# $|\int_{C} f(z)\, dz|\leq \frac{\pi R}{R^{2}-a^{2}}$, $R>a$

Could someone help me through this problem?

Let $C$ be an open (upper) semicircle of radius R withits center at the origin, and consider $$\displaystyle\int_{C} f(z)\, dz.$$ Let $\displaystyle f(z)=\frac{1}{z^{2}+a^{2}}$ for real $a>0$. Show that $$|f(z)|\leq \frac{1}{R^{2}-a^{2}}~\text{for }R>a,$$ and $$\left|\displaystyle\int_{C} f(z)\, dz \right|\leq \frac{\pi R}{R^{2}-a^{2}}~\text{for }R>a$$

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Consider the reverse triangle inequality – Alex Apr 25 '12 at 0:10

## 2 Answers

For $z\in C$, we have $|z|= R>a$. Hence, we have $$|f(z)|=\frac{1}{|z^2+a|^2}\leq\frac{1}{|z|^2-a^2}=\frac{1}{R^2-a^2}.$$ Now, using this inequality, we have $$\left|\int_{C} f(z)\, dz\right|\leq \left|\int_{C} |f(z)|\, dz\right|\leq\frac{1}{R^2-a^2}\left|\int_{C}\, dz\right|=\frac{\pi R}{R^{2}-a^{2}}$$ since $C$ is an open (upper) semicircle of radius R with its center at the origin.

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We have $|z^2 + a^2| \geq |z|^2 - |a|^2 = R^2 - a^2$ for $z\in C$. Taking reciporicals, we get the first inequality

For the second, we have by definition of a path integral that

$$|\int_C f(z) dz| = |\int_0 ^\pi f(Re^{i\theta}) Re^{i\theta} d\theta| \leq \int_0 ^\pi R|f(Re^{i\theta})|d\theta \leq \frac{\pi R}{R^2 - a^2}$$

using the first inequality.

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