1
$\begingroup$

This problem asks about a few integrals over the open curve $C$, which is the bottom half of a circle centered about the origin with radius $\epsilon\rightarrow 0$.

The first integral was $\int_{\tiny{C}} dz\frac{1}{z}$, which I found to be $\pi\,i$ since it should be 1/2 of the same integral if $C$ was a full circle.

The second integral is $\int_{\tiny{C}} dz\frac{1}{z-i}$. I'd like to say this is equal to 0, since the singularity at $i$ is away from $C$, but then again $C$ doesn't enclose anything. Can I do a contour deformation to include $i$? If so, how?

The third integral is $I = \int_{\tiny{C}} dz\frac{1}{z(z-i)}$ Couldn't I just do $I = -\frac{1}{i}\int_{\tiny{C}}dz\frac{1}{z} + \frac{1}{i}\int_{\tiny{C}}dz\frac{1}{z-i}$ and use the answers to the first two integrals?

Thanks a ton.

$\endgroup$
  • $\begingroup$ Actually, partial fraction expansion is perfectly suitable here. $\endgroup$ – Mark Viola Nov 23 '16 at 19:53
1
$\begingroup$

HINT:

For the integral $\int_C \frac1{z-i}\,dz=\int_{\pi}^{2\pi}\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\,d\phi$, what is the length of the contour and what is the maximum of the magnitude of the integrand on $C$ as $\epsilon \to 0$?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Note that we can write $$\begin{align}\left|\int_C \frac1{z-i}\,dz\right|&=\left|\int_{\pi}^{2\pi}\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\,d\phi\right|\\\\&\le \int_{\pi}^{2\pi}\left|\frac1{\epsilon e^{i\phi}-i}\,i\epsilon e^{i\phi}\right|\,d\phi\\\\&=\int_{\pi}^{2\pi}\frac{\epsilon}{|\epsilon e^{i\phi}-i|}\,d\phi\\\\&\le\int_{\pi}^{2\pi}\frac{\epsilon}{|1-\epsilon|}\,d\phi\\\\&=\frac{\pi\epsilon}{|1-\epsilon|}\end{align}$$Therefore, $\lim_{\epsilon \to 0}\int_C \frac1{z-i}\,dz=0$.

$\endgroup$
  • $\begingroup$ The length of the contour is $pi\,\epsilon$ but it goes to 0. And as $\epsilon\rightarrow 0$, wouldn't the maximum magnitude be 0? Is the hint implying that this integral is $=0$? $\endgroup$ – Spuds Nov 23 '16 at 20:03
  • $\begingroup$ Yes, the second integral is indeed $0$. The maximum magnitude is $\frac{1}{1-\epsilon}$ $\endgroup$ – Mark Viola Nov 23 '16 at 20:23
  • $\begingroup$ You did help, thank you very much. $\endgroup$ – Spuds Dec 5 '16 at 0:05
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Dec 5 '16 at 0:35
  • $\begingroup$ Of course, sorry I should've done that sooner. $\endgroup$ – Spuds Dec 5 '16 at 3:01

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.