Prove that $\int_\gamma \frac{cos(z)}{z}dz \le 2\pi e^2$ where gamma is a circle of radius 2 around the origin

Question

Prove that $\int_\gamma \frac{cos(z)}{z}dz \le 2\pi e^2$

Been stuck on this for a while. We need to use:

$$\int_\gamma f(z)dz \le l(\gamma)M $$

Where $ l(\gamma)$ is the length of the path and M is the maximum norm: $$M=max_{t\in [a,b]}|f(\gamma (t))| $$

. I've found that I can parameterize the path as follows:

$$\gamma = 2e^{it}\;0\le t\le 2\pi $$.

And the length of the path, l, is given as $4\pi $, but I'm unable to find the norm since I end up with an exponent inside the cos. I'd appreciate a hint

Answer 1

I think you have to show that $|\int_\gamma \frac{cos(z)}{z}dz| \le 2\pi e^2$.

For $|z|=2$ we have $|e^{iz}| \le e^{|z|}=e^2$, hence

$|\frac{\cos z}{z}|=\frac{1}{2} \cdot \frac{1}{2}|e^{iz}+e^{-iz}| \le \frac{1}{2} \cdot \frac{1}{2}(e^2+e^2)=\frac{1}{2}e^2$.

Can you proceed ?

Remark: with Cauchy's formula we get

$\int_\gamma \frac{cos(z)}{z}dz= 2 \pi i \cos(0)=2 \pi i $

Answer 2

You can simply use Cauchy’s integral formula to obtain $\int_{