2. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)2(
ﻣوﺗﺴﻤﯿﺎت ﺼﻄﻠﺤﺎت
ﻟﯿﻜﻦz = x + i yﻣﺮﻛﺒﺎ ﻋﺪدا
(1اﻟﺤﻘﯿﻘﻲ ﻟﻠﻌﺪد ﯾﻘﺎلxأﻧﮫ)اﻟﺤﻘﯿﻘﻲ اﻟﺠﺰءReal part(ﺑﺎﺧﺘﺼﺎر ﻟﮫ وﯾﺮﻣﺰ:Re ( z ) = x
(2اﻟﺤﻘﯿﻘﻲ ﻟﻠﻌﺪد ﯾﻘﺎلyأﻧﮫ)اﻟﺘﺨﯿﻠﻲ اﻟﺠﺰءImaginary part(ﺑﺎﺧﺘﺼﺎر ﻟﮫ وﯾﺮﻣﺰ:Im ( z ) = y
(3إذاﻛﺎنy = 0ﻋﻨﺪﺋﺬ ﻓﯿﻜﻦz = xوﯾﺼﺒﺢzﺣﻘﯿﻘﯿﺎ ﻋﺪدا
ﺑﺎﻟﺼﻮرة ﻣﺮﻛﺒﺎ ﻋﺪدا ھﻮ ﺣﻘﯿﻘﻲ ﻋﺪد ﻛﻞ أن أيz = x + 0iاﻟﺤﻘﯿﻘﯿﺔ اﻷﻋﺪاد ﻣﺠﻤﻮﻋﺔ ﺗﻜﻮن ﻟﺬا
اﻟﻤﺮﻛﺒﺔ اﻷﻋﺪاد ﻣﺠﻤﻮﻋﺔ ﻣﻦ ﺟﺰﺋﯿﺔ ﻣﺠﻤﻮﻋﺔ ھﻲ.
(4إذاﻛﺎنx = 0ﺑﺎﻟﺸﻜﻞ اﻟﻤﺮﻛﺐ اﻟﻌﺪد ﻓﯿﺼﺒﺢz = y iاﻟﻤ اﻟﻌﺪد ﻋﻠﻰ وﯾﻄﻠﻖاﺳﻢ اﻟﺤﺎﻟﺔ ھﺬه ﻓﻲ ﺮﻛﺐ:
ﺑﺤﺖ ﻣﺮﻛﺐ ﻋﺪدPure imaginary number
(5اﻻﻋﺘﯿﺎدي أو اﻟﺠﺒﺮي اﻟﺸﻜﻞz = x + i yأﺧﺮى ﺻﯿﻐﺔ ﻟﮫھﻲz = ( x , y )ﺑﺎﻟﺼﯿﻐﺔ ﺗﺴﻤﻰ
اﻟﺪﯾﻜﺎرﺗﯿﺔﻟﯿﻤﺜﻞ اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪدzﻓﻲ وﺣﯿﺪة ﻧﻘﻄﺔااﻟﻤﺮﻛ ﻟﻤﺴﺘﻮىﺐ( Complex plane).
ﻣﺜﻼ:ﺑﯿﺎﻧﯿﺎ ﻣﻤﺜﻠﺔ اﻟﺘﺎﻟﯿﺔ اﻟﻤﺮﻛﺒﺔ اﻷﻋﺪاد:2 +5i , - 5 +i , 4 – 3i , -5i
Real axis
Imaginaryaxis
- 5 i
- 5+ i
2 + 5 i
4 – 3 i
3. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)3(
ﻣﺜﻼ:اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪدzﺟﺪ اﻟﺘﺎﻟﻲ:Im(z) , Re(z)
1 1 11) 2 3 Re( ) 2 ; Im( ) 3z i z z
2 2 22) 2 3 Re( ) 3 ; Im( ) 2z i z z
3 3 33) 2 3 Re( ) 2 ; Im( ) 3z i a b i z a z b
4 4 44) 5 Re( ) 0 ; Im( ) 5z i z z
5 5 55) 6 Re( ) 6 ; Im( ) 0z z z
6 5 66) Re( ) ; Im( )
a
h h h
a i b b
z z z
ﺍﳌﺮﻛﺒﺔ ﺍﻷﻋﺪﺍﺩ ﻋﻠﻰ ﺍﻟﻀﺮﺏ ﻭﻋﻤﻠﻴﺔ ﺍﳉﻤﻊ ﻋﻤﻠﻴﺔ ﺧﻮﺍﺹ
أﻷﻋﺪاد ﺧﺬ, ,z w v
1(ﺗﻮﻓﺮاﻟﺠﻤﻌﻲ اﻟﻤﺤﺎﯾﺪ اﻟﻌﻨﺼﺮواﻟﻮﺣﯿﺪ)( additive unitﺣﯿﺚ اﻟﺼﻔﺮ وھﻮ:z + 0 = z.
2(ﻣﺮﻛﺐ ﻋﺪد ﻟﻜﻞzاﻟﺠﻤﻌﻲ ﻧﻈﺮﯾﮫ ﯾﻮﺟﺪ- zوﯾﺤﻘﻖ:Additive inverses z + (-z)= 0
3(ﻋﻠﻰ ﺗﺠﻤﯿﻌﯿﺔ اﻟﺠﻤﻊ ﻋﻤﻠﯿﺔﻣﺠﻤﻮﻋﺔاﻷﻋﺪاداﻟﻤﺮﻛﺒﺔ:Addition is associative
z w v z w v
4(ﻋﻤﻠﯿﺔاﻟﺠﻤﻊاﻟﻤﺮﻛﺒﺔ اﻷﻋﺪاد ﻣﺠﻤﻮﻋﺔ ﻋﻠﻰ ﺗﺒﺪﯾﻠﯿﮫAddition is commutative
z w w z
5(ﻋﻠﻰ اﻟﻀﺮب ﻋﻤﻠﯿﺔ ﻓﻲاﻷﻋﺪاداﻟﻤﺤﺎﯾﺪ اﻟﻌﻨﺼﺮ ﯾﺘﻮﻓﺮ اﻟﻤﺮﻛﺒﺔاﻟﻮﺣﯿﺪاﻟﻮاﺣﺪ وھﻮ:
Multiplicative unit z · 1 z
6(ﻣﺮﻛﺐ ﻋﺪد ﻟﻜﻞz،0z ھﻮ اﻟﻀﺮﺑﻲ ﻣﻌﻜﻮﺳﮫ ﯾﻮﺟﺪ
1
z
وﯾﺤﻘﻖ:
1 Multiplicative z · z 1inverse
7(ﺗﺠﻤﯿﻌﯿﺔ اﻟﻤﺮﻛﺒﺔ اﻷﻋﺪاد ﻋﻠﻰ اﻟﻀﺮب ﻋﻤﻠﯿﺔ:
( Multiplication is associative) z w v z ( w v).
4. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)4(
8(ا اﻷﻋﺪاد ﻋﻠﻰ اﻟﻀﺮب ﻋﻤﻠﯿﺔﺗﺒﺪﯾﻠﯿﮫ ﻟﻤﺮﻛﺒﺔ:
Multiplicationis commutative z w wz
9(اﻟﺠﻤﻊ ﻋﻤﻠﯿﺔ ﻋﻠﻰ ﺗﻮزﯾﻌﯿﺔ اﻟﻀﺮب ﻋﻤﻠﯿﺔ:
Distributivity of multiplication over addition z w v z w z v
ﻣﻦ:(1 , (2 ,(3 , (4ﻧﺴﺘﻨﺞأن:( , )ﺗﺒﺪﯾﻠﯿﮫ زﻣﺮة ﺗﻤﺜﻞ....( 11 )
ﻣﻦ:(5 , (6 ,(7 , (8ﻧﺴﺘﻨﺞأن:( , )
ﺗﺒﺪﯾﻠﯿﮫ زﻣﺮة ﺗﻤﺜﻞ......(12)
ﻣﻦ(9وﻣﻊ(11)و(12)ﻧﺴﺘﻨﺘﺞ( , , ) ﺣﻘﻼواﻟﻤﺮﻛﺒﺔ اﻷﻋﺪاد ﺣﻘﻞ ﯾﺴﻤﻰ.
اﻟ اﻷﻋﺪاد ﻋﻠﻰ ﯾﻄﺒﻖ اﻟﺤﻘﯿﻘﯿﺔ اﻷﻋﺪاد ﻧﻈﺎم أن ﻧﺴﺘﻨﺘﺞ ﺳﺒﻖ ﻣﻤﺎﻤﺮﻛﺒﺔ.
21 1i i
4p اﻟﻜﺒﯿﺮة ﻟﻠﻘﻮى وﺧﺎﺻﺔ اﻟﺘﺎﻟﯿﺔ اﻟﻘﺎﻋﺪة وﺿﻊ وﻧﺴﺘﻄﯿﻊ:ﻟﺘﻜﻦ
وﻟﺘﻜﻦqﻗ ﺧﺎرج ﺗﺴﺎويﺴﻤﺔpﻋﻠﻰ4اﻟﺒﺎﻗﻲ وﻟﯿﻜﻦrاﻟﻘﻮل ﯾﻤﻜﻦ اﻟﺒﺎﻗﻲ ﻧﻈﺮﯾﺔ وﺑﻤﻮﺟﺐ
. ., 4 0 4i e p q r where r
44 4 4( ) 1 1p q r q qr r r rHence i i i i i i i i i
34 32 2 4 8 2
1: ( ) 1ex i i i i
63 4 15 3 3 2
2: ( ) 1ex i i i i i i i
1992 4 498
3: ( ) 1ex i i
ﻗﻮىiاﻟﺼﺤﯿﺤﺔ)The powers of i(
5. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)5(
18 18 18 20 2
4: 1 1ex i i i i i
31 32 33 34 31 32 33 34
31 32 33 34 4
31 32 33 34 36
5 4 3 2 4
5: ( ) 1
( ) ,
( ) ; 9
1
q
ex i i i i i i i i
i i i i i q
i i i i i q
i i i i i i
2
1i i
0i i
7 7 7 7 7
4 2
6:( 4 ) ( 4 1) (2 ) 2
128 128
ex i i
i i i i
اﻟﻘﻮل ﯾﻤﻜﻦ اﻟﺴﺎﺑﻘﺔ اﻷﻣﺜﻠﺔ ﻣﻦ:إذاﻛﺎنnﻓﺎن ﺻﺤﯿﺤﺎ ﻋﺪدا 1,1, ,n
i i i
وأن:
4 4 1 4 2 4 3 mod 4
1 ; ; 1 ; ;n n n n n n
i i i i i i i i
Modulo operation finds the remainder of division of one number by another.
n mod 4ﺗﻌﻨﻲ)اﻟﺒﺎﻗﻲ(اﻟﻌﺪد ﻟﻘﺴﻤﺔnﻋﻠﻰ4ﻣﺜﻼ ، 25 mod 4 1 , 25 mod 7 4
ﯾﻠﻲ ﻟﻤﺎ اﻧﺘﺒﮫ:اﻟﺘ اﻟﻌﺒﺎرةﺧﺎﻃﺌﺔ ﺎﻟﯿﺔ
1 1 ( 1) ( 1) 1 1
واﻟﺼﺤﯿﺢ:
2
1 1 1i i i
ﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺗﺴﺎوي ﺗﻌﺮﯾﻒEquality of Complex Numbers
1 2,let z a bi z c di
a bi c di a c and b d
أي:1 2 1 2Re( ) Re( ) Im( ) Im( )z z and z z
6. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)6(
Ex:
ﻗﯿﻤﺔ ﺟﺪx , yاﻟﻌﻼﻗﺔ ﺗﺤﻘﻖ واﻟﺘﻲ اﻟﺤﻘﯿﻘﯿﯿﻦ:x2
- 2xy i – y2
= 4i – 3
اﻟﺤﻞ:(1اﻟﻘﯿﺎﺳﯿﺔ ﻟﻠﺼﯿﻐﺔ اﻟﻤﻌﺎدﻟﺔ ﻃﺮﻓﻲ ﺣﻮل a b iاﻟﻤﺮﻛﺐ ﻟﻠﻌﺪدﯾﻠﻲ وﻛﻤﺎ:
2 2
( – ) ( 2 ) – 3 4 x y xy i i
(2ﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺗﺴﺎوي ﺗﻌﺮﯾﻒ ﻃﺒﻖ:
2 2
1 2Re( 3 ..........( 1) Re( ) )x y ez z q
1 2 2 4
2
.......
