𝑦𝑧
= 2 cos (α+ β+ 𝛾 /2 ) [ D96,D04]

43. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 43
21. If x -1/x = 2i sin 𝜃 , y -1/y = 2i sin 𝛟 show that
𝑥
𝑚
𝑦𝑛 +
𝑦𝑛
𝑥
𝑚 = 2cos (
𝜃
𝑚
−
ϕ
𝑛
) [M05]
22. If
1 1 1
x 2isin , y 2isin , z 2i sin
x y z
        show that
1
xyz 2cos( )
xyz
     
23. If
1 i
z
2 2
  then by using De Moivre’s theorem simplify    
1010
z z [M89]
24. If n is the + ve integer, show that
(i)    
n n n 2
1 i 1 i ( 2) cos
4
n  
     
 
(ii)    
n n
n 1
1 i 3 1 i 3 (2) cos
3
n  
     
 
(iii)    
n n
n 1
3 i 3 i (2) cos
6
n  
     
 
25. If α, β are the roots of quadratic equation x2
- 2x+ 4 = 0, then
(i) Prove that αn
+ βn
= 2n+1
cos (
𝑛𝜋
3
) [ M82, M88,M95,M03 ]
(ii) Find the value of α15
+ β15
Ans : -216
[ D81, M93 ]
26. Find all the values of
1/4
2 3
1
i
i
 
 
 
[D85]
Ans :  
1/4 2 2
13 cos sin
4 4
k k
i
         
    
    
where  1
tan 1/ 5 
 , k = 0,1,2,3
27. Solve : (i) 6
x i 0  [D94] A:
5 5 3 3
cos i sin , cos i sin , cos i sin
12 12 12 12 4 4
          
          
     
(ii) 5
x 3 i  [D96]
28. (i) x7
+ x4
+ x3
+ 1 = 0 [D88,M95] Ans: -1, 1/2 ±
1 3
2
,
1
2
±
1
2
,
−1
2
±
1
2
(ii) x10
+ 11x5
+ 10 = 0 [D95] Ans: (-10)1/5
, -1, cos
𝜋
5
± i sin
𝜋
5
, cos (
3𝜋
5
) ± i sin (
3𝜋
5
)
(iii) x9
- x5
+ x4
- 1 = 0. [M95] Ans: ± -1, ± i, cos
𝜋
5
± i sin
𝜋
5
, cos
3𝜋
5
± i sin
3𝜋
5
(iv) x14
+ 127x7
- 128 = 0 [M99] Ans: 2 [ cos (2k+ 1)
𝜋
7
+ i sin (2k+ 1)
𝜋
7
] k = o to 6
(v) x7
+ x4
+ i (x3
+1) = 0 Ans: -1, 1/2 ± i
3
2
, ± ( cos
𝜋
8
- i sin
𝜋
8
), ± ( cos
3𝜋
8
+ i sin
3𝜋
8
),
29. Solve
(i) x4
- x2
+ 1 = 0 [M96] Ans : ±
3
2
, ±
𝑖
2
.
(ii) x4
- x3
+ x2
- x+ 1 = 0. Ans : cos
𝜋
5
± i sin
𝜋
5
, cos
3𝜋
5
± i sin
3𝜋
5
,
30. Find the continued product of all the values of
(i) [ 1+ i]2/3
Ans : 2i [D92]

44. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 44
(ii) [ 1+ i]1/5
Ans : 1+ I [M95,M05]
(iii) (1+ i 3 )1/4
Ans : - ( 1+ i 3 )
31. Show that the nth
roots of unity are given by 2 3 4 1
1, , , , ,...... n
     
where 𝜆 = cos 2𝜋/𝑛 + i sin
2𝜋/𝑛. Show that continued products of the all these nth
roots is (-1)n+1
32. Prove that nth
roots of unity are in geometric progression. Also find sum of nth
root of
unity.[8D07]
33. Find the roots of  
33
z z 1  and show that the real part of all the roots is -1/2
34. Solve  
33
z i z 1  [6D05]
Hint :
3
z
i cos 2 k i sin 2 k
z 1 2 2
 
