TITLE: Evaluating an complex integral
QUESTION [0 upvotes]: Let $f(z) = \frac{1}{(1-z)^3} $. Evaluate $\int_{|z-1|= \frac{1}{2}} f(z) dz $
We know $|z-1|=\frac{1}{2}$ is parametrized by the curve $\gamma(t) = \frac{1}{2}e^{it} + 1 $ where $t \in [0,2 \pi]$. We know $f(z)$ is analytic on the inside on this curve. We show $\gamma$ is homotopic to a point that lies this curve. Let $H(s,t) = \frac{s}{2}e^{it} + 1 $. Notice $H(0,t) = 1$. $H(1,t) = \gamma(t)$ and $H(s,0) = H(s,2 \pi)$. Therefore, $\gamma$ is homotopic to the point $z=1$. It follows by Homotopic version of Cauchys theorem that 
$$ \int_{|z-1|= \frac{1}{2} } f(z) dz = 0 $$
is this a correct approach ?

REPLY [0 votes]: I am afraid that you have to do it directly.
Let $\gamma(t) = \frac{1}{2}e^{it}+1$ where $t \in [0, 2 \pi]$.
$$\int_\gamma \frac{1}{(1-z)^3}dz = \int_0^{2 \pi}\frac{8}{-e^{3it}}(\frac{i}{2}e^{it}) = 4i \int_0^{2\pi}e^{-2it}dt = -2e^{-2it}|_0^{2\pi}= 0$$
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