Find the integral in the complex plane I'm having some trouble computing these integrals, they're on the practice final, but no solutions given. I'm hoping to get some help here. 
Calculate the following
Integral of 


*

*$(z \cdot \cos(\frac{1}{z}))dz$ for a circle of radius $50$ centered at $2$, traversed once in counterclockwise direction

*$\frac{100! \cdot (e^{iz})}{(z+1)^{100} }dz$ for a circle of radius $5$, centered at $0$, traversed once in counterclockwise direction

*$\frac{z\cdot cos(z)}{(2z-pi)^2}dz$ for a circle of radius $4$, centered at $0$, traversed once in counterclockwise direction

 A: All of these function can be expanded into a laurent around their (single) pole series relatively easy.
The residual at a pole $p$ can be found from the laurent series expansion of $f$ around $p$, i.e. $$
  \textrm{Res }(f,p) = c_{-1} \text{ if } f(z) = \sum_{k=-\infty}^\infty c_k (x - p)^k \text{ around $p$.}
$$
One can then use that $$
  \oint_{\gamma} f(z) \,dx = i2\pi\sum_{k=1}^n\textrm{Res }(f,p_k)
$$
where $\gamma$ is a positively oriented (i.e, counter-clockwise) simple closed curve, and $p_k$ the poles of $f$ inside of $\gamma$.
1: We know that $\cos z = 1 - \frac{1}{2!}z^{-2}  \ldots$ It follows that $z\cos\frac{1}{z} = z - \frac{1}{2!}z^{-1} + \ldots$, i.e that $Res(z\cos\frac{1}{z},0) = -\frac{1}{2}$, and therefore that $\oint_{B_{50}(2)} z\cos\frac{1}{z} \,dz = -i\pi$.
2: Expanding $e^{iz}$ around $p=-1$, using that $\frac{d^k}{dx^k}e^{iz} = i^ke^{iz}$ and that $e^{-i} = \cos 0 + i\sin(-1)$, yields $e^{iz} = \sum_{k=0}^\infty \frac{i^{k+1}\sin(-1)}{k!}(x + 1)^k$. From that, it follows that $$
  \frac{100!e^{iz}}{(z+1)^{100}} = \ldots + \frac{100!i^{100}\sin(-1)}{99!}(z+1)^{-1} + \ldots
$$
and thus that $\textrm{Res }\left(\frac{100!e^{iz}}{(z+1)^{100}}, -1\right) = 100\sin(-1)$. Therefore, $\oint_{B_{5}(0)} \frac{100!e^{iz}}{(z+1)^{100}} \,dz = i200\pi\sin(-1)$.
3: I'll leave that to you. You'll need to compute a few terms of the taylor expansion of the nominator around the pole (where is it?), but not very many.