 
     
         
     
Ans : o
2
1
c t
2
x i

  where  4k 1
6

   & k = 0, 1, 2
35. Obtain the solution of the equation  
6 6
x 1 x 0  
Hint:    
6
1
= -1= cos 2k 1 i sin 2k 1x
x
 
 
    
 
Ans: c
2
1
ot
2
x i

  where  2k 1
6

   & k = 0, 1, 2, 3, 4, 5
36. Solve    
5 5
x 1 32 x 1   where k = 0,1, 2, 3, 4. Ans:
2 2 2
cos sin
5 5
2 2 2
cos sin
5 5
k k
i
x
k k
i
 
 





37. If cos sin
3 3
r r r
x i
    
    
   
, then 0 1 2 3........x x x x i  .State true of false. Ans: True [M03]
38. If arg (z+ 1) =
6

and arg (z- 1) =
2
3

find z. Ans:
1 i 3
2

[M97,00,D01 5M 08 ]
39. Find z if amp (z+ 2i) =
4

, amp (z- 2i) =
3
4

Ans : z = 2+ i0
40. If
2 i 4i
a i 1 i


 
represents a point on the line 3x+ y = 0 in Argand’s diagram, find a.
Ans : a= 1 or 3/4
41. Find two complex numbers whose sum is 4 and product is 8. Ans : z1 = 2+ 2i, z2 = 2- 2i [M96]
42. If 1 2z cos i sin ,z cos i sin ,       where ,
2

    . Find polar form of
2
1
1 2
1
1
z
iz z


.
Hint : Divide N & D by z1 Ans : r ( cos i sin )  where cos , sec
4 2
r
  

   
   
  

45. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 45
43. (a) Express 2 2
1 1
( ) ( )x iy x iy

 
in the form a+ bi. Find value of a & b in terms of x and y.
(b) If  x iy
a ib
c id

 

, prove that 2 2 2 2 2 2 2
( ) ( ) / ( )x y a b c d  
44. If 2 2
x y 1  , Prove that
1 x iy
x iy
1 x iy
 
 
 
45. Prove that      
m/2nm/n m/n 2 2 1
x iy x iy 2 x y cos ( tan )
m y
n x

    
Hint: Let x iy r ( cos i sin )    where 2 2
r x y  and 1
tan
y
x
   
  
 
