Answer :

Solution :

${\oint }_C^{}\frac{\cosh 3z}{2z}dz$ where C represents unit circle i.e. radius is unity.

The above equation can be written in standard form i.e. ${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz$

Therefore $f(z)=\frac{\cosh 3z}{2}$ and a = 0.

The pole of the given function is at z = 0, and lie inside the circle.

$f(z)=\frac{\cosh 3z}{2}=\frac{e^{3z}+e^{-3z}}{2\times 2}=\frac{e^{3z}+e^{-3z}}{4}$

At z = 0,

$f(0)=\frac{e^0+e^{-0}}{4}=\frac{2}{4}=\frac{1}{2}$

Cauchy's Integral Formula :

${\oint }_c^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)$

${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz=2\pi i\times f(0)$

${\oint }_C^{}\frac{\frac{\cosh 3z}{2}}{z-0}dz=2\pi i\times \frac{1}{2}=\pi i$