Im( ) Im( )
...( 2)
xy
y eq
x
z z
ﻗﯿﻤﺔ ﺗﻌﻮضyاﻷوﻟﻰ اﻟﻤﻌﺎدﻟﺔ ﻓﻲ:
2
( )
2 4 2
2
4 2
2 2
4
3 ..........( 1) 4 3
3 4 0
( 4)( 1)
x
x eq x x
x
x x
x x
2 2
4 0 4 4 4 1 2x x x i ﻞ ﺗﮭﻤ or
2 2
1 0 1 1
2
1 2
1 2
x x x
but y
x
When x y
When x y
ﻣﺮﻛﺒﲔ ﻋﺪﺩﻳﻦ ﻭﻃﺮﺡ ﲨﻊ:AAddddiittiioonn aanndd SSuubbttrraaccttiioonn ooff CCoommpplleexx NNuummbbeerrss
ﻋﺪدا ﻣﺮﻛﺒﺎ ﻓﺈن a + b i , c +d i ﻛﺎن إذاﻣﻦ ﻛﻞ
( ) ( ) = (a+c)+(b+d) a b i c d i i Sum:اﻟﻤﺠﻤﻮع
7. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)7(
( ) ( ) = ( )+( - ) ﺮح اﻟﻄ
= ( ) ( )
:a bi c di a bi c di
a c b d i
subtraction
ﺍﳌﺮﻛﺒﺔ ﺃﻷﻋﺪﺍﺩ ﺿﺮﺏ)Multiplying Complex Numbers:(
اﻟﺤﻘﯿﻘﯿﺔ ﻟﻸﻋﺪاد اﻟﺤﺪود ﻛﺜﯿﺮات ﻓﻲ اﻟﻀﺮب ﻋﻤﻠﯿﺔ ﻧﻔﺲ ھﻲ اﻟﻤﺮﻛﺒﺔ ﻟﻸﻋﺪاد اﻟﻀﺮب ﻋﻤﻠﯿﺔ إن.
1:Ex
2
( )( ) ( ) ( )
( ) ( )
a b i c d i a c d i b i c d i
ac ad i b c i b d i
ac ad i b c i b d
ac b d ad b c i
2
2 : ( 2 4 )(3 2 ) 6 4 1 2 8
6 4 1 2 8
2 1 6
E x i i i i i
i i
i
2
( 4 1)(3 9) ( 2 1)(3 3 ) 6 6 3 3 9 3i i i i i i 2 :Ex
2 2
3 : ( 2 4 ) 4 16 16 12 16E x i i i i
2 24 4 2 2
2 2
4 : (1 4 ) (1 2 ) (1 2 ) 1 4 4
( 3 4 ) 9 2 4 16 7 24
E x i i i i
i i i i
16 8
21 1 1
5: ( )
1 1
i i
Ex
i i
2
2i i
1 2
2i i
8
8 82
( ) ( 1) 1
2
i
i
2
1i
3 3 2 2
2 2
2 2
6 : (1 2 ) (1 2 ) (1 2 ) (1 2 ) (1 2 ) (1 2 )
(1 4 4 )(1 2 ) (1 4 4 )(1 2 )
( 3 4 )(1 2 ) ( 3 4 )(1 2 )
3 6 4 8 3 6 4 8
11 2 11 2 4
E x i i i i i i
i i i i i i
i i i i
i i i i i i
i i i
8. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)8(
2
7 : (2 )( ) 3Ex i x iy
i
22 2
(2 )( ) 3 2 2 2i x i y x y i x i y i
i i
2
( i )
2 2
( ........ ) ( ...... )
( )2 2)
2 3
3 ( 2
x yy i x i i
x
i
y x iy
1 2 7
1 2 7 1 2 7 4 3 7 8
9 : ,
( ) ( )
1
n
n n n
E x i n
i i i i i i
i i
ﺍﳌﺮﻛﺐ ﺍﻟﻌﺪﺩ ﻣﺮﺍﻓﻖComplex conjugate
ﺗﻌﺮﯾﻒ:
ﻛﺎن إذاz x yi ﻣﺮاﻓﻖ ﻓﺈنzﺑﺎﻟﺼﻮرة ﯾﻜﺘﺐz z x yi
وأن:
1) ; 2)z z z z
اﻟﻌﺪد أن ھﻮ ﻟﻠﻤﺮاﻓﻖ اﻟﮭﻨﺪﺳﻲ واﻟﺘﻔﺴﯿﺮوﻣﺮاﻓﻘﮫ اﻟﻤﺮﻛﺐﻣﺤﻮر ﻣﻊ ﻣﺘﻨﺎﻇﺮانx-axisاﻟﺘﺎﻟﻲ اﻟﺸﻜﻞ ﻓﻲ ﻛﻤﺎ
اﻟﺘﺮاﻓﻖ ﺧﻮاص واﻟﯿﻚ:
9. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)9(
7(zﺣﻘﯿﻘﯿﺎ ﻋﺪداإذاوإذاﻓﻘﻂ
8(ﺑﺎﻟﻌﻼﻗﺔ ﯾﺮﺗﺒﻄﺎن واﻟﺘﺨﯿﻠﻲ اﻟﺤﻘﯿﻘﻲ اﻟﺠﺰأﯾﻦ:Re( ) ; Im( )
2 2
z z z z
z z
i
9(إذاﻛﺎن z x y i ﻓﺈن:
2 2
z z x y
ﻣﺜﺎل ﻣﻦ اﻟﺘﺎﻟﯿﺔ اﻷﻣﺜﻠﺔ9ﻣﺜﺎل اﻟﻰ19اﻻﻋﺘﯿﺎدﯾﺔ ﺑﺎﻟﺼﯿﻐﺔ ﻣﻘﺎدﯾﺮھﺎ أﻛﺘﺐ)اﻟﻘﯿﺎﺳﯿﺔ(اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪد:
2
2
2 3
9 :
3
2 3 3 2(3 ) 3 (3 ) 6 2 9 3
3 3 9 9 1
9 7 9 7
10 10 10
i
E x
i
i i i i i i i i
i i i
i
i
S am ple
3 2 3 2
10:
2 2
i i
Ex
i i
2
( i
2
3 2 2 3 3
) 1
2 2 2
i i i
i
وا اﻟﺒﺴﻂ ﺑﻀﺮب اﻟﺴﺆال ﺣﻞ ﯾﻤﻜﻦ واﻟﻤﻘﺎم ﺑﻤﺮاﻓﻖ ﻟﻤﻘﺎم.
2 2
2 2
2 1
11:
3 1
2 2 3 4 2 12 6 4 2
3 3 4 2 4 2 16 4
10 10 10 10 1 1
20 20 20 2 2
i i
Ex
i i
i i i i i i i i
i i i i i i
i
i i
2
2
7 3
12:
1 12
7 3 1 7 3 1 2 3 7 14 3 3 2 3
1 2 31 12 1 1 2 3 1 2 3
13 13 3
1 3
13
Ex
i i i i i
ii i
i
i
10. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)10(
3
2
3 3 3 3
2
2 2
2
3
13:( )
1
3 1 3 3 2 4
( ) ( ) ( ) (1 2 )
1 1 1 2
(1 2 )(1 2 ) (1 2 )(1 4 4 ) (1 2 )( 3 4 )
3 4 6 8 11 2
i
Ex
i
i i i i i i
i
i i i
i i i i i i i
i i i i
1
2
2
14: (4 3 ) (2 1)
4 3 1 2 4 8 3 6 10 5
2
1 2 1 2 1 4 5
Ex i i
i i i i i i
i
i i i
2 2
2 2
1 1
15 :
(2 ) (2 )
1 1 1 1
4 4 4 4 3 4 3 4
3 4 (3 4 ) 3
(3 4 )(3 4 )
Ex
i i
i i i i i i
i i
i i
4 3i
2
4 8
0
9 16 25
i
i
i
2 2
(1 ) (1 )
16:
1 1
1
i i
Ex
i i
2
2i i 1
1 i
2
2i i 2 2
1 1 1
2 (1 ) 2 (1 ) 2
(1 )(1 )
i i
i i i
i i i i i
i i
2
2 2i i 2
2
2 4
2 0
1 2
i
i
i
2 2
2 2
2 3
1 7 )
1 2
2 3 2 1 3 2
. ( ) ( )
1 2 1 1 2 2
2 2 6 3 2 1 3 5 5
1 4 2 5
1 3 5 5 3
( ) ( )
2
1
2
1
2
2 5 5 2
3
( ) ( )
2
i i
E x x y
i i
i i i i i i
an s x y x y
i i i i i i
i i i i i i i i
x y x y
i i
i x i y x i y i
xy i
x y
x y
11. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)11(
7 1
6 9
71 7 0 1 70
69 70 1
7 0
2
1
18 :
1
1 1 1 1
: 1 1
11 1 1
1 1 1
(1 )
1 1
i
E X A M PL E
i
i i i i
S o lutio n i i
ii i i
i i
i
i i
2
2i i
7 0
2
70
7 0 m o d 4 2
2
1
2
2 2 2 2 2 0
2
i
i
i i i
2 73
2 73
2 3 4 5 6 7 68 69 70 71 72 73
19:1 .....
: 1 .....
(1 ) ( ) ..... ( )
EXAMPLE i i i
Solution i i i
i i i i i i i i i i i i i
ھﻮ اﻟﺤﺪود ﻋﺪد ان73+1=74ﻋﻠﻰ ﺗﻘﺴﻢ ﺣﺪا4ﻓﯿﻨﺘﺞ18واﻟﺒﺎﻗﻲ2اﻟﻰ اﻟﻤﻘﺪار ﯾﻘﺴﻢ ﻟﺬا ،18
اﻟﺤﺪ ﻣﻦ ﺑﺪءا ﻣﺘﺘﺎﻟﯿﺔ ﻟﺤﺪود ﻣﺠﻤﻮﻋﺔﻧﺎﺗﺞ اﻷولاﻷﺧﯿﺮﯾﻦ اﻟﺤﺪﯾﻦ ﻟﯿﺒﻘﻰ ﺻﻔﺮ ﺗﺴﺎوي ﻣﺠﻤﻮﻋﺔ ﻛﻞ.
ﯾﺴﺎوي اﻟﻨﺎﺗﺞ:
18 1872 73 4 4
1i i i i i i
آﺧﺮ ﻧﻮع ﻣﻦ أﻣﺜﻠﺔ:ﺟﺬراھﺎ اﻟﺘﻲ اﻟﺜﺎﻧﯿﺔ اﻟﺪرﺟﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ ﺟﺪM , Lأن ھﻞ ﺑﯿﻦ ﺛﻢM , Lﻣﺘﺮاﻓﻘﺎن.
