0
$\begingroup$

Here's a complex analysis question I'm fighting with.

Let $\gamma$ be the arc of the circle $|z|=2$ that lies in the first quadrant. Show that $$\left| \int_{\gamma} \frac{dz}{z^2+1} \right| \leq \frac{\pi}{3}.$$

My current trials have been as follows:

$\gamma= 2\mathrm{e}^{it}, \ t \in [0,\pi/2]$. So $$\left| \int_{\gamma} \frac{dz}{z^2+1} \right| = \left| \int_0^{\pi} \frac{1}{2 \mathrm{e}^{2it}+1} 2i \mathrm{e}^{it} \, \mathrm{d}t \right| = 2 \left| \int_1^i \frac{1}{2x^2+1} \, \mathrm{d}x \right|$$ where the last equality comes from the substitution $x=\mathrm{e}^{it}$. Performing this integral doesn't get me closer to $\pi/3$ at all!

My other thought was to use the inequality $$\left| \int_{\gamma} \frac{dz}{z^2+1} \right| \leq \int_{\gamma} \left| \frac{1}{z^2+1} \right|\, \mathrm{d}z.$$ However, letting $z=x+iy$ and finding the modulus also gives some yuck. Can you help?

EDIT: Using the estimate $$\left|\int_{\gamma} u(z) \, \mathrm{d}z \right| \leq \left( \max_{z\in \gamma} |u(z)| \right) \text{length}(\gamma)$$ where $u$ is a continuous function on the range of $\gamma$, we get that $$\left| \int_{\gamma} \frac{dz}{z^2+1} \right| \leq \frac{1}{3} \text{length}(\gamma)$$ by using the inequality $\left| z^2 +1 \right| \geq \left| |z^2|-1 \right|=3$. It is clear that $\text{length}(\gamma)=\pi$, and the estimate follows.

Thanks guys!

$\endgroup$
  • $\begingroup$ Have you heard of residues? en.wikipedia.org/wiki/Residue_%28complex_analysis%29 $\endgroup$ – Antoine Sep 22 '13 at 17:33
  • $\begingroup$ No not yet, but I'm willing to learn! $\endgroup$ – Numbersandsoon Sep 22 '13 at 17:34
  • 1
    $\begingroup$ @Antoine, but $\gamma$ is not a closed curve. How can you use residues here? $\endgroup$ – njguliyev Sep 22 '13 at 17:38
  • $\begingroup$ I am so sorry. I haven't read the definition of gamma carefully. I can't use them. Should I remove my first comment? $\endgroup$ – Antoine Sep 22 '13 at 17:40
  • $\begingroup$ No worries!! :) $\endgroup$ – Numbersandsoon Sep 22 '13 at 17:41
8
$\begingroup$

Hint: $|z^2+1|\ge \left||z^2|-1 \right|= 3$.

$\endgroup$
  • $\begingroup$ Nice! How did that first inequality come about? I sense some use of the reverse triangle inequality $\endgroup$ – Numbersandsoon Sep 22 '13 at 17:36
  • 1
    $\begingroup$ For any complex numbers $z_1$ and $z_2$ we have $|z_1+z_2| \ge ||z_1|-|z_2||$. This is proved by using the triangle inequality. $\endgroup$ – njguliyev Sep 22 '13 at 17:39

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.