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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 !!!!!