How to use the ML inequality?

Question

Using the ML inequality how would I prove $$\left|\oint_C \frac{e^{2z}}{6z^5}\, dz\right|< \frac{\pi\cdot e^2}3$$ where $C$ denotes the unit circle described anticlockwise?

I understand that when you use the ML inequality we must calculate where $f(z)$ is bounded (denoted $M$). But how do I go about in calculating this?

Answer 1

Parameterize $z:=e^{i\theta}$ and rewrite the integral as $$\oint_{|z|=1}\frac{e^{2z}}{6z^5}\,dz=\int^{2\pi}_0\frac{e^{2e^{i\theta}}}{6e^{5i\theta}}ie^{i\theta}\,d\theta$$ Therefore $$\Big|\int^{2\pi}_0\frac{e^{2e^{i\theta}}}{6e^{5i\theta}}ie^{i\theta}\,d\theta\Big|\leqslant \int^{2\pi}_0\frac{|e^{2e^{i\theta}}|}{|6e^{5i\theta}|}|ie^{i\theta}|\,d\theta= \int^{2\pi}_0\frac{e^{2\cos\theta}}{6}\,d\theta\leqslant\int^{2\pi}_0\frac{e^2}{6}\,d\theta=\frac{e^2\pi}{3}$$

Answer 2

Here $L=2\pi$ and or $z=\cos(t)+i\sin(t)$ with $t\in[0,2\pi)$, $$\left|\frac{e^{2z}}{6z^5}\right|=\frac{e^{2\cos(t)}}{6}\leq \frac{e^{2}}{6}:=M.$$ Hence, by ML inequality (see What is ML Inequality property of complex integral) $$\left|\int_{|z|=1} \frac{e^{2z}}{6z^5}\, dz\right|\leq \int_{|z|=1}\left|\frac{e^{2z}}{6z^5}\right|\, |dz|\leq ML=\frac{e^{2}}{6}\cdot 2\pi.$$

Answer 3

Brutally applying the triangle inequality and recalling that $|e^z| \leq e^{|z|}$ gives $$ \left| \int \frac{e^{2z}}{6z^5}dz \right| = \leq \int \left| \frac{e^{2z}}{6z^5} \right| dz \leq \int \frac{e^{2|z|}}{6|z|^5} dz = 2\pi \frac{e^2}{6} < \pi e^2. $$