# Complex integral Q

I have this question:

Let C be an open (upper) semicircle of radius R with its centre at the origin, and consider $\int_C f(z) \, dz$ where $f(z)=\frac 1{z^2 + a^2}$ for real $a > 0$. Show that $$\left|f(z)\right| \ \le \ \ \frac 1{R^2-a^2} \qquad\textrm{and} \qquad \left|\int_C f(z) \, dz\right| \le \frac {\pi R}{R^2 - a^2} \, .$$

I'm not sure how to work out $\left|f(z)\right|$ - any help appreciated!

-

## 1 Answer

Use the inequality: $|z^2+a^2| \geq |z|^2-|a|^2$.

Then $|f(z)|=|\frac{1}{z^2+a^2}|=\frac{1}{|z^2+a^2|}\leq\frac{1}{|z|^2-|a|^2}\leq\frac{1}{R^2-a^2}$. The last inequality comes from the fact that $|z|<R$ ("inside" $C$).

Now $|\int_C f(z) dz| \le \int_C |f(z)| dz \leq$(length of $C$)$\times ||f(z)||$$\leq \frac {\pi R}{R^2 - a^2}$, where $||f(z)||=\sup\{|f(z)|: z \in C^*\}$ and $C^*$ is the image of $C$.

-
The last inequality comes from not $|z|<R$, but from $|z|=R$. – GEdgar Mar 16 '13 at 16:22
Great answer, thank you very much! – Sanya Mar 16 '13 at 17:14
Your welcome. I hope it is clear enough. – user39280 Mar 16 '13 at 17:25