How to use the ML inequality? 
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?
 A: 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)
$$\left|\int_{|z|=1} \frac{e^{2z}}{6z^5}\, dz\right|\leq \int_{|z|=1}\left|\frac{e^{2z}}{6z^5}\right|\, |dz|\leq ML=\frac{e^{2}}{6}\cdot 2\pi.$$
A: 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 
$$\Big|\int^{2\pi}_0\frac{e^{2e^{i\theta}}}{6e^{5i\theta}}ie^{i\theta}\,d\theta\Big|\leqslant \int^{2\pi}_0\frac{|e^{2e^{i\theta}}|}{|6e^{5i\theta}|}|ie^{i\theta}|\,d\theta= \int^{2\pi}_0\frac{e^{2\cos\theta}}{6}\,d\theta\leqslant\int^{2\pi}_0\frac{e^2}{6}\,d\theta=\frac{e^2\pi}{3}$$
A: Brutally applying the triangle inequality and recalling that $|e^z| \leq e^{|z|}$ gives
$$
\left|
\int \frac{e^{2z}}{6z^5}dz
\right|
= \leq
\int \left|
\frac{e^{2z}}{6z^5}
\right| dz
\leq
\int
\frac{e^{2|z|}}{6|z|^5} dz
= 2\pi \frac{e^2}{6}
< \pi e^2.
$$
