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!

  • $\begingroup$ 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
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    $\begingroup$ @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

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

  • $\begingroup$ 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
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    $\begingroup$ 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

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