Correct Answer - Option 2 : πi
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:
\(\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)\)
\(\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times f(0)\)
\(\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times \frac{1}{2}=\pi i\)
