Answer to Question #236289 in Chemical Engineering for Lokika

Question

5) Evaluate ∮c z³ Cos 1/z Where C is |z|=2.

Explain the problem with step by step process?

Answer 1

$|z| =2$ means that the limits of integration are from 0 to 2

Hence

$\oint_Cz^3\cos(\frac{1}{z})dz = \int_0^2 z^3\cos(\frac{1}{z})dz\\ \int \:uv'=uv-\int \:u'v\\ Let \space u = z^3 \implies du = 3z^2dz; dv= \cos (\frac{1}{z})dz \implies v=z\cos \left(\frac{1}{z}\right)+\text{sin}\left(\frac{1}{z}\right)\\ \implies z^4\cos \left(\frac{1}{z}\right)+\text{sin}\left(\frac{1}{z}\right)- \int 3z^3\cos \left(\frac{1}{z}\right)+\text{sin}\left(\frac{1}{z}\right)\\ \implies z^4\cos \left(\frac{1}{z}\right)+\text{sin}\left(\frac{1}{z}\right)- 3z^3\int \cos \left(\frac{1}{z}\right)+\int \text{sin}\left(\frac{1}{z}\right)\\ \implies [z^4\cos \left(\frac{1}{z}\right)+\text{sin}\left(\frac{1}{z}\right)- 3z^3 \sin \left(\frac{1}{z}\right)-\frac{1}{z^2}\text{cos}\left(\frac{1}{z}\right)]|_0^2\\ [2^4\cos \left(\frac{1}{2}\right)+\text{sin}\left(\frac{1}{2}\right)- 3*2^3 \sin \left(\frac{1}{2}\right)-\frac{1}{2^2}\text{cos}\left(\frac{1}{2}\right)]\\-[0^4\cos \left(\frac{1}{0}\right)+\text{sin}\left(\frac{1}{0}\right)- 3*0^3 \sin \left(\frac{1}{0}\right)-\frac{1}{0^2}\text{cos}\left(\frac{1}{0}\right)]\\ =15.6534$