MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

15. Let z = 1 + i. Since f(z) = 1
z 2 , we have
For z = x + iy,
f(z) =
f(z) = f(1 + i) =
1
(x + iy) 2 = 1
x 2 − y 2 + 2xyi =
and so we can say
16-19 Continuity:
Ref =
1
(1 + i) 2 = 1
1 + 2i − 1 = 1 2i = −2i
(2i)(−2i) = −2i
4 = − i 2 .
x 2 − y 2 − 2xyi
(x 2 − y 2 + 2xyi)(x 2 − y 2 − 2xyi) = x2 − y 2 − 2xyi
(x 2 − y 2 ) 2 + 4x 2 y 2
x 2 − y 2
(x 2 − y 2 ) 2 + 4x 2 y 2 , Imf = −2xyi
(x 2 − y 2 ) 2 + 4x 2 y 2 .
16. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
[Re(z 2 )]/|z| 2 = r 2 . cos 2θ/r 2 = cos 2θ,
this function isn’t continuous at z = 0.
17. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
[Im(z 2 )]/|z| = r 2 . sin 2θ/r = r. sin 2θ r→0 −→ 0,
this function is continuous at z = 0.
18. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ) and Re( 1 z
|z| 2 .Re( 1 z ) = r2 .r. cos θ/r 2 = r. cos θ r→0 −→ 0,
r. cos θ
) = . Because
r 2
this function is continuous at z = 0.
19. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
this function is continuous at z = 0.
20-24 Derivative:
(Imz)/(1 − |z|) = r. sin θ/(1 − r) =
20. ( z2 −9
z 2 +1 )′ = (2z).(z2 +1)−(z 2 −9).(2z)
(z 2 +1) 2 = 2z3 +2z−2z 3 +18z
(z 2 +1) 2 = 20z
(z 2 +1) 2 .
21. [(z 3 + i) 2 ] ′ = 2.(z 3 + i).3z 2 = 6z 5 + 6z 2 i.
22. ( 3z+4i
1,5iz−2 )′ = 3.(1,5iz−2)−(3z+4i).1,5i = 4,5iz−6−4,5iz+6 = 0.
(1,5iz−2) 2
(1,5iz−2) 2
23. (
i
(1−z) 2 ) ′ = −i.2.(1−z).(−1)
(1−z) 4 = 2i
(1−z) 3 .
24. (
z 2
(z+i) 2 ) ′ = 2z.(z+i)2 −z 2 .2.(z+i)
(z+i) 4 = 2zi
(z+i) 3 .
r r→0
. sin θ −→ 0,
1 − r
13