3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$$\int\limits_\gamma\frac{z^3+3}{z(z-i)^2}dz=\oint\limits_{\gamma_1}\frac{\frac{z^3+3}z}{(z-i)^2}dz-\oint\limits_{\gamma_2}\frac{\frac{z^3+3}{(z-i)^2}}zdz=$$

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

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

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Right! Thanks @Timbuc $\endgroup$ – ELEC Oct 7 '14 at 15:38
  • $\begingroup$ My pleasure, @ELEC $\endgroup$ – Timbuc Oct 7 '14 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.