Originally Posted by

**Deadstar** Right isn't so much a question as an 'am i doing this right'.

Evaluate $\displaystyle \int_\delta \frac{z^2}{z - 1} dz$ where $\displaystyle \delta$ is the circle with centre 0 and radius 2, traversed in the usual anti-clockwise sense.

And my answer...

The function $\displaystyle \frac{z^2}{z - 1}$ is holomorphic on the complex plane except at z=1.

So, using Cauchys integral formula we get $\displaystyle f(1) = \frac{1}{2 \pi i} \int_\delta \frac{f(z)}{z - 1} dz$ with $\displaystyle f(z) = z^2$.

Hence, $\displaystyle \int_\delta \frac{f(z)}{z - 1} = 2 \pi i f(1) = 2 \pi i 1^2 = 2 \pi i$.

This right?