Evaluate $\int_{|z-1| = \frac{1}{2}} \frac{dz}{(1-z)^3}$

Question

integration theorem cauchy riemann I hope someone can help me:

$$(ii)\;\;\int_{|z-1| = \frac{1}{2}} \frac{dz}{(1-z)^3}$$ $$(iii)\;\int_{|z+1| = \frac{1}{2}} \frac{dz}{(1-z)^3}$$
Image.

ii) I understand it is a circle of radius 5/4 and the integral is zero iii) I have problem to identify what is the circumference. it can not be radio -3/4

I'm new here. sorry for my English

Answer 1

Hints - Fill in details:

$\;|z+1|=\frac12\;$ = the circle of radius $\;\frac12\;$ and center at $\;-1\;$ , and thus the second integral is zero by Cauchy's Theorem .

We cannot dispose so fast of (ii) as the function isn't analytic inside the circle $\;|z-1|=\frac12\;$ (why you think the radius is $\;\frac54\;$ ??) as it has a triple pole at $\;z=1\;$ , but evaluate the corresponding Laurent series:

$$\frac1{(1-z)^3}=-\frac1{(z-1)^3}+0\cdot\frac1{(z-1)^2}+0\cdot\frac1{z-1}+\ldots$$

...and again the integral equals zero.