# Math Help - Complex Integration

1. ## Complex Integration

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

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

And my answer...

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

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

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

This right?

If so, how come the answer doesnt depend on $\delta$? Does it not matter what the radius of the circle is?

2. Hello,
Originally Posted by Deadstar
Right isn't so much a question as an 'am i doing this right'.

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

And my answer...

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

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

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

This right?
Yes

If so, how come the answer doesnt depend on $\delta$? Does it not matter what the radius of the circle is?
It does depend on it.
Because the pole z=1 is within the circle, you get the result.
If it wasn't (say you considered $\delta$ as the circle of center 5 and radius 1), then the integral would have been 0.
Look here : Cauchy's integral formula - Wikipedia, the free encyclopedia (they say 'Then for every a in the interior of D')