Using Cauchy's Integral Formula on a simple closed contour Hello I'm trying to evaluate the following two integrals where C is the unit circle centered at origin, but I encounter the same problem in both of them and can't think of what to do.
$$1)\ \oint_{C} \frac{1}{(2z-1)^2}$$
for this problem I let $f(z)= \frac{1}{2z-1}$ so I know the integral is equal to $\frac{1}{2}2\pi i f(\frac{1}{2})$ but here is where I have trouble understanding if I evaluate $f(\frac{1}{2})$ it is undefined so how would I go about this? Am I making $f(z)$ the wrong thing? I know the final answer is $0$ just don't understand how to get there.
$$2)\ \oint_{C}e^{z^{2}} (\frac{1}{z^{2}} - \frac{1}{z^{3}}) $$
For this problem I tried dividing this into two integrals and giving each on its $f(z)$ but again the singularity point is at $0$ so I would have to evaluate $f(0)$ for $f(z)=\frac{e^{z^{2}}}{z}$ and $f(z)=\frac{e^{z^{2}}}{z^{2}}$ this has a solution of $-2\pi i$ so again I don't know what do do when $f(z)$ is undefined. Thank you guys in advanced for the help or any kind of hints. I'm trying to master this to pass my final.
 A: Cauchy's integral formula only works when the function $f$ is holomorphic inside the contour you are integrating over. However, you can still use the formula here as long as you are careful what you are calling $f$. For the first one, consider Cauchy's Integral formula for the derivative of $f$:
$$ f'(z_0)= \frac{1}{2 \pi i} \int_{C}\frac{f(z)}{(z-z_0)^2}dz$$
We can make the argument of your first integral look like the one above by noting:
$$ \frac{1}{(2z - 1)^2} = \frac{1}{[2(z - 1/2)]^2} = \frac{1/4}{(z - 1/2)^2}$$
So, if we take $f(z)=1/4$, a constant function, then we can use Cauchy's Integral formula because $f(z)=1/4$ is holomorphic everywhere, so in particular it is holomorphic inside the unit disk. The result is:
$$ \int_{C}\frac{1/4}{(z-1/2)^2}dz = 2\pi i*f'(1/2)$$
where again $f(z)=1/4$.
The second integral you present can be done in a similar manner.
A: Hint: For (1), your $f$ has a primitive in the plane minus the origin.
For (2), expand the numerator in its Taylor series around $z=0$.
