Contour integration with pole

Question

I wanted to integrate

$$\oint \frac{\sqrt{z+5}}{z^5}$$

around a circular contour radius $1$ center $0$.

So the function has pole at 0. How can I proceed from here?

Answer 1

Cauchy's integral formula is

$$f^{(n)} (z_0) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}} dz$$ so $$ \oint \frac{\sqrt{z+5}}{z^5} dz = \frac{2\pi i}{4!} \,\left.\frac{d^4}{dz^4}(z+5)^{1/2}\right|_{z=0}$$

$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{2\pi i}{4!} \, \left. \frac{15}{16(5+z)^{7/2}}\right|_{z=0}$$

$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{\pi i}{64\cdot 5^{5/2}} $$