Let C represent the unit circle centered at origin in the complex plane, and complex variable, z = x + iy. The value of the contour integral ${\oint }_C^{}\frac{\cosh 3z}{2z}dz$ (where integration is taken counter clockwise) is

Concept:

Cauchy's Integral Formula:

For Simple Pole:

If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then

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

$\cosh z=\frac{e^z+e^{-z}}{2}$

Calculation:

Given:

${\oint }_C^{}\frac{\cosh 3z}{2z}dz$ 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:

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

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

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