<s>[INST] compute the integral $\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$ Compute this integral: 
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
my solution is I used derivative of Cauchy integral formula, which is 
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int \frac{f(z_0)}{(z-z_0)^{n+1}}$$
Then, I got
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$ 
$$ = \int_{|z|=1} \frac{(z-2)^3}{(2z-1)^3}dz$$
$$ = \frac{2 \pi i}{2!}\left(\frac{1}{2} - 2\right)^3 = \frac{27\pi i}{8}$$
That's what I get from the Cauchy integral formula. So, am I correct for this approach. If not, can someone please show me how to get the right one? Thank you 
 [/INST] Notice,
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz=\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ We see that $z=2$ & $z=\frac{1}{2}$ are two points out of which point $z=\frac{1}{2}$ is a pole of third order as it is lying inside the unit circle $|z|=1$. Hence, $f(z)=(z-2)^3$
hence, using cauchy integral formula, we have $$\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ $$=\frac{1}{8}\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2f(z)}{dz^2}\right]_{z=\frac{1}{2}}$$ $$=\frac{1}{8}\left(\frac{2\pi i }{2!}\right)\left[\frac{d^2}{dz^2}(z-2)^3\right]_{z=\frac{1}{2}}$$  $$=\frac{\pi i}{8}\left[6\left(z-2\right)\right]_{z=\frac{1}{2}}$$ $$=\frac{3\pi i}{4}\left[\frac{1}{2}-2\right]$$ $$=\frac{3\pi i}{4}\left[\frac{-3}{2}\right]$$ $$=\color{blue}{-\frac{9\pi i}{8}}$$
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