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 \(\mathop \oint \nolimits_C^{} \frac{{\cosh 3z}}{{2z}}dz\) (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:

\(\mathop \oint \nolimits_C^{} \frac{{\cosh 3z}}{{2z}}dz\) where C represents unit circle i.e. radius is unity.

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

Therefore \(f(z)=\frac{\cosh3z}{2}\) and a = 0.

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

\(f(z)=\frac{\cosh3z}{2}=\frac{e^{3z}+e^{-3z}}{2\times2}=\frac{e^{3z}+e^{-3z}}{4}\)

At z = 0,

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

Cauchy's Integral Formula:

\(\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)\)

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

\(\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times \frac{1}{2}=\pi i\)