[M80]
46. If z x iy  , prove that
2 2
2 2
x y
z / z /
x y
z ( )z 


47. If z a ( cos i sin )   , prove that z / z / z 2 cos 2z  
48. Prove that 1
1
1z
z



49. If
22
1 1z z   . Prove that z lies on imaginary axis where z is a complex number. [5D07]
Part II: Exponential form of Complex Number
1. If z = x+ iy and 𝑒 𝑧2
= a+ ib. Find the a and b.
Hint : a+ ib = 𝑒 𝑧2
= 𝑒(𝑥+𝑖𝑦)2
= 𝑒 𝑥2−𝑦2+12𝑥𝑦
Ans : a = 𝑒 𝑥2−𝑦2
cos 2xy, b = 𝑒 𝑥2−𝑦2
sin 2xy
2. If r1 𝑒 𝑖𝜃1 + r2 𝑒 𝑖𝜃2 = R 𝑒 𝑖𝜃
, find R and 𝛟.
Ans : R = 𝑟1
2
+ 𝑟2
2
+ 2𝑟1 𝑟2 cos(𝜃1 − 𝜃2) , 𝛟 = tan-1
(
𝑟1 𝑠𝑖𝑛 𝜃1+ 𝑟2 sin 𝜃2
𝑟1 𝑐𝑜𝑠 𝜃1+ 𝑟2 cos 𝜃2
)
3. If p = a+ ib, q = a- ib where a and b are real then prove that pep
+ qeq
is real.
4. Prove that (1- 𝑒 𝑖𝜃
)-1/2
+ (1- 𝑒 𝑖𝜃
)-1/2
= ( 1+ cosec 𝜃/2)1/2
. [M04,8M06]
5. Prove that ( 1- sec 𝜃/2 )1/2
= ( 1+ 𝑒 𝑖𝜃
)-1/2
- ( 1+ 𝑒 𝑖𝜃
)-1/2
6. Show that
𝑠𝑖𝑛𝜃
2
+
𝑠𝑖𝑛2𝜃
22 +
𝑠𝑖𝑛3𝜃
23 + …………..=
2𝑠𝑖𝑛𝜃
5−4𝑐𝑜𝑠𝜃
[D89,M93]
7. Solve the equation 7 cosh x + 8 sinh x = 1 for real values of x. Ans : - log 3
8. If tanh x = 1/2, find sinh 2x and cosh 2x Ans : 4/3, 5/3
9. If x = tanh-1
(0.5). show that sinh 2x = 4/3 [M-99] Hint : sinh 2x = 2 tanh x/ 1- tanh2
x
10. Prove that tanh ( log 3 ) = 1/2. Hint: use definition of tanhx.
11. Prove that 16 sinh5
x = sinh 5 x – 5 sinh 3x + 10 sinh x.
12. Prove that 32 (cosh6
x- 10 ) = cosh 6x+ 6 cosh 4x+ 15 cosh 3x.
13. If cosh6
x= a cosh 6x + b cosh 4x + c cosh 2x + d, prove that 5a+ 5b+ 3c- 4d = 0
14. Prove that 2
2
1
=
1
1
1
1
1 cos
cosh x
h x



[M96]
15. Prove that (i) [
1+ tan ℎ𝑥
1−tanh 𝑥
]n
= cosh2nx + sinh2nx [ D99]

46. INFOMATICA ACADEMY
CONTACT: 9821131002/ 9076931776 46
(ii) (cos hx – sin hx)n
= cosh nx – sinh nx [D01]
16. Prove that [
cosh 𝑥+sinh 𝑥
cosh 𝑥−sinh 𝑥
]n
= cosh 2nx + sinh 2nx
17. If log ( tan x) = y, prove that (i) sinh ny = 1/2 (tann
x – cotn
x) [D04,M05]
(ii) 2 cosh ny cosec 2x = cosh (n+ 1) y + cosh (n- 1) y [M05]
18. If sin (𝜃+ i𝛟) = 𝑒 𝑖𝛼
, prove that sin 𝛼 = ± cos2
𝜃 = ± sinh2
𝛟 [D81,82]
19. If cosh (𝜃+ i𝛟) = 𝑒 𝑖𝛼
, prove that sin2
𝛼 = sin4
𝛟 = sinh4
𝜃
20. If sin (𝜃+ i𝛟) = R (cos α + I sin α) prove that R2
=
1
2
(cos 2𝛟–cos 2𝛳) and tan α=tanh𝛟.cot𝛳 [M86]
21. If cos (x+iy) = eiπ/6
, Prove that (i) 3sin2
x-cos2
x = 4sin2
x.cos2
x
(ii) 3sinh2
y + cosh2
y = 4sinh2
y.cosh2
y
22. If log [cos(x-iy)] = α + iβ, prove that α =
1
2
log
cosh2 cos2
2
y x 
  
and find β. [M84, D92]
23. If sin-1
(α+iβ) = λ + iμ. Prove that sin2
λ and cosh2
μ are the roots of the equations
x2
– (1+ α2
+ β2
)x + α2
= 0
24. Let P(z) where z = sin(α+iβ). If α is variable, show that the locus of the P(z) is an ellipse
2 2
2 2
1
cosh sinh
x y
 