2
: 3 4 1 4 4
(3 4 ) (1 4 ) 3 12 4 16 19 8
,
20: 3 4 ; 1
4
Ex
solution M L i i
M L i i i i i i
M L ﺘﺮاﻓﻘﯿﻦ ﻏﯿﺮﻣ
M i L i
ھﻲ اﻟﺜﺎﻧﯿﺔ اﻟﺪرﺟﺔ ﻣﻦ اﻟﻤﻌﺎدﻟﺔ:
2
( ) 0x M L x M L
ھﻲ اﻟﻤﻄﻠﻮﺑﺔ ﻓﺎﻟﻤﻌﺎدﻟﺔ ﻟﺬا:
2
4 19 8 0x x i
12. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)12(
2 1 : 5 1 ; 5 1
: 5 1s o lu tio n M
E x M i L i
L i
5 1i
2
2
1 0
,
(5 1) (5 1) 2 5 1 2 6
1 0 2 6 0
i
M L ﺘﺮاﻓﻘﯿﻦ ﻏﯿﺮﻣ
M L i i i
X i X T h e e q u atio n
5 5 9 2
22 : ,
1 3 4
i i
Ex M L
i i
اﻟﺤﻞ:ﻣﻦ ﺑﺎﻟﺘﺨﻠﺺ اﻟﺠﺬرﯾﻦ ﻧﺒﺴﻂiﺑﺎﻟﻤﻘﺎم
2
2
5 5 1 3 5 15 5 15 20 10
1 3 1 3 1 9 10
20 10
2
10 10
i i i i i i
M
i i i
i i
2
2
9 2 4 36 9 8 2 34 17
4 4 16 17
34 17
2
17 17
i i i i i i
L
x i i
i i
1) 2M L i 2 i
2
4
2) (2 )(2 ) 4 5M L i i i
M , Lھﻲ واﻟﻤﻌﺎدﻟﺔ ، ﻣﺘﺮاﻓﻘﺎن:
2
4 5 0x x
اﻷﺳﺌﻠﺔ ﻣﻦ آﺧﺮ ﻧﻮع:
ﻣﺜﺎل23:ﯾﺴﺎوي ﻣﺠﻤﻮﻋﮭﻤﺎ اﻟﻠﺬﯾﻦ اﻟﺘﺮﺑﯿﻌﯿﺔ اﻟﻤﻌﺎدﻟﺔ ﺟﺬرا ﺟﺪ2ﺿﺮﺑﮭﻤﺎ وﺣﺎﺻﻞ=17.
اﻟﺤﻞ:اﻟﺠﺬرﯾﻦ ﻟﯿﻜﻦm , n
أن وﺑﻤﺎ:2 ; 17 ,m n m n m n are conjugate
:Let m a bi n a bi
m n a bi
a bi
2 2 2 2 2
2 2
2 2 2 1
( )( )
17 1 4
1 4 , 1 4
a a a
m n a bi a bi a b i a b
b b
i i ﺔ ﺟﺬرااﻟﻤﻌﺎدﻟ
13. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)13(
ﻣﺜﺎل24(ﺟﺬرﯾﮭﺎ أﺣﺪ واﻟﺘﻲ اﻟﺤﻘﯿﻘﯿﺔ اﻟﻤﻌﺎﻣﻼت ذات اﻟﺘﺮﺑﯿﻌﯿﺔ اﻟﻤﻌﺎدﻟﺔ ﺟﺪ5 2i
اﻟﺤﻞ:ﻓﺈن ﻟﺬا ﺣﻘﯿﻘﯿﺔ أﻋﺪاد اﻟﻤﻌﺎﻣﻼت أن ﺑﻤﺎﻣﺘﺮاﻓﻘﺎن اﻟﺠﺬرﯾﻦ
إذناﻟﺠﺬرانھﻤﺎ:5 2 , 5 2i i اﻟﺤﻞ أﻛﻤﻞ.
25:Ex
أن ﺑﺮھﻦ3 iاﻟﻤﻌﺎدﻟﺔ ﺟﺬري أﺣﺪ ھﻮ:
2
8 16 2 0x x i اﻵﺧﺮ اﻟﺠﺬر ﺟﺪ ﺛﻢ.
اﻟﺤﻞ:اﻟﺠﺬرﯾﻦ ﻧﻔﺮض:m , nوأنm = 3 + iأن ﻓﯿﺠﺐ ﻟﻠﻤﻌﺎدﻟﺔ ﺟﺬرا ھﻮ ﻋﺪد أي ﯾﻜﻮن وﻟﻜﻲ ،
ﯾﺤﻘﻘﮭﺎ:
أن أي3 + iﺟﺬرﯾﮭﺎ أﺣﺪ ﻓﮭﻮ اﻟﻤﻌﺎدﻟﺔ ﺣﻘﻖ.
أن ﻧﻼﺣﻆ اﻟﻘﯿﺎﺳﯿﺔ اﻟﻤﻌﺎدﻟﺔ ﻣﻊ اﻟﻤﻌﻄﺎة اﻟﻤﻌﺎدة ﻣﻘﺎرﻧﺔ وﻣﻦ:
2
2
8 16 2 0
( ) 0
8 ; 16 2 3
3 8 5
x x i
x m n x m n
m n m n i But m i
i n n i ﺬراﻵﺧﺮ اﻟﺠ
26:Eq
اﻟﻤﻌﺎدﻟﺔ ﺟﺬري أﺣﺪ ﻛﺎن إذا
2
12 6 0x x x i c ﻗﯿﻤﺔ ﻓﺠﺪ اﻵﺧﺮ أﻣﺜﺎل ﺛﻼﺛﺔ ھﻮc
اﻟﺤﻞ:اﻟﺠﺬرﯾﻦ ﻧﻔﺮض:m , nوأنm = 3nاﻟﻘﯿﺎﺳﯿﺔ ﻟﻠﺼﯿﻐﺔ اﻟﻤﻌﺎدﻟﺔ ﻧﺤﻮل ،:
2
2
(12 16 ) 0
( ) 0 12 16
3 12 16 4 12 16 3 4
3(3 4 ) 9 12
x i x c
x m n x m n m n i
n n i n i n i ﺬراﻻول اﻟﺠ
m i i ﺬراﻟﺜﺎﻧﻲ اﻟﺠ
ﻃﺮﯾﻘﺘﯿﻦ اﻟﺤﻞ ﻟﺒﺎﻗﻲ:ﻓﻨ اﻟﻤﻌﺎدﻟﺔ ﯾﺤﻘﻖ اﻟﺠﺬرﯾﻦ أﺣﺪ اﻷوﻟﻰﻗﯿﻤﺔ ﻋﻠﻰ ﺤﺼﻞc
واﻟﺜﺎﻧﯿﺔ:اﻟﺠﺬرﯾﻦ ﺿﺮب ﺣﺎﺻﻞ=cاﻟﺜﺎﻧﯿﺔ وﻟﻨﻄﺒﻖ:
2
(3 4 )(9 12 ) 27 36 36 48 21 72c i i i i i i
2
2
( ? ) 8( ? ) 16 2 0
(3 ) 8( 3 ) 16 2 0
9 6
i
i i i
i
2
24 8i i 16 2i 0
9 1 24 16 0 0 0 ﺎﺋﺒﺔ ﻋﺒﺎرةﺻ
14. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)14(
اﺧﺮ ﻧﻮع:ﻟﻠﻌﺪد أﻟﻀﺮﺑﻲ اﻟﻨﻈﯿﺮ اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪد اﻻﻋﺘﯿﺎدﯾﺔ وﺑﺎﻟﺼﯿﻐﺔ ﺟﺪ:
1
2
27 : 2 4
1 4 2 4 2 4 2
: 4 2
4 2 4 2 16 4 20
1 1
5 10
Ex i
i i i
solution let z i z
i i i
i ﺮﺑﻲ اﻟﻨﻈﯿﺮاﻟﻀ
2
.
2 8 :
1
2
1E x z
i
so l z
i
2
( i 1
2
1 1 2 1 2
) 1 2
1 2 1 2 1 4
1 2 1 2
5 5 5
i i
i z
i i i
i
i ﺮﺑﻲ اﻟﻨﻈﯿﺮاﻟﻀ
1
2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2
.
9 :
2
2
2 2
a bi a bi
sol z
a bi a bi
a abi b i a
a bi
Ex z
a b
abi b a b ab
i
a b i a b a b a b
i
ﻧﻮعاﺧﺮ:30:Ex
ﻟﺘﻜﻦ:
7 11
,
1
i x i
a b
i x y i
ﻣﻦ ﻛﻼ وﻟﯿﻜﻦa , bﻗﯿﻤﺔ ﺟﺪ ، ﻣﺘﺮاﻓﻘﯿﻦx , y
اﻟﺤﻞ:
2
2
7 1 7 7 6 8
3 4
1 1 1 2
i i i i i i
a i
i i i
b = 3- 4iﻷنbﻟـ ﻣﺮاﻓﻖa
11
3 4
x i
i
x yi
2
(3 4 )( ) 11 3 3 4 4 11
3 3 4 4 11
2 3 4 4 11 0
( 2 4 ) (3 4 11) 0 0
i x yi x i x yi xi yi x i
x yi xi y x i
x yi xi y i
x y y x i i
15. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)15(
ﻋﻠﻰ ﻧﺤﺼﻞ ﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺗﺴﺎوي ﺗﻌﺮﯾﻒ وﻣﻦ:
ﺗﺪرﯾﺐ 1 1
ﻣﻦ اﻷﺳﺌﻠﺔ1اﻟﻰ14ﻟﻠﺼﯿﻐﺔ اﻟﻤﻘﺪار ﺣﻮل:
18(أن ﺑﺮھﻦ:
w w
c
z z
[Hint: ﺐ اﻛﺘz =a+b i , w = c+d i
ﻟﻠﺼﯿﻐﺔ ﺣﻮلa b iاﻟﻤﻘﺎدﯾﺮﻣﻦ19اﻟﻰ31
3
19 (2 )(3 2 ) 6 1
1
20
3
x y i x y i i
i
i
4 4
1 2 3 49
39
37
8 7 8 7 7 8
2 3
(21)
1 2
(22) 2 2
(23) 1 .....
