Calculate integral $$\oint_{\gamma} \frac{z+1}{z^4+2iz^3} dz$$ where $\gamma$ is parameterization of circle $B(0,1)$ along one positive rotation.

I did something like this with Cauchy. \begin{align} \oint_{\gamma} \frac{z+1}{z^4+2iz^3}dz&=\oint_{\gamma} \frac{z+1}{z^3(z+2i)} =\oint_{\gamma} \frac{\dfrac{z+1}{z+2i}}{(z-0)^{2+1}}dz \\ &=\frac{2i\pi}{n!}f^{''}(z_0) \\ &=\frac{2i \pi}{2!}\left(\frac{2-4i}{(0+2i)^3}\right) \\ &=\frac 12 \pi i-\frac 14 \pi \end{align}

with this one I'm not that sure

Calculate integral $$\oint_{\gamma} \frac{z^3+3}{z(z-i)^2}dz $$ Where $\gamma$ is 8-like curve where the rotation is positive around $(0,i)$ and negative around $(0,0)$.

Isn't the integral zero over $\oint_{\gamma_1}(\text{positive part})$ + $\oint_{\gamma_2}(\text{negative part})=0$. So is it two integrals were $\oint_{\gamma_2}=-\oint_{\gamma_1}$.

And the integral would be \begin{align}\oint_{\gamma_1}&=\frac{\dfrac{z^3+3}{z}}{(z-i)^2}dz \\ &=\frac{2\pi i}{1!}\left( \frac{2(-i)^3 -3}{(-i)^2}\right) \\ &=2\pi i (3-2i)\\ &=4\pi+6\pi i \\ \oint_{\gamma_2}&= -4\pi-6\pi i \end{align}

Did I get these question correct? I'm not 100% on the second question.

  • $\begingroup$ The first one looks correct: I did the calculation and that's what I got. For the second one: you mean $\;\gamma\;$ is like $\;8\;$ intersecting the $\;y$- axis on, say $\;(0\,,\,1/2)\;$ ? $\endgroup$ – Timbuc Oct 7 '14 at 12:55
  • $\begingroup$ Yeah. That's exactly how I figure the shape is. $\endgroup$ – ELEC Oct 7 '14 at 13:19

The second one, assuming I guessed correctly what the path is, I'd do as follows: let $\;\gamma_1\;$ be a circle (with radius $\;\le 1/2)\;$) around $\;i=(0,1)\;$ in the positive direction, and $\;\gamma_2\;$ a circle (with radius $\;\le1/2\;$ in the negative direction around the original, and the union of both paths above is the whole $\;\gamma\;$ , then


$$2\pi i\left[ \frac d{dz}\overbrace{\left(\frac{z^3+3}z\right)}^{z^2+\frac3z}{}_{z=i}-\left(\frac{z^3+3}{(z-i)^2}\right)_{z=0}\right]=2\pi i\left[\left(2z-\frac3{z^2}\right)_{z=i}-\frac3{(-i)^2}\right]=$$

$$2\pi i(2i+3+3)=-4\pi+12\pi i$$

Check the above: either you or I have a mistake somewhere.

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    $\begingroup$ That makes sense. The curves have different singularity points right $(0,i)$ and $(0,0)$? So you have to form the cauchy a bit differently for the two integrals. $\endgroup$ – ELEC Oct 7 '14 at 13:25
  • $\begingroup$ Indeed so, @ELEC..and not only that: $\;z=0\;$ is a simple pole, whereas $\;z=i\;$ is a double one. $\endgroup$ – Timbuc Oct 7 '14 at 14:35
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    $\begingroup$ Right! Thanks @Timbuc $\endgroup$ – ELEC Oct 7 '14 at 15:38
  • $\begingroup$ My pleasure, @ELEC $\endgroup$ – Timbuc Oct 7 '14 at 15:42

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