# 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

1. $(z \cdot \cos(\frac{1}{z}))dz$ for a circle of radius $50$ centered at $2$, traversed once in counterclockwise direction
2. $\frac{100! \cdot (e^{iz})}{(z+1)^{100} }dz$ for a circle of radius $5$, centered at $0$, traversed once in counterclockwise direction
3. $\frac{z\cdot cos(z)}{(2z-pi)^2}dz$ for a circle of radius $4$, centered at $0$, traversed once in counterclockwise direction
• I didn't think of using residues.. I was just using the formula where integral of f(z)dz = integral of f(z(t))(dz/dt)dt
– Kiwi
Apr 14, 2014 at 23:33
• Well @Kiwi, and what did you get? It doesn't look very easy that way... Apr 14, 2014 at 23:36
• well, at the moment i'm quite stuck. but i guess i'll continue reading to find out what to do.
– Kiwi
Apr 14, 2014 at 23:39

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, $\oint_{B_{5}(0)} \frac{100!e^{iz}}{(z+1)^{100}} \,dz = i200\pi\sin(-1)$.