  . Also show that x2
cosec2
α – y2
sec2
α = 1 if β is variable.
25. If sinh (x+ iy) = eiπ/3
, prove that (i) 3cos2
y – sin2
y = 4sin2
y cos2
y
(ii) 3sinh2
x + cosh2 x = 4sinh2
x.cosh2
x
26. If u+ i v = cosh ( 𝛼+ i 𝜋/4 ).Find the value of u2
– v2
Ans : 1/2 [D96,D03]
27. If x+ iy = 2 cosh (𝛼+ i 𝜋/3), prove that 3x2
- y2
= 3
28. If x = 2 sin 𝛼 cosh β, y = 2 cos 𝛼 sinh β
Show that (i) cosec(𝛼 − 𝑖 β ) + cosec (𝛼 + 𝑖 β ) =
4𝑥
𝑥2+ 𝑦2
(ii) cosec(𝛼 − 𝑖 β ) - cosec (𝛼 + 𝑖 β ) =
4𝑖𝑦
𝑥2+ 𝑦2
29. If tan(
𝜋
6
+ 𝑖𝛼) = x+ iy, prove that x2
+ y2
+ 2x/ 3 = 1. [M96]
30. If cot (
𝜋
6
+ 𝑖𝛼) = x+ iy, prove that x2
+ y2
- 2x/ 3 = 1
31. Show that tan
u iv
2
 
 
 
=
sin u i sin h v
cos u cosh v


32. If tan h (𝛼 +i β ) = x+ iy, prove that x2
+ y2
- 2x cot 2 𝛼= 1, x2
+ y2
+ 2y coth 2 β + 1 = 0.
33. If cot (𝛼 +i β ) = i. Prove that β =
𝜋
4
, 𝛼 = 0
34. If 𝛼 +i β = tan h ( x + i
𝜋
4
), prove that 𝛼2
+ β2
= 1 [M97]
35. If tan h (a+ ib )= x+ iy, Prove that x2
+ y2
- 2x coth 2 𝛼 + 1 = 0 & x2
+ y2
+ 2y coth 2 β - 1 = 0
36. If tan 𝛼 = tan x. tanh y, tan β = cot x. tanh y, Show that tan (𝛼 + β) = sin h 2 y. cosec 2x
37. If tan y = tan 𝛼 tanh β and tan z = cot 𝛼 tanh β. Prove that tan(y+ z) = sin h 2 β. cosec 2 𝛼.
38. Separate into real and imaginary parts, (i) sec (x+ iy) (ii) tanh (x+ iy)

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Ans : (i) 2 (
𝑐𝑜𝑠 𝑥 cos ℎ𝑦+𝑖 sin 𝑥 sin ℎ𝑦
cos 2𝑥+cosh 2𝑦
) (ii)
sinh 2𝑥+𝑖 𝑠𝑖𝑛2𝑦
cosh 2𝑥+cos 2𝑦
39. Show that (i) sinh-1
x = cosh-1
( 1 + 𝑥2) [D02,M04,3M07]
(ii) tanh-1
(𝛟) = sinh-1
(
𝜙
1−𝜙2
) [D90,D01,3M06]
(iii) Prove that tanh-1
(sin 𝜃) = cosh-1
(sec 𝜃).
40. Show that sech-1
(sin 𝜃) = log (cot 𝜃/2)
41. Show that sinh-1
(tan x) = log [ tan (
𝜋
4
+
𝑥
2
) ] [M96]
42. Prove that cosech-1
z = log (
1+ 1+𝑧2
𝑧
).Is defined for all values of z ? [D03]
43. Show that cos-1
z = - i log ( z± 𝑧2 − 1 )
44. If cosh-1
a + cosh-1
b = cosh-1
x, then prove that a 𝑏2 − 1 + b 𝑎2 − 1 = 𝑥2 − 1.
45. If cosh-1
(x+ iy) + cosh-1
(x- iy) = cosh-1
a, prove that 2(a- 1) x2
+ 2(a+ 1) y2
= a2
- 1.
46. If A+ iB = C tan (x+ iy), prove that tan 2x =
2𝐶𝐴
𝐶2−𝐴2−𝐵2
47. Separate tan-1
(cos𝜃 + i sin 𝜃 ) into real and imaginary parts [M81,D86,M87,D95]
48. If tan (𝜃 + i𝛟) = cos 𝛼 + i sin 𝛼, show that 𝜃 =
𝑛𝜋
2
+
𝜋
4
, 𝛟 = ¼ log (
1+𝑠𝑖𝑛𝛼
1−𝑠𝑖𝑛𝛼
) [M93]
49. If tan (𝜃 + i 𝛟 ) = 𝑒 𝑖𝛼
show that 𝜃 = ( n+ 1/2) 𝜋/2 and 𝛟 = 1/2 log tan (𝜋/4 + 𝛼/2 ) [D83,93]
50. Separate into real and imaginary parts : tan-1
(a+ iy)
or Prove that tan-1
(a+ iy) = 1/2 tan-1
( 2a/1- a2
- y2
) + i/4 log
(1+𝑦)2+𝑎2
(1−𝑦)2+𝑎2 [D02]
51. Prove that one value of tan-1
(x+ iy/x- iy) is 𝜋/4 + 𝑖/2 log x+ y/x- y where x > y > 0. [D80]
52. If tan (x+ iy) = i, x, y ∈ R . Show that x is indeterminate and y is infinite.
Hint : tan(x- iy) – I, then tan 2x=tan[ (x+ iy)+(x- iy)] & tan 2iy = tan [(x+ iy)-(x-iy)]
53. If tan ( u+ iv) = x+ iy then prove that curves u = constant and v = constant are families of circles.
Part III: Logarithmic Form Of Complex Number
1. Show that  
2
2
2(1
2
i)l
1
(log2) log2
4 1og 1 i 6 4
1
(log2)
4 16
i 
 
