MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

7-15: Complex Aritmetic Let z 1 = 2 + 3i and z 2 = 4 − 5i.
7. (5z 1 + 3z 2 ) 2 = [5.(2 + 3i) + 3.(4 − 5i)] 2 = [(10 + 15i) + (12 − 15i)] 2 = (22) 2 = 484
8. z 1 .z 2 = (2 − 3i).(4 + 5i) = 23 − 2i
9. Re( 1
1
−5−12i
) = Re(
z1
2 −5+12i
) = Re(
169
) = Re(− 5
169 − 12
169 i) = − 5
169
10.• z2 2 = (4 − 5i)2 = −9 − 40i ⇒ Re(z2 2) = −9
11. z 2
z 1
• (Rez 2 ) 2 = 4 2 = 16
12. • z 1
z 2
= 4−5i
2+3i = (4−5i)(2−3i)
4+9
= −7−22i
13
= − 7
13 − 22
13 i
= (2+3i)
(4−5i) = 2−3i
4+5i = −7−22i
41
= − 7
41 − 22
41 i
• ( z 1
z 2
) = ( 2+3i
4−5i ) = ( −7+22i
41
) = (− 7
41 + 22
41 i) = − 7
41 − 22
41 i
13. (4z 1 − z 2 ) 2 = [4.(2 + 3i) − (4 − 5i)] 2 = (8 + 12i − 4 + 5i) 2 = (4 + 17i) 2 = −273 + 136i
14. • z 1
z 1
= 2−3i
2+3i = −5−12i
4+9
= − 5
13 − 12
13 i
• z 1
z 1
= 2+3i
2−3i = −5+12i
4+9
= − 5
13 + 12
13 i
15. z 1+z 2
z 1 −z 2
= (2+3i)+(4−5i)
(2+3i)−(4−5i) = 6−2i
−2+8i = −28−44i
68
= − 7
17 − 11
17 i
16-19: Let z = x + iy.
16. • Im(z 3 ) = Im((x + iy) 3 ) = Im((x 3 − 3xy 2 ) + i.(3x 2 y − y 3 )) = 3x 2 y − y 3
• (Imz) 3 = (Im(x + iy)) 3 = y 3
17. 1 z = 1
x−iy = x+iy = x y
+ i ⇒ Re( 1 x 2 +y 2 x 2 +y 2 x 2 +y 2 z ) = x
18. At first we will find (1 + i) 8 :
(1 + i) 2 = 2i,
(1 + i) 3 = (1 + i) 2 .(1 + i) = 2i.(1 + i) = −2 + 2i,
(1 + i) 4 = (1 + i) 3 .(1 + i) = (−2 + 2i).(1 + i) = −4,
(1 + i) 5 = (1 + i) 4 .(1 + i) = (−4).(1 + i) = −4 − 4i,
(1 + i) 6 = (1 + i) 5 .(1 + i) = (−4 − 4i).(1 + i) = −8i,
(1 + i) 7 = (1 + i) 6 .(1 + i) = (−8i).(1 + i) = 8 − 8i,
(1 + i) 8 = (1 + i) 7 .(1 + i) = (8 − 8i).(1 + i) = 16.
x 2 +y 2
Then, Im((1 + i) 8 .z 2 ) = Im(16.z 2 ) = Im(16.(x 2 − y 2 + i2xy)) = Im(16(x 2 − y 2 ) + i32xy) = 32xy
19. Re( 1
z 2 ) = Re(
1
(x−iy) 2 ) = Re(
1
x 2 −y 2 −i2xy ) = Re( x2 −y 2 +i2xy
x 4 +2x 2 y 2 +y 4 ) =
x 2 −y 2
x 4 +2x 2 y 2 +y 4 .
13.2
1-8 Represent the followings in the polar form:
1. Let z = 3 − 3i. Then |z| = √ 3 2 + (−3) 2 = √ 9 + 9 = √ 18 = 3 √ 2 and since tan θ = 3
−3 = −1,
we get
⇒ θ = 7π 4 ⇒ z = 3√ 2(cos( 7π 4 ) + i. sin(7π 4 )).
3