MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

1. We use Theorem 1 on the page 658.
f(z) = cosh 3z is analytic in D.
f n (z 0 ) = n! ∮
2πi C
Because n + 1 = 5, we need to know f (4) (z):
z 5 0 = 0 ⇒ z 0 = 0 ∈ D
f(z)
(z − z 0 )
∮C
n+1 ⇒ f(z)
(z − z 0 ) n = f n (z 0 ).2πi
, n = 1, 2, ...
n!
f (1) (z) = 3 sinh 3z, f (2) (z) = 9 cosh 3z, f (3) (z) = 27 sinh 3z, f (4) (z) = 81 cosh 3z
Hence, we get
∮
C
cosh 3z
z 5 dz = f 4 (0).2πi 81 cosh 0.2πi
= = 81.1.2πi
4! 4.5.2.1 4.3.2.1 = 27πi
4
2. z 0 = πi
2 ∈ D. f(z) = sin z is analytic in D. We need to know f (3) (z):
f (1) (z) = cos z, f (2) (z) = − sin z, f (3) (z) = − cos z
Hence, we obtain
∮
C
sin z
(z − πi dz = f (3) ( πi
2 ).2πi
2 )4 3!
=
πi
− cos(
2 ).2πi =
3!
πi
− cos(
2 ).πi = − cosh( π 2 ).πi
3
3
3. z 0 = πi
2 ∈ D and f(z) = ez cos z is analytic in D.
f(z) = e z cos z ⇒ f (1) (z) = e z cos z − e z sin z = e z (cos z − sin z)
So,
∮
C
e z cos z
(z − π dz = f (1) ( π 2 ).2πi = e π π
2 (cos
2 )2 1!
2 − sin π 2 ).2πi = −e π 2 .2πi
4. z 0 = 0 ∈ D and f(z) = cos z is analytic in D.
f (1) (z) = − sin z, f (2) (z) = − cos z, f (3) (z) = sin z,
f (4) (z) = cos z, f (5) (z) = − sin z, f (6) (z) = − cos z,
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