 

 [M98]
2. Find the value of (i)  2log 3  (ii)  log 5 Ans : (ii)  log 5 i 2n 1  
(iii)    log 1 i log 1 i   Ans : log 2 i (2 )n
3. Solve for z if
z
e 1 i 3 
4. log( ) 2 ( ) ( )
2 2
i i
e e log cos i       
    
[D03]
5. Prove that
2
(1 ) (2 )i
log e log cos i
   
6. Prove that  log log i log
2 2
i 
   .

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7. Show that
1x i
log i (2tan x – )
x i
 
 
 
8. Prove that 2 2
2
.
x iy xy
tan i log
x iy x y
  
     
[D82]
9. Show that    11 cosh2 cos2
log cos x iy log – i tan tanx. tanhy
2 2
y x  
     
 
10. Show that      1
log sin x iy / sin x iy 2i tan cot x. tanh y
     [ M97,M04,4M07]
11. If  log cos x iy a ib   , prove that (i)
2a
2e cosh2y cos2x  (ii) tanb tan x.tanh y 
12. Separate into real and imaginary parts :
(i)  
2 3i
1 i

 Ans:
log 2 3 /2 3 3
e cos – i.sin –
2 2 log 2 2 2 log 2
       
    
    
(ii)
1
i i
Ans:
/2
e cos i .sin
2 2
   
  
(iii)
i
(sin i cos )  Ans: 2
e

 
(iv)  
1 i
1 i

 Ans:
 
   8x 1 /4 1 1
2 e cos 8x 1 – i.sin 8x+1
4 2log2 4 2log2
       
      
    
13. Separate into real and imaginary parts  
(1 3)
1 3
i
i

 (consider principal values only) [D91,M04]
Ans : ( / 3)
2 ( 3 2 / 3) ( 3 2 / 3)e cos log isin log
     
14. Prove that the real value of principal of 
log i
1 i is
2
8
cos
4log 2
e


[M92,D02]
15. Prove that the general values of  
i
1 i tan 

 is    (2 )
cos log cos i sin log cosx
e  
 
  
Hence find the principal value. [D01,D03,D04]
16. If
. . inf
i
i ad ii
i
i  

  , show that
 4m 12 2
e

   
 

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CONTACT: 9821131002/ 9076931776 49
Know What You Don’t Know !!!!!