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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
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  • $\begingroup$ 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 $\endgroup$
    – Kiwi
    Apr 14, 2014 at 23:33
  • $\begingroup$ Well @Kiwi, and what did you get? It doesn't look very easy that way... $\endgroup$
    – DonAntonio
    Apr 14, 2014 at 23:36
  • $\begingroup$ well, at the moment i'm quite stuck. but i guess i'll continue reading to find out what to do. $\endgroup$
    – Kiwi
    Apr 14, 2014 at 23:39

1 Answer 1

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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.

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