MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

=⇒ f is analytic. Then by Cauchy’s integral theorem ∫ dz
3z−π = 0.
6.
f(z) = sec( z 1
2
) =⇒ f(z) =
cos( z 2 )
cos( z 2
) = 0 =⇒ z = (2k + 1)π, k ∈ Z
But these points lie outside the unit circle. Then f is defined and differentiable on C and inside C.
=⇒ f is analytic. Then by Cauchy’s integral theorem ∫ sec( z 2
)dz = 0.
12-17 :
12.
a) f(z) = 1
z 2 +4
z 2 + 4 = 0 =⇒ z 2 = −4 =⇒ z = ±2i.
These points lie outside the region.
=⇒ f is analytic.
=⇒ ∫ dz
z 2 +4 = 0.
b) f(z) = 1
z 2 +4
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