1
(24)
1
(25) ,n n n
i i
x y
i i
i i
i i i i
i
i
i i i n
2
1
24 4 2
25 1 3
26 3
27 3
28
1
29
2
i
i
i
i
x iy
x iy
i
i
2 2
7 3
30
1 12
2 2
31
1 2 1 2
i i
i i
2 4 0 .........( . 1) 2
3 4 11 0 .......( .2)
8 4
x y eq
y x eq
y x
0
3 4y x
11 0
11 11 0 1
2 4 0 2
y y
x x
16. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)16(
.(32)أن ﺑﺮھﻦ2 4 iاﻟﻤﻌﺎدﻟﺔ ﺟﺬري أﺣﺪ ھﻮ
2
4 20 0x x
33ﻛﺎن إذا
7 13
,
2 4
i i
m n
i i
1(أن ﺑﺮھﻦ,m nﻣﻦ ﻛﻼ ﺟﺪ ﺛﻢ ﻣﺘﺮاﻓﻘﺎن:
2 2 3 3
, , ,m n m n m n m n
2(ﺟﺬراھﺎ اﻟﺘﻲ اﻟﺘﺮﺑﯿﻌﯿﺔ اﻟﻤﻌﺎدﻟﺔ ﺟﺪ,m n
.(34)ﺑﺎﻟﺼﯿﻐﺔ اﻟﺘﺎﻟﯿﺔ ﻟﻸﻋﺪاد اﻟﻀﺮﺑﻲ اﻟﻨﻈﯿﺮ ﺟﺪa b iﻣﻦ ﻛﻼ:
1
, 1 3i i i
: 35اﻟﻤﺘﻄﺎﺑﻘ ﺻﺤﺔ ﺑﺮھﻦﺔ:
1 tan
cos2 sin 2
1 tan
i
i
i
ﺣﯿﺚ
2
,
2
n
n
أﻣﺜﻠﺔ:ﻳﻠﻲ ﳑﺎ ﻛﻼ ﺃﻛﺜﺮ ﺃﻭ ﻣﺮﻛﺒﲔ ﻋﺪﺩﻳﻦ ﺿﺮﺏ ﳊﺎﺻﻞ ﺣﻠﻞ:
ﯾﺴﺘﻐﻞ أﺳﺌﻠﺔﻓﯿﮭﺎ2
1i اﻟﺤﺪﯾﻦ أﺣﺪ ﯾﻀﺮب ﻓﺮق اﻟﻰ ﻣﺠﻤﻮع ﻓﻠﺘﺤﻮﯾﻞ× 2
i
2
2
2
31: 9 2
. 9 25 (3 5 ) 3
5
( 5 )
E
sol z x i x i
x z x
x i
2
2 2 2
2
. 4 9 (2 3 )(2 3
32: 9
)
4E
sol z x y i x yi x y
x z x y
i
4
2 2 2 2 2
2
. ( 9)
33: 5 36
( 4) ( 9)( 4 ) ( 3)( 3)( 2 )( 2 )sol z x x x x i x x x i x
Ex z x x
i
2
2 2
. (4 ) 9 (4 3
34: (4
)(4
) 9
3 )sol z x i x i
Ex z x
x i
2
. 16 9 16 9
35: 25
(4 3 )(4 3 )sol z i
Ex z
i i
2
2
. 81 36 81 36 (9 6 )(9 6 )
9 13 9(9 4) 9(9 4 ) 9(3 2
36:
)(3 2 )
117
sol z i i i
or z i
Ex z
i i
17. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)17(
3 3 2 2
3
2 2
. 8 (2 )(
37: 8
4
4 2 )
(2 )(4 2 1) ( 12 )( 2 )
sol z x i x i x x i i
x i x x i x i x
x
i
E z x i
x
3 3
3
3 3 3 2
3
2
21 1 1
. (216 ) (216
1
38: 5
) (6 )(36 6 )
4 4 4
1
(6 )( 6 )
4
1
4
4
36
sol z x i y x i
Ex
y x y i x xy i i
x y i x y
z x i y
x i
2
2
2
39: 6
. 6 16 ( 8 )( 2 )
16E
s
x z x
ol z x
xi
xi i x i x i
2 2
2
2
2
2 2
1
. 6 25 ( 6) 9
2
6 9 9 25
( 3) 16
( 3) 16 ( 3 4 )( 3 4
40: 6 25
)
sol z x x added and subtract
x
E
x
x
x i x i x
x z x x
i
2 2
2
2
2 2
2
1
. 8 ? 52 ( 8)
41
16
2
8 16 16 52
( 4) 36
( 4) 36 ( 4 6 )
: 8
( 4 )
2
6
5
sol z x x Added and subtract
x x
x
x i x i
Ex z x x
x i
2 2
2
2
1 1
. 1 ( 1)
42 :
2 4
1 1
1
4 4
1
sol z x x Added and su
Ex a z x x
btract
x x
18. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)18(
2
2 2
2 2 2 2 2
2 2 2
2
4
1 3
( )
2 4
1 3 1 3 1 3
( ) ( )( )
3: 4 9 25
2 4 2 2 2 2
. 4 9 25 (2 3 25 )(2 3 25 )
Ex z x
x
x i x i x i
sol z x i Sin y x i Sin y x i Sin y
Sin y
اﻷوﻟﻰ اﻟﺪرﺟﺔ ﻣﻦ ﻋﻮاﻣﻞ ﺿﺮب ﻟﺤﺎﺻﻞ ﺣﻠﻞ:
3
44 : 8Ex b x
23 2 2
2 2
.: 8 2 2 4 2 2 1 1 4 2 1 3
2 1 3 2 1 3 1 3
Sol x x x x x x x x x
x x i x x i x i
أﺧﺮى ﻓﻜﺮة:ﺑﺎﳌ ﺍﻟﻀﺮﺏ ﺑﺪﻭﻥﺑﺼﻴﻐﺔ ﺍﻛﺘﺐ ﺮﺍﻓﻖa + biﺍﳌﺮﻛﺐ ﺍﻟﻌﺪﺩ:
2
5
44:
3
i
Ex z
i
اﻟﺤﻞ:اﻟﻤﻘﺎم ﻟﯿﻜﻦz = 3 + i;2 2
9 1 10z z a b
وﺗﻨﺲ ﻻﻣﻦاﻟﻤﺮﻛﺐ اﻟﻌﺪد وﺣﺪة ﺗﻌﺮﯾﻒ
2 2
1 1i i
ﻧﻀﺮبوﻣﻘﺎم ﺑﺴﻂاﻟﻌﺪد)5i(ﻓﻲ10
25
(10) (9 1) (9 ) (3 ) (3 )
10 2 2
3 3 3
i i i
i i i
z
i i i
2
3
i
i
21 1 1 3
(3 ) (3 ) (3 1)
2 2 2 2 2
i
i i i i i
4
45:
2 4 3
i
Ex z
i
اﻟﺤﻞ:اﻟﻤﻘﺎم z = 4 + 3 i
2 2
16 9 25 z z a b
19. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)19(
24 4
(25) (16 9) (4 3 ) (4 3 )(16 9 )
25 25 2
4 3 4 3 4 3
i i i
i ii
z
i i i
4
25
4 3
i
i
24 4 4 12 16
(4 3 ) (4 3 ) (4 3)
25 25 25 25 25
i
i i i i i
ﻧﻮعﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺗﺴﺎوي ﺗﻌﺮﯾﻒ ﻋﻠﻰ آﺧﺮ:
ﻗﻴﻤﺔ ﺟﺪx , yﺍﳊﻘﻴﻘﻴﲔﺍﳌﻌﺎﺩﻟﺔ ﳛﻘﻘﺎﻥ ﻭﺍﻟﻠﺬﻳﻦ:
2 1 3 2 1
.: (
2 3
) ( )
1 1 2 2
1
46:
1 2
i
i i
Ex x
i i i
sol x y
i i i i i
y
i i i
2
( i )
2 2
2 2
2 2 6 3 2
( ) ( )
1 4
i i i i i i
x y i
i i
1 3 5 5 1 3
( ) ( ) ( ) (1 )
2 5 2 2
1 3
2 2
i i
x y i i x i y i
x x i y y i i
ﻟﻠﺼﯿﻐﺔ اﻟﻤﻌﺎدﻟﺔ ﻃﺮﻓﻲ ﻧﺤﻮل:a + b i
3
2
3
( ) ( ) 0
2
1
1
( )
2
1
2
x y
x
x i y i i
x y iy i
ﺗﻌﺮﯾﻒ وﻣﻦﻋﻠﻰ ﻧﺤﺼﻞ ﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺗﺴﺎوي:
2
1 2
1
Re ( . . )
Im ( . .
1
Re ( . . ) 0 .....( 1)
2
3
Im ( . . ) 1 .....( 2)
2
1
1
)
1
2
z
z
z
z
R H S x y eq
R H S x y eq
x x
y
L H S
L H S
20. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)20(
2
2 21 1
.
1
.47 : (1 3
(1 6 9 ) 1 3 3
1 1
) (1 )(1 3 )
1
i
Ex x
i i
sol x i i y i
i
i i
i i
y i i
i
1
2
2i i
1 2
1 2
( 8 6 ) 4 2
2
6 4 2 ( 8 ) ( 6 ) 4 2
1
Re Re 4
2
1
Im Im
8
6 2 6( ) 2 5
2
8z z
z z
x i y i
i x i y i y x y i i
y
and x x
y
y
y x
2
2
2
3
6 3 2
. (1 ) (1 )
2 2
6 6 3 2
(
6 3
.48: (1 )
2
1
4
i i
sol i i
x yi i i
i i i
x yi i
i
Ex i
x yi i
2
2i i
2
) (1 )
6 5 5
2 2
5
6 6
1 2 2 1
6 6 1
1 1 1
6 6
3 3 3 ; 3
2
i
i
i i
x yi
i i i
x yi x yi
i
x yi x yi
i i i
i
x yi i x y
ﻟﻠﻨﺘﺎﺋﺞ ﺟﺪوﻻ ﻧﻈﻢ:اﻟﻄﺒﻊ ﻟﻤﺴﺎﺣﺔ اﺧﺘﺼﺎرا ﻟﻠﻨﺘﺎﺋﺞ ﺟﺪاول اﻧﻈﻢ وﻟﻢ
-3x
3y
21. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)21(
1 2
1 2
2
2
. 2 2 4 1 8
2 2 8 2 2 3 8
(2 2 ) 3 8
Re Re 3 ......( 1)
Im Im 2 2 8 4 ....(
.49: ( 2 )(
2
2 ) 1
4 .
8
3
)
z z
z z
sol x y x i i y i i
x i i y i xy x i i y i
x y x y i i
Ex x i
x y eq
an
y i i
x
d x y x
y
y xy eq
اﻟﻤﻌﺎد ﺗﺤﻞﺑﺎﻷوﻟﻰ اﻟﺜﺎﻧﯿﺔ اﻟﻤﻌﺎدﻟﺔ ﺑﺘﻌﻮﯾﺾ آﻧﯿﺎ ﻟﺘﯿﻦ:
2 2
(4 ) 3 4 3 4 3 0
( 3)( 1) 0 3 1 4
1 4 1 3 ; 3 4 3 1
x x x x x x
x x x or x But y x
when x y when x y
5
2 2 3 2 23
0
2 2 2 2 5
2 4 2 6 3 2 23
0
5 5 5
2 4 2 (6 3 2
2 3 23
50: 0
2 2
) 23 0
4 3 2
5
2 2 6 0
8 2
3
2
6
x y i i x y i i
i i i i
x x i y i y x x i y i y
x x
x yi x yi
i y i y x xi y i y
x i yi x i y i y
x y y i x i
Ex
i i
x y x
23 0 ( 8 23) ( 6 2 ) 0 0x y y x i i
1 2
1 2
(2)
Re Re 8 23 0 ......( 1)
Im Im 6 2 0 3 .....( 2)
8(3 ) 23 0 23 23 0 1 3
z z
z z
x y eq
and y x x y eq
y y y y then x
2
51: ( ) 4
4 2 0 2x yi x yi i
Ex
x and y
x yi
2
4 4 0
52 :( ) 16
x yi x and y
Ex x yi
22. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)22(
2 2 2 2
2
2
2 2
2 8 2 8
( ) (2 ) 0
53:( ) 8
8
x xyi y i i x
Ex x yi
xyi y i
i
x y xy i i
1 2
1 2
2 2
Re Re 0 ......( 1)
4
Im Im 2 8 .....( 2)
z z
z z
x y eq
and xy y eq
x
2
2 4 2 2
2
16
0 16 0 ( 4)( 4) 0
2 2
x
ﻞ ﯾﮭﻤ
x x x x
x
x y
1 2
1 2
2
2
2 2
2 2
2 2 2 2 2
( )
2 4 2 4 2
2
2 2
2 12 16 2 12 16
( ) (2 ) 16
Re Re ......( 1)
8
Im Im 2 16 .....( 2)
64
12 64 12 12 64 0
5
( 16
4 : ( ) 12 16
12
12
)( 4
z z
z z
x
ي
x xy i y i i x xy i y i
xy i i
eq
and xy y eq
Ex x yi
x
x x x x x
x
x
x
x
x
i
y
y
8
) 0 4
4 2 ; 4 2
ﻞ ھﻤ
x but y
x
when x y when x y
- 44x
2- 2y
2
2 2
2
3 3 (3 )
. ( ) ( )
3 3 10
3 3 1
( )
10 10 10
3 1 3 1
10
3
.55 : ( )
3
10 10 10
i
Ex
i i i
sol x y i x y i
i i
i
x y i x y i
x y i
i
W hen x y or W e x y
i
h n
23. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)23(
2
2 2 2
2
. 2
.56: 2 (2 )( 4
(2 )( 4 ) 2
)Ex x yi i x y
sol x yi i x y i x yi
(2 ) ( 2 )i x yi ( 2 )
1 2
1 (2 )( 2 ) 2
2 2
2 2 1 2 1
2 2 ;
5 5 5 5 10
x yi
i
i x yi x yi
i i
i
x yi x yi i x y
1
2
2 2
2 2 2 2 2
2 2
7 4 2 14 7 8 4 10 15
. 2 3
2 2 5 5
2 3 2 3
( ) 2 ( ) 4 3
2 4 3 2 4 3
( ) (
7 4
.57 : ( ) 2
2 ) 4 3
Re
2
z
i i i i i
sol m n i i
i i
m n i i m and n
i
Ex x y i m n i W hen m
B ut x y i m n i x y i i
x
n
x yi y i i x xyi y i
x y i
i
i
i
x y
2
1 2
2
2 2
(4 )
2 4 2 4 2
2
2 2
Re 4 ......( 1)
3
Im Im 2 3 .....( 2)
2
9
4 4 9 16 4 16 9 0
4
3 3
(2 9)(2 1)
2 2
0
3 3
2
z
z z
x
ﻞ ﯾﮭﻤ
x y eq
and x y y eq
x
x x x x x
x
x x x but y
x
w hen x y
2 2
3
2
1
;
2
3 1
2 2
w hen x y
24. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)24(
ﻛﻞiﺍﻟﻔﺮﺿﻴﺔ ﻃﺮﻳﻘﺔ ﻧﺴﺘﻌﻤﻞ ﺍﳉﺬﺭﺍﻟﱰﺑﻴﻌﻲ ﲢﺖ
10
.
.58: ( )
6 8
10
6 8
sol x y
E y i
i
x x
i
i
ﻧﻔﺮض:8 6 ,i a bi a b ﻋﻠ ﻧﺤﺼﻞ اﻟﻄﺮﻓﯿﻦ وﺑﺘﺮﺑﯿﻊﻰ:
1 2
1 2
2
2 2 2 2 2
2 2
( )
2 4 2 4 2 2 2
2
( ) 8 6 2 8 6 ( ) (2 ) 8 6
Re Re 8 ......( 1)
3
Im Im 2 6 .....( 2)
9
8 9 8 8 9 0 ( 1)( 9) 0
z z
z z
a
ﻞ ﯾﮭﻤ
a bi i a abi b i a b ab i i
a b eq
and ab b eq
a
a a a a a a a
a
3
1
1 3 ; 8 6 1 3
1 3 ; 8 6 1 3
a but b
a
when a b then i i
when x b then i i
اﻷوﻟﻰ اﻟﺤﺎﻟﺔ:ﻋﻨﺪﻣﺎ8 6 1 3i i ﺗﺼﺒﺢ اﻟﻤﻌﺎدﻟﺔ:
10
1 3
x yi
i
10 1 3 10
1 3 1 3
i
x yi
i i
(1 3 )
10
i
1 3 1 , 3i then x y
اﻟﺜﺎﻧﯿﺔ اﻟﺤﺎﻟﺔ:ﻋﻨﺪﻣﺎ8 6 1 3i i ﺗﺼﺒﺢ اﻟﻤﻌﺎدﻟﺔ:
10
1 3
x yi
i
10 1 3 10
1 3 1 3
i
x yi
i i
( 1 3 )
10
i
1 3 1 , 3i then x y
2
2 14 53
59: 1
2 7
x x
EXAMPLE x xy i
x i
2 2
2 2 2
.: 14 53 14 49 49 53
7 4 7 4
7 2 7 2
sol x x x x
x x i
x i x i
25. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)25(
ﺑﺎﻟﻤﻌﺎدﻟﺔ ﯾﻌﻮض اﻟﺘﺤﻠﯿﻞ ﻧﺎﺗﺞ:
2
7 2
1
x i
x xy i
7 2
2 7
x i
x i
2 2
2
2 2
1 7 2 6 2
6 2
Re : 6 6 0 3 2 0
3 2
Im : 2
2 1
3 2 / 3
x xy i x i x xy i x i
x xy i x i
x x x x x x
x or x
xy
when x y
when x y
( . 60Exﻟﻠﻌﺪﺩ ﺍﻟﱰﺑﻴﻌﻴﺎﻥ ﺍﳉﺬﺭﺍﻥ ﺟﺪ:
8
1 3
اﻟﺤﻞ:اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪد اﻟﻘﯿﺎﺳﯿﺔ ﻟﻠﺼﯿﻐﺔ اﻟﻌﺪد ﻧﺤﻮل:
8 8 8 1 3 8(1 3)
2 2 3
41 3 1 1 3 1 3 1 3
i i
i
i i
ﻧﻔﺮض2 2 3 i x yi اﻟﻄﺮﻓﯿ وﺑﺘﺮﺑﯿﻊﻦﻋﻠﻰ ﻧﺤﺼﻞ:
2
( ) 2 2 3x yi i
2 2 2 2 2
2 2
2 2 2 3 2 2 2 3
( ) (2 ) 2 2 3
x xyi y i i x xyi y i
x y xy i i
1 2
1 2
2
2 2
( )
2 4 2 4 2
2
2 2
Re Re 2 ......( 1)
3
Im Im 2 2 3 .....( 2)
3
2 3 2 2 3 0
3
( 3)( 1) 0 3
3 1 ; 2 2 3 3
3 1 ; 2 2 3 3
z z
z z
x
ﻞ ﯾﮭﻤ
x y eq
and x y y eq
x
x x x x x
x
x x x but y
x
w hen x y then i i
w hen w hen x y then i i
ﻣﻼﺣﻈﺔ1:ﻋﺎﻣﺔ ﺑﺼﻮرة اﻟﺠﺬور انﻻﺣﻘﺎ دﯾﻤﻮاﻓﺮ ﻣﺒﺮھﻨﺔ ﺑﻤﻮﺟﺐ ﺳﺘﺤﻞ.
26. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)26(
ﻣﻼﺣﻈﺔ2:ﻟﻠﻌﺪد اﻟﺘﺮﺑﯿﻌﻲ اﻟﺠﺬر ﻧﺠﺪ أن ﯾﻤﻜﻦz x yi اﻟﻄﺮﯾﻘﺔ ھﺬه وﺗﺴﺘﻌﻤﻞﻟﻠﺘﺤﻘﻴﻖﻏﯿﺮ ﻷﻧﮭﺎ
اﻟﺒﺎﺋﺲ ﻣﻨﮭﺠﻨﺎ ﻓﻲ ﻣﻮﺟﻮدةﻛﺴﺆال اﺳﺘﻨﺘﺎﺟﮭﺎ ﯾﻤﻜﻦ ﺣﯿﺚ.وھﻲﻟﻸﻃﻼﻉ
The modulus or absolute value of a complex number z ﻧﺠﺪﺍﳌﺮ ﺍﻟﻌﺪﺩ ﻣﻘﻴﺎﺱﻛﺐ
ﻟﯿﻜﻦ=r=|z|ﺣﯿﺚ2 2
r x y ،ر=اﻟﻤﻘﯿﺎس
اﻟﺘﺮﺑﯿﻌﻲ اﻟﺠﺬر ﻹﯾﺠﺎد اﻟﻘﺎﻋﺪة ﻧﻄﺒﻖ
2 2( )
r x y
x iy i
r x
ﻣﺜﻼ:8 6i
اﻟﺤﻞ:2 2
64 36 10r x y ﺣﯿﺚ8, 6x y
10 8 6 6
8 6 9 3
2 2(10 8) 36
i i i i
.61Ex:ﻟﻠﻌﺪﺩ ﺍﻟﱰﺑﻴﻌﻴﺎﻥ ﺍﳉﺬﺭﺍﻥ ﺟﺪ4 3i
اﻟﺤﻞ:اﻟﻤﻘﯿﺎس=16 9 5 r ﺣﯿﺚx = 4 , y = 3
5 4 3 3 1
4 3 ( ) ( )
2 2(5 4) 2 2
i i i
آﺧﺮ ﺗﻄﺒﯿﻖﺗﺴﺎ ﻓﻲﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ وي
62(EX.اﻟﻤﺮﻛﺐ اﻟﻌﺪد ﻛﺎن إذا 1 , 2اﻟﻤﻌﺎدﻟﺔ ﺟﺬري اﺣﺪ ھﻮ2
2 2 7 0x x bx a
ﻗﯿﻤﺔ ﻓﺠﺪa , bاﻟﺤﻘﯿﻘﯿﯿﻦ.
اﻟﺤﻞ:أن ﺑﻤﺎاﻟﻤﺮﻛﺐ اﻟﻌﺪد1 – 2iاﻟﻤﻌﺎدﻟﺔ ﯾﺤﻘﻖ ﻓﮭﻮ ﻟﺬا اﻟﻤﻌﺎدﻟﺔ ﺟﺬري أﺣﺪ ھﻮ:
2 2
2
2 2 7 0 2( ? ) 2( ? ) ( ? ) 7 0
2(1 2 ) 2(1 2 ) (1 2 ) 7 0
2(1 4 4 ) 2(1 2 ) (1 2 ) 7 0
6 8 2 4 2 7 0 15 4 2 0
( 15 ) ( 4 2 ) 0 0
4 2 0 2 15 0 15 2 0
x x bx a b a
i i b i a
i i b i a
i i b bi a b a i bi
b a b i i
b b and b a a a
17
.63Ex:ﻟﻠﻌﺪﺩ ﺍﻟﱰﺑﻴﻌﻴﺎﻥ ﺍﳉﺬﺭﺍﻥ ﺟﺪ32
اﻟﺤﻞ:
4 2 32 1 32 32z i z z z
27. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ)27(
اﻟﺴﺆاﻟﯿﻦ ﻓﻲ)1-2(اﻟﻤﺮﻛﺐ اﻟﻌﺪد ﺟﺪzﯾﺤﻘﻖ واﻟﺬي:
1:4 7 3 ( ) ; 2:3 2 4 ( ) :3 4Q z i i z Q z z i z ans i
ﺑﺎﻟﺼﯿﻐﺔ اﻛﺘﺐ ﺑﺎﻟﻤﺮاﻓﻖ اﻟﻀﺮب ﺑﺪوناﻟﻌﺪد اﻟﻤﺮﻛﺐ ﻟﻠﻌﺪد اﻻﻋﺘﯿﺎدﯾﺔ:
25
3 :
2 4
i
Q
i
ﻣﻦ ﻛﻼ أﻛﺜﺮ أو ﻣﺮﻛﺒﯿﻦ ﻋﺪدﯾﻦ ﺿﺮب ﻟﺤﺎﺻﻞ ﺣﻠﻞ:4:Q
4 4 4 2
2 2 3 3
90 45 - 16y + 12x
8
- 27 +
25
a b c x d x
e x x f x g x i h x
; ; ; -64
- 8 +12 ; -6x + 25 ; ; 5 i
اﻷﺳﺌﻠﺔﻣﻦ5اﻟﻰ16ﻗﯿﻤﺔ ﺟﺪx , yاﻟﺘﺎﻟﯿﺔ ﺗﺤﻘﻖ واﻟﺘﻲ: 2
5: 2 3 4Q x y i i
2 2 2
2
2
2 1 2
6:( ) 16; 7: 1 8:
3
8 4 2
9: 10: 4 1 8
8 4
12 52
11: 6 ; 12: (
4 6
i
Q x yi Q x y i Q x y a bi x y i
i
i
Q x y i Q x y i x y i i
i i
x x
Q x y i x Q xy
x i
2 2
2
4 8
2 )
1 1
5 25 4(1 5 )1 200
13: 3 2 8 14: 3 2 ; 15:
4 3 2
16: 2 7 5 whene z = -1= 2i
x y i
i i
i
Q x i y i Q x y Q x y i
i i
Q z z x y i i
ﻣﻦ ﻟﻜﻞ اﻟﺘﺮﺑﯿﻌﯿﺎن اﻟﺠﺬران ﺟﺪ:17 : Q
100 7 3
: ; B: ; : 11-24 3 ; E : -i
16 3 1 12
: 4 3 ; : 16 12 ; : 4 2 3 2
A D
F i G i H i
ﺗﺴﺘﻄﯿﻊ ھﻞإﺟﺎﺑﺔاﻟﻔﺮﻋﯿﻦG , HﺑﺎﻻﻋﺘﻤﺎدﻋﻠﻰإﺟﺎﺑﺔF؟
ﺗﺪﺭﻳﺐ)1 - 2(
28. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ
28
ﺍﳌﻌﺎﺩﻻﺕ ﺣﻞ
ﻓﻲاﻟﺘﺎﻟﯿﺔ اﻟﻤﻌﺎدﻻت ﻣﻦ ﻛﻞ ﺣﻞ ﺟﺪ:
2
2
. 8 8 2 2
2 2 , 2 2
.58: 8 0
sol x x i
S i
Ex
i
x
2
2
.59: 8
. 8 8
0
ﯿﺔ ﻓﺮﺿ
sol x
Ex x i
i x i
2
2 2
2 2
: 8 ( ) 8
( ) (2 ) 8
0 ...(1)
4
2 8 ...(2)
ﺎﻟﻄﺮﻓﯿﻦ ﺑﺘﺮﺑﯿﻌ
L et i a bi a bi i
a b ab i i
a b
ab b
a
ﻋﻠﻰ ﻧﺤﺼﻞ وﺑﺎﻟﺘﻌﻮﯾﺾ:
2
2 4 2 2
2
, 2 2
16
0 16 0 ( 4)( 4) 0 2
4 4
2
2
8 (2 2 )
2 2
ﻞ ﯾﮭﻤ
a
i
a a a a a
a
but b b
a
Hence i i
Hence S i
4
2 2 2 2
2
. ( 64)( 9) 0 (
.60
64 8) (
: 55 576 0
9 3)
8, 8, 3, 3
sol z z z z or z z i
S
q z
i i
E z
2
.61: 4 8 4 0Ex i z i z i
ﻓﻲ ﻃﺮﻓﯿﮭﺎ ﺑﻀﺮب اﻟﺜﺎﻧﯿﺔ اﻟﺪرﺟﺔ ﻣﻌﺎﻣﻞ ﻣﻦ ﻧﺘﺨﻠﺺ ﻟﻠﺤﻞ)-i(ﻋﻠﻰ ﻓﻨﺤﺼﻞ:
2
4 8 4 0z z i
ﺑﺎﻟﻘﺎﻧﻮن ﺗﺤﻞ اﻟﺜﺎﻧﯿﺔ اﻟﺪرﺟﺔ ﻣﻦ ﻣﻌﺎدﻟﺔ ﻛﻞ)اﻟﺪﺳﺘﻮر: (
29. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ
29
2
2
1 , 4 , 8 4 ; 4 16 4(8 4) 32
4 4 32 4 4 2
2 2 2
2 2 2 ﯿﺔ ﻓﺮﺿ
a b c i D b ac i i
b b ac i i
z i
a
اﻟﺘﺮﺑﯿﻌﻲ اﻟﺠﺬر أوﻻ ﻧﺠﺪ:
2
2 2
2 2
: 2 ( ) 2
( ) (2 ) 2
0 ...(1)
1
2 2 ...(2)
ﺎﻟﻄﺮﻓﯿﻦ ﺑﺘﺮﺑﯿﻌ
L et i a bi a bi i
a b ab i i
a b
ab b
a
2
2 4 2 2
2
( )
1
0 1 0 ( 1)( 1) 0 1
1 1
1 2 (1 )
1
ﻞ ﯾﮭﻤ
a
a a a a a
a
but b b hence i i
a
اﻟﻔﺮﺿﯿﺔ ﻗﺒﻞ ﻟﻤﺎ ﻧﻌﻮد ﺛﻢ:2 2 2z i
2 2 2 2 2(1 )
2 2 2 4 2 (1) 2 2 2 2 (2)
4 2 , 2
z i i
z i i The root or z i i The root
Hence S i i
.(62)Exaأن اﺛﺒﺖ1ﺟ أﺣﺪ ھﻮﺬوراﻟﻤﻌﺎدﻟﺔ
3
2 1 0z z اﻵﺧﺮﯾﻦ اﻟﺠﺬرﯾﻦ ﺟﺪ ﺛﻢ.
اﻟﺤﻞ:اﻟﻌﺪد ﯾﻜﻮن ﻟﻜﻲ1ﯾﺤﻘﻘﮭﺎ أن ﻓﯿﺠﺐ اﻟﻤﻌﺎدﻟﺔ ﺟﺬور أﺣﺪ.
ﺻﺎﺋﺒﺔ ﻋﺒﺎرة
3
1: 2 1 0 1 2 1 0 0 0z z z
z = 1.
أﺻﺒﺢx – 1اﻟﺤﺪود ﻣﺘﻌﺪد ﻋﻮاﻣﻞ أﺣﺪ ھﻮﺗﺮﺗﯿﺐ ﺑﻌﺪ اﻟﻄﻮﯾﻠﺔ اﻟﻘﺴﻤﺔ ﻧﺴﺘﻌﻤﻞ اﻷﺧﺮى اﻟﻌﻮاﻣﻞ وﻹﯾﺠﺎد
ﺗﻨﺎزﻟﯿﺎ ﺗﺮﺗﯿﺒﺎ اﻟﺤﺪود.
اﻟﻘﺴﻤﺔ ﺧﺎرج أن ﻧﻼﺣﻆ اﻟﻄﻮﯾﻠﺔ اﻟﻘﺴﻤﺔ ﺑﻌﺪ
ھﻮ2
2 2 1z z
اﻟﻤﻌﺎدﻟﺔ ﺗﻜﻮن ﻟﺬا: 2
1 2 2 1 0Z z z
اﻣﺎZ – 1= 0اﻟﻰ ﯾﺆديZ = 1اﻷول اﻟﺠﺬر
2
3
2 2 1
21
z z
zZ
3
1
2
z
z
2
2
2
2 z
z
2
1
2 z
z
2
1
1
0
z
Z
Z
30. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ
30
ﺃﻭ
2
2 2 1 0z z اﻟﺜﺎﻧﯿﺔ اﻟﺪرﺟﺔ ﻣﻦ ﻣﻌﺎدﻟﺔ وھﺬه
اﻟﻄﺮﻓﯿ ﯾﻀﺮبﻦ×( - 1 )ﻋﻠﻰ ﻓﻨﺤﺼﻞ:
2
2 2 1 0z z
ﺑﺎﻟﺪﺳﺘﻮر ﺗﺤﻞ.a = 2 , b = 2 , c = 1
واﻟﺜﺎﻟﺚ اﻟﺜﺎﻧﻲ اﻟﺠﺬران
2
2 4 4(2)(1)4 2 4 2 2 1 1
2 2(2) 4 4 2 2
b b ac i
z i
a
ﺍﻟﺘﻜﻌﻴﺒﻴﺔ ﺍﳉﺬﻭﺭ
( .62Exﺟﺪاﻟﺘﻜﻌﯿﺒﯿﺔ اﻟﺠﺬور8 iﺑﯿﺎﻧﯿﺎ اﻟﺠﺬور وﻣﺜﻞ.
2
( )
3 3 3 3
2 2
2
2
0
1
8 8 8 0
( 2 ) ( 2 4 ) 0
2 0 2 (1)
2 4 0 1 , 2 , 4
4 2 4 16 2 12 2 2 3
3
2 2 2 2
: 2 ( 2, 0) 1
3 (
i
L et z i z i z i
z i z z i i
z i z root
Or z z i a b i c
b b ac i i i
z i
a
T HE roots z root
z i
2
2 2
3,1) 2
3 ( 3 ,1) 3
| | 2
root
z i root
z r a b
ﺑﺪاﯾﺘﮫ ﻣﺘﺠﮭﺎ ﺗﺸﻜﻞ اﻷﺻﻞ ﻧﻘﻄﺔ ﻣﻊ ﻧﻘﻄﺔ وﻛﻞ أرﻛﺎﻧﺪ ﻣﺴﺘﻮى ﻓﻲ ﻧﻘﺎط ﺛﻼث ھﻮ اﻟﺠﺬور ﻟﮭﺬه اﻟﺒﯿﺎﻧﻲ اﻟﺘﻤﺜﯿﻞ
وﻃﻮﻟﮫ اﻷﺻﻞ ﻧﻘﻄﺔ2ﻣﺘﺘ ﻣﺘﺠﮭﯿﻦ ﺑﯿﻦ اﻟﺰاوﯾﺔ وﻗﯿﺎس ، ﻃﻮل وﺣﺪةﺗﺴﺎوي ﺎﻟﯿﯿﻦ
2
3
ﻗﻄﺮﯾﺔ ﻧﺼﻒ زاوﯾﺔ
ﻧﻘﻄﺔ ﻣﺮﻛﺰھﺎ داﺋﺮة ﻣﻦ ﻧﻘﻂ ھﻲ اﻟﺜﻼﺛﺔ واﻟﻨﻘﺎطاﻷﺻﻞاﻟﻤﻌﺎدﻟﺔ ﺟﺬور ﻣﻦ ﺟﺬر ﻛﻞ ﻣﻘﯿﺎس ﻗﻄﺮھﺎ وﻧﺼﻒ
ﯾﺴﺎوي واﻟﺬي2 2
| | 2z a b .
ﻣﻼﺣﻈﺔ:ﺑﺎﺳﺘﺨﺪام اﻟﺠﺬور ھﺬه إﯾﺠﺎد ﯾﻤﻜﻦ
اﻟﺘﻲ دﯾﻤﻮاﻓﺮ ﻣﺒﺮھﻨﺔﻻﺣﻘﺎ ﺳﺘﺄﺗﻲ.x
x = 1
r = 2
y
2z
0z
1z
2
3
2
3
2
3
31. ﺍﻟﻨﺎﺻﺮﻱﻣﻮﺳﻰﻛﺎﻣﻞ
31
ﻣﻦ اﻷﺳﺌﻠﺔQ1اﻟﻰQ6ﻓﻲ اﻟﻤﻌﺎدﻻت ﺣﻞ ﺟﺪ
2 2 2
2 3
2
2 2
4 5 6
1: 8 7 1 0 ; : 5 5 3i ; Q : 7 13 1 0
3
:z 2 4 3 0 ; : 4 1 3 0 ; : 12 5 0
1
Q x i x Q i z i z i
i
Q z i z Q z z i Q z i
i
7:Qﻛﺎن اذا2اﻟﺜﻼﺛﺔ اﻟﺠﺬور أﺣﺪ ھﻮﻟﻠﻤﻌﺎدﻟﺔ3 2
8 4 8 0z Z Z اﻵﺧﺮﯾﻦ اﻟﺠﺬرﯾﻦ ﻓﺠﺪ.
8:Qﺟﺪ55 48i ا ُوﻓﻲ اﻟﻨﺎﺗﺞ ﺳﺘﺨﺪمإﯾﺠﺎداﻟﻤﻌﺎدﻟﺔ ﺟﺬري:2
(1 2 ) 13(1 ) 0z i z i
9:Qاﻟﻤﺮﻛﺐ اﻟﻌﺪد ﺟﺪzﯾﺤﻘﻖ واﻟﺬي
22
18 24 0z z i
: 10Qا اﻟﺠﺬور ﺟﺪﻟﺘﻜﻌﯿﺒﯿﺔﻟﻠﻌﺪدiﺑﯿﺎﻧﯿﺎ اﻟﺠﺬور وﻣﺜﻞ.
1
.: 1 , ; 2 1 , 4 ; 3 4 ,3 ; 4 1 , 1 5 ;
8
5 3 , 2 ; 6 2 3 , 2 4 ; (7) 3 ; (8) 38- , 1 5 ,
Ans i i i i i i i i
i i i i i i i
2 3
1 3
9 4 3 , 4 3 ; 10 ,
2 2
i
i i i i
ﺗﺪﺭﻳﺐ( 1 – 3 )
32. ALNASSIRYKAMIL32
ﺍﻟﺼﺤﻴﺢ ﻟﻠﻮﺍﺣﺪ ﺍﻟﺘﻜﻌﻴﺒﻴﺔ ﺍﳉﺬﻭﺭ
3 3 2
2
2
1 1 0 ( 1)( 1) 0
1 0 1
1 0 ; 1
4 1 1 4 1
2
3
2
13
2 2
Let x x x x x
Either x x
or x x a b c
b b a ic
x
a
1 1z 0 1z 0 1z
2
1 3
2 2
z i 1z 2
1z
3
1 3
2 2
z i
2
2z 2z
2
1 , ,
1 3
2 2
i 2 1 3
2 2
i
1 3
2 2
i 2 1 3
2 2
i
Omegaw
1
2
,
2
2 2
1
2
2
1 3
( ) ( )
2 2
1 3 3 1 3
4 2 4 2 2
z i
i i i z
43. KAMIL ALNASSIRY43
ﺍﳌﺮﻛﺒﺔ ﻟﻸﻋﺪﺍﺩ ﺍﳍﻨﺪﺳﻲ ﺍﻟﺘﻤﺜﻴﻞGeometrical representation of Complex Numbers
ﺃﺭﮔﺎﻧﺪ ﳐﻄﻂArgand diagram
Real axis
Imaginary axis
ﺍﻟﻘﻄﺒﻲ ﺍﶈﻮﺭPolar axis
ﺍﳌﺴﺘﻮﻱ ﰲ ﻗﻄﺒﻴﺔ ﺇﺣﺪﺍﺛﻴﺎﺕPolar coordinate in the plane
r
Polar axis
za , bz( , )r
r
2 22 2 2
cos ; sin
1tan so t
so
na
a r b r or
r a r a b
b b
a a
b
44. KAMIL ALNASSIRY44
Example6 2 , 3 4
1 2
z i z i
1 2 1 2,z z z z
1 21) z z
1 2 (6 2 ) (3 4 ) (6 3) (2 4) 9 2z z i i i i
1 26 2 (6 , 2) , 3 4 (3, 4)z i m z i n
Om1z
n2z
m , n , O , hh
h = ( 9 , - 2 )1 2 9 2z z i
1 22) z z
1 2 1 2( ) (6 2 ) ( 3 4 ) (6 3) (2 4) 3 6z z z z i i i i
1
2
6 2 (6,2)
3 4 ( 3,4)
m z i
n z i
h
1 2(3,6) 3 6 3 6h i z z i
45. KAMIL ALNASSIRY45
ﺍﳌﺮﻛﺐ ﻟﻠﻌﺪﺩ ﺍﻟﻘﻄﺒﻴﺔ ﺍﻟﺼﻮﺭﺓPolar form of the complex number
(cos sin )z r i
z r cis ccosinessine
r( )z x y i
2 2
| |z r x y r > 0|z|
complex numberThe modulus of
1 2,z z
1 2 1 2
1 1
2 2
1 2 1 2
1) | | | | | |
| |
2)
| |
3) | | | | | |
z z z z
z z
z z
z z z z
ﺍﳌﺮﻛﺐ ﺍﻟﻌﺪﺩ ﻣﺴﺘﻮﻯ ﰲ ﺑﻨﻘﻄﺔ ﳝﺜﻞ ﺍﳌﺮﻛﺐ ﺍﻟﻌﺪﺩ ﺃﻥﻫﻲ ﳐﺘﻠﻔﺔ ﺻﻴﻎ ﻭﻟﻪ:
1ﺩﻳﻜﺎﺭﺗﻴﺔ ﺻﻴﻐﺔrectangular co-ordinates( , )x yArgand
2ﺍﻋﺘ ﺻﻴﻐﺔﺟﱪﻳﺔ ﺃﻭ ﻴﺎﺩﻳﺔRectangular or Cartesian formz x i y
3ﺻﻴﻐﺔﺍﺣﺪﺍﺛﻴﺎﺕﻗﻄﺒﻴﺔ( , )
ﺮة اﻟﻤﺨﺘﺼ
r 1
tan
y
x
4ﻗﻄﺒﻴ ﺻﻴﻐﺔﺔﻭﻣﺜﻠﺜﻴﻪ(cos sin )z r i
5ﺃﺳﻴﺔ ﺻﻴﻐﺔEuler's formula
i
z r e
e
46. KAMIL ALNASSIRY46
ﺍﳌﺮﻛﺐ ﺍﻟﻌﺪﺩ ﺳﻌﺔAmplitude (Argument)of Complex Number
A nti clockwise rotatin
ﺍﳌﺮﻛﺐ ﻟﻠﻌﺪﺩ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺴﻌﺔz a bi (Principal value of an argument )
( )Arg zz
, 20
( )arg z( ) 2 ;arg z n n
AAa
Principal value of an argument
The principal value of anargument is the valuewhich lies between and
ﺍﻷﻭﻝ ﺍﻟﺮﺑﻊ ﰲ( )Arg z ﺍﻟﺜﺎﻧﻲ ﺍﻟﺮﺑﻊ ﰲ( )Arg z
ﺍﻟﺮﺍﺑﻊ ﺍﻟﺮﺑﻊ ﰲ( )Arg z ﺍﻟﺜﺎﻟﺚ ﺍﻟﺮﺑﻊ ﻭﰲ( )Arg z
47. KAMIL ALNASSIRY47
, 0 , 0z a bi a b
1( , )z a b
2
2 2
z r a b
cos , sin
a a
r r
tan
b
a
3, ,
3 4 6
(cos ,sin )
1 3
( , )
2 2
3 1
( , )
2 2
1 1
( , )
2 2
3
6
4
z
Z
z
Z
z
Z
z
Z
2
48. KAMIL ALNASSIRY48
4a or b
a = b = 0
z = 0
ﺇﺣﺪﺍﺛﻴﺎﺗﻪﺍﻟﻘﻄﺒﻴﺔﺍﳌﺜﻠﺜﻴﺔ
: 89)Example2 2 3i
1
2 3 2 ( 2 3 , 2 ) 2ed
z i quadrant
Then
22 2 2 2
( 2 3) 2 4r x y
3
( 2 3 , 2 ) 3 1
( , )
4 2 2
z
r
co s , sin
2 3 3 2 1
co s , sin
4 2 4 2
3 1
,
2 2 6
5
( )
6 6
x y
r r
B u t A rg z
5 5
(4, ) (4, 2 ) , k
6 6
z or z k
5 5
4(cos sin )
6 6
z i
(1,0)
0
(1,0)
(0,1)
2
3
(0,-1)
2
49. KAMIL ALNASSIRY49
2 : (90)z i Example
a = 0
2 2
2 (0, 2)
4 2
z o i
z r x y
2
cos 0 , sin 1
2
3
(cos ,sin ) (0, 1) int
2
x y
r r
circle
u
3 3
2 , or 2 , 2
2 2
z z k
3 3
2(cos sin )
2 2
z i
7 3
: ( 91)
1 12
Examplez
2
2 2
7 3 7 1 3 7 3 1 2 3
1 12 1 1 12 1 2 3 1 2 3
7 14 3 3 6 13 13 3
1 3
1 12 13
1 3 (1 , 3 ) ( , ) 4 2
| | 1 3 2
th
i i
i i
i i i
i
i
z i Quadrant
z r x y
z
cos , sin
1 3
cos , sin
2 2
1 3
,
2 2 3
5
( ) 2 2
3 3
x y
r r
But A rg z
50. KAMIL ALNASSIRY50
5
arg( ) 2 2 ,
3
z n k k
5 5
2 , 2 , 2
3 3
z or z k
5 5
2 cos sin
3 3
z i
42 14
: ( 92)
2
i
Example
i
z
2
2 2
42 14 2 84 42 28 14 70 70
14 14
2 2 4 1 5
| | 196 196 196 2 14 2
14 14 ( 14,14 ) ( , ) 2ed
i i i i i i
i
i i
z r x y
z i Quadrant
z
cos , sin
14 1 14 1
cos , sin
14 2 2 14 2 2
1 1
,
42 2
3
( )
4 4
x y
r r
But A rg z
ﺍﺍﻻﻋﺘﻴﺎﺩﻳﺔ ﺃﻭ ﺍﳉﱪﻳﺔ ﻟﺼﻴﻐﺔﺍﻟﻘﻴﺎﺳﻴﺔ ﺃﻭ:14 14z i
ﺍﻟﺪﻳﻜﺎﺭﺗﻴﺔ ﺍﻟﺼﻴﻐﺔ: 14 ,14z
ﺍﻟﻘﻄﺒﻴﺔ ﺍﻻﺣﺪﺍﺛﻴﺎﺕ:
3 3
14 2 , 14 2 , 2
4 4
z or z k
ﺍﳌﺜﻠﺜﻴﺔ ﺍﻟﺼﻴﻐﺔ)ﺍﻟﻘﻄﺒﻴﺔ(:
3 3
14 2 cos sin
4 4
z i
ﺍﻷﺳﻴﺔ ﺍﻟﺼﻴﻐﺔ
3
4
14 2
i
z e
51. KAMIL ALNASSIRY51
.(93)Ex
11
6
8
11
6
3 1
(cos ,sin ) ( , )
2 2
11 11 3 1
(cos sin ) 8(cos sin ) 8( ( ) )
6 6 2 2
4 3 4
z r i i i
i
.(94)Ex
3
4
12
3
4
1 1
(cos ,sin ) ( , )
2 2
3 3 1 1
(cos sin ) 12(cos sin ) 12( ( ) )
4 4 2 2
1 1 1 1 1
12 2 ( ( ) ) 12 2( ) 6 2 6 2
2 22 2 2
z r i i i
i i i
.(95)Ex
27
2
7
mod (2 )
27 24 3 3 3
rg( ) : 12
2 2 2 2
A z
3
( ) (0, 1) (cos ,sin )
2
(cos sin ) 7(0 1 ) 7
Arg z
z r i i i
6a i2
3
a
2
3
1 3
( , ) (cos ,sin )
2 2
2 2 1 3 1 3
(cos sin ) ( )
3 3 2 2 2 2
: 6
1 3
6
2 2
z r i r i r r i
z a i
hence r r i a i
B ut
52. KAMIL ALNASSIRY52
1 2
1 2
3
Im( ) Im( ) 6 2 2
2
1 1
Re( ) Re( )
2 2
z z r r
z z r a a
2 2 2
ﻣﻬﻤﺔ ﻗﻮﺍﻋﺪ:
1 2
1 2 1 2
1
1 2
2
1) arg( ) arg( ) arg( ) mod 2
2) arg arg( ) arg( ) mod 2
3) If 0 and is any integer then
arg( ) arg( ) mod 2
If z and z are two non zero complex numbers then
z z z z
z
z z
z
z n
nz n z
7
mod 2
11 77
2) arg ( 3 ) 7
:96
2 2 3 5 5 5
1) arg arg (2 2 3 ) arg( 1 )
1 3 4 12
arg( 3 ) 7
6 6
5 5
(12 )
6 6
i
E
i
i
i
x
i
i
54. KAMIL ALNASSIRY54
ﺍﻟﻘﻄﺒﻴﺔ ﺑﺎﻟﺼﻴﻐﺔ ﻋﺪﺩﻳﻦ ﻭﻗﺴﻤﺔ ﺿﺮﺏ
Products and Quotients of Complex Numbers in Polar Form
Let z1 and z2 be complex numbers, where
1 2 1 2 1 2 1 2 1 2 1 2
1 2 1 21 2
cos sin1 2 1 2
cos cos sin sin (sin cos cos sin
cos( ) sin( )
:
z z r r i
r r
This mean
i
s
ﻣﺮﻛﺒﲔ ﻋﺪﺩﻳﻦ ﻗﺴﻤﺔ
1 1 1 2 2 2
1 2 1 2 1 2
: (cos sin ) / (cos sin )
/ [cos ( ) sin ( )]
ﻤﺔ ﻗﺴ
r i r i
r r i
The quotient
1 1 1
2 2 2
1
1 2 1 2
2
:
| |
1.
| |
2. arg arg( ) arg( ).
This means
z z r
z z r
z
z z
z
55. KAMIL ALNASSIRY55
ﻣﱪﻫﻨﺔﺩﳝﻮﺍﻓﲑDe Moivre’s theorem
n
cos sin cos sin .
n
i n i n
97 : Prove that (cos sin ) cos sin ,n
Ex i n i n n
cos( ) cos( ) sin( ) sin ( )n n and n n
. . (cos sin ) cos sin( )
nn
L H S i i
cos sin( ) cos sin( )
n
i n i n
cos sin( ) cos sin( ) . .n i n n i n R H S
(cos sin )z r i
(cos sin )n n
z r n i n
56. KAMIL ALNASSIRY56
6
2 2
6
6 6 6
98 : Prove that (1 3 ) 64
. W rite z in polar form : z = r(cos + i sin )
r = 1 3 2
5
arg( ) 2 use De M oivre's Theorem
3 3
5 5 5 5
(1 3 ) ( ) 2(cos sin 2 cos( 6) sin( 6)
3 3 3 3
Ex i
S ol
a b
z then
i z i i
mod 2
64( cos10 sin 10 ) 64( cos 0 sin 0) 64i i
6
66 6 3 21 3
(1 3 ) 2( ) 2( ) 64 64( ) 64
2 2
i i
24
1 1 1 1 1
2 2 2 2 2
1
1 2
2
1 3
.99) Pr 1
3
: 1 3 | | 2 ; arg( )
3
3 | | 2 ; arg( )
6
| | 2
| | 1 ; arg( ) arg( ) arg( )
| | 2 3 6 3
i
Ex If z ove that z
i
Sol let z i z r z
let z i z r z
z
z z z z
z
1
1 2 1 2
2
24
24
mod 2
| |
cos( ) sin( ) cos sin
| | 6 6
cos sin cos 24 sin 24
6 6 6 6
cos4 sin 4 cos0 sin 0 1 (0) 1
z
z i i
z
then z i i
i i i
( .100)Ex
6 4
( 2 6 ) ( 1)i i
57. KAMIL ALNASSIRY57
1 1 1 1
1
6 4
6 6
6
6
1
.:
5
2 6 | | 2 6 2 2 ; arg( ) 2
3 3
5 5
(cos sin ) 2 2 (cos sin )
3 3
5 5 5 5
( ) 2 2 (cos sin ) 2 2 cos sin
3 3 3 3
5 5
512 cos( *6) sin(
( 2 6 ) ( 1
*6) 512(cos10
3 3
)Sol Let z
Let z i z z
z r i i
z i
i
i
i
i
mod 2
sin10 ) 512(cos0 sin0) 512i i
2 2 2 2
2
3
1 1 | | 1 1 2 ; arg( )
4 4
3 3
(cos sin ) 2 (cos sin )
4 4
Let z i i z z
z r i i
4
4
4
2
mod 2
6 4
1 2
3 3 3 3
( ) 2 (cos sin ) 2 cos 4 sin 4
4 4 4 4
4(cos3 sin3 ) 4(cos sin ) 4
( ) ( ) 512( 4) 2048
z i i
i i
z z z
.102: cos sin :
1 1
1) 2cos 2) 2 sin
1 1
3) 2cos 4) 2 sinn n
n n
Ex If z i then prove that
z z i
z z
z n z i n
z z
1 11
1) cos sin (cos sin )
cos sin cos( ) sin( )
cos sin
z z z i i
z
i i
i
cos sini 2cos
58. KAMIL ALNASSIRY58
1 11
2 ) cos sin (cos sin )
cos sin cos( ) sin( )
cos sin cos sin
cos
z z z i i
z
i i
i i
sin cosi 2 sin
1
3 ) (cos sin ) (cos sin )
cos sin cos ( ) sin ( )
cos sin
n n n n n
n
i
z z z i i
z
n i n n i n
n i n
cos sinn i n 2cos n
1
4 ) (cos sin ) (cos sin )
cos sin cos ( ) sin ( )
cos sin cos sin
cos
n n n n n
n
z z z i i
z
n i n n i n
n i n n i n
n
sin cosi n n sin 2 sini n i n
ﺗﻌﺮﻳﻒ:ﻛﺎﻥ ﺍﺫﺍzﻋﺪﺩﺍﻭﺃﻥ ﻣﺮﻛﺒﺎnﻓﺈﻥ ﻣﻮﺟﺒﺎ ﺻﺤﻴﺤﺎ ﻋﺪﺩﺍ
1
n
zﻟﻠﻌﺪﺩ ﺍﻟﻨﻮﻧﻲ ﺍﳉﺬﺭ ﻫﻮz
n(cos sin )z r i
1 1 1
2 2
(cos sin ) cos ( ) sin( )n n n k k
z r i r i
n n
k0 , 1 , 2 , 3 , ….., (k -1 )ﺣﻴﺚkﺍﳉﺬﺭ ﺭﺗﺒﺔ
k0 , 1 , 2
k0 , 1 , 2 , 3 , 4 , 5
59. KAMIL ALNASSIRY59
Ex103)- 32
5
1 1 1
5 5 5
1
5
(cos sin ) , , 32 32 , arg( )
32 32(cos sin )
2 2
32 (cos sin ) (2 ) cos ( ) sin( )
5 5
2 2
2 cos ( ) sin( )
5 5
L et z r i z z r z
T hen z i
k k
z i i
k k
z i
123
2
n
n
( 104)Ex4 4 3i
4 4 3i z
1
3
z
2 2
4 4 3 (4, 4 3 ) 4 2
| | 16 48 8
4 1 5
cos 2
8 2 3 3 3
5 5
(cos sin ) 2 (cos sin )
3 3
th
i Quadrant
z r a b
a
r
z r i z i
0
1
2
3
4
0 : 2(cos sin )
5 5
3 3
1: 2(cos sin ) 2
5 5
2 : 2(cos sin ) 2 3
7 7
3 : 2(cos sin ) 4
5 5
9 9
4 : 2(cos sin ) 5
5 5
w hen k z i ﺬراﻷول اﻟﺠ
w hen k z i root
w hen k z i root
w hen k z i root
w hen k z i root
60. KAMIL ALNASSIRY60
0
1
1 1
3 3
1
3
5 5
2 2
5 5 3 38 (cos sin ) 2 cos ( ) sin ( )
3 3 3 3
5 6 5 6
2 cos ( ) sin ( )
9 9
5 5
: 2 (cos sin ) (1)
9 9
11 11
: 2 (cos sin ) (2)1
9
0
9
when k
when k
wh
k k
z i i
k k
z i
z i Root
z ot
e
i Ro
2
17 17
: 2 (cos sin ) (3)
9
2
9
n z i Rootk
( 105)Exu
2
3
( 2 2 3)u i
(cos sin ) ; 2 2 3
( 2, 2 3) 2 .
| | 4 4 3 4
2 1 2
cos
4 2 3 3 3
2 2
4 ( cos sin )
3 3
nd
Let z r i z i
z qua
z r
z i
1 1
22 1 3 3
2 23 3
3
3
3
0
1
2
3
2 2 4 4
4 ( cos sin ) 4 ( cos( ) sin( )
3 3 3 3
4 4
2 2
3 316 cos sin
3 3
4 6 4 6
16 cos sin
9 9
4 4
0 : 16 (cos sin )
9 9
1: 16
z z i i
k k
i
k k
z i
when k z i ﺬراﻷول اﻟﺠ
when k z
3
3
2
10 10
(cos sin )
9 9
16 16
2 : 16(cos sin )
9 9
i ﺬراﻟﺜﺎﻧﻲ اﻟﺠ
when k z i ﺬراﻟﺜﺎﻟﺚ اﻟﺠ
61. KAMIL ALNASSIRY61
( 106)Ex
6
64 0z i
1
6 6
664 0 64 64z i z i z i
2 2
1
11 6
66
1
2
3 3 3
64 ; 64 ; arg( ) 64(cos sin )
2 2 2
3 3 3 3 1 3
64 64(cos sin ) 2(cos sin ) ,,, ( 2 )
2 2 2 2 6 2
3 4 3 4
2(cos sin )
12 12
2(cos sin ) 2 2
4 4
7
2(cos sin
1
0
1
2
i g r a b g i
z i i i k
k k
i
z i i
z i
let k
let k
3
4
5
6
7
)
12
11 11
2(cos sin )
12 12
5 5
2(cos sin ) 2 2
4 4
1
2
3
4
9 19
2(cos sin )
12 12
23 23
2(cos sin )5
12 12
z i
z i i
z
let k
let k
let k
let k
i
z i
2 2
6 3n
60ه
0z
5z
4z
1z
2z
3z
Real axis
Imaginary axis
62. KAMIL ALNASSIRY62
3
1
:( 106)
1 3
i
Let z Ex
i
z
5 5
sin , cos
12 12
3 2
. : 1 1 (1Sol i i 2
2i i
2
3
1
1 1
1
2 2
1 1
1
2
)(1 ) 2 2 2 2
1 2 2 1 3 2 2 3 2 2 3 2 3 2 2 3 2
4 4 41 3 1 3 1 3
2 2
2,2 2 .
3
tan 1
4 4 4
| | 2 2
3 3
2 2(cos sin )
4 4
1 (1, 3 ) 1 .
nd
st
i i i i
i i i i i
z i ﺔ ﺟﺒﺮﯾ
i i i
Let z i
z Quad
r z a b
z i
Let z i Quad
2
2 2
2 2
2 2
2
1
2
| | 2
3
tan 3
1 3
2(cos sin )
3 3
3 3
2 2(cos sin )
3 3 5 54 4 2 cos( ) sin( ) 2(cos sin )
4 3 4 3 12 122(cos sin )
3 3
r z a b
z i
iz
z i i
z i
2 3 2 2 3 2
4 4
5 2 3 2 5 2 3 2 6 2
: 2 cos cos
12 4 12 44 2
5 2 3 2 3 1 2
5 5
2(cos sin )
12
6 2
: 2 sin
12 4 42
12
2
Re.
Im.
2
ﺔ ﻣﺜﻠﺜﯿ
But
z i
parts
parts
z i ﺔ ﺟﺒﺮﯾ
63. KAMIL ALNASSIRY63
( 107)Ex
9 4
cos sin cos sini i
9 4 5 4 4
45
: cos sin cos sin cos sin cos sin cos sin
cos sin cos sin cos sin
cos5
Solution i i i i i
i i i
2 2
sin5 cos sin cos5 sin5i i
sin sin ; cos cos
49 4 9
cos sin cos sin cos sin cos( ) sin( )
cos9 sin9 cos( 4 ) sin( 4 ) cos(9 4 ) sin(9 4 )
cos5 sin5
i i i i
i i i
i
( 108)Ex
5
3
cos2 sin 2
cos3 sin 3
i
i
5
3
cos2 sin2 cos10 sin10
cos(10 9) sin(10 9) cos sin
cos9 sin9cos3 sin3
i i
i i
ii
( 109)Ex1 26 cos sin ; 8 cos sin
4 4 2 2
z i z i
1 2z z
1
2
6
cos( ) sin(
8 4 2 4 2
z
i
z
3 3
cos( ) sin( ) = cos( ) sin( )
4 4 4 4 4 4
i i
( 109)Ex
3 2
1 0z z z
4 3 2
1 1 1z z z z z 4
1 0z 1
4 2 2 2 2 2
1 1 1 1 1 1 0z z z z z i z z z i z i
1 ,z z i 11, ,i i
4 2
1 0z z 6 2 4 2
1 1 1 0z z z z
64. KAMIL ALNASSIRY64
: 1Q
3 1
2 2
z i
32
z
: 2Qsin3 , cos3 sin , cos
: 3Q
1 3 1 3
3 2
1 2 2
i
z i i
i
zz 4
a + b i
: 4Qa
a+ b i
1 1 2 2 = 2 2 ; 2 1 3 4 3 4 8 3 8
1 4 4 3
3 ; 4 2 3 2
1 3
i i i i i
i i
i i
i i
b
5 5 5 5
45
8 6
(1) 1 3 1 3 32 ; (2) 1 3 1 3 32 3
3 2 cos6 sin 6 16 3 16 ; 4 2 cos75 sin75 2 2 3
5 1 16 ; 6 1 8 ;
i i i i i
i i i i
i i i
20
3 3
9
4 3
1 3 1 3
7
2 2 2 2
1 3 1 33 1
8 ; 9 ; 10 1
2 2 82 2 1 3
i i
i i ii
i i
i i
: 5Q
9 11
11
9
1 cos sin cos sin cos2 sin 2
cos sin
2 cos20 sin 20
cos sin
i i i
i
i
i
: 6Q
2
2
1
(1, ) Prove that : tan
1
z
If z i
z
ﺗﺪﺭﻳﺐ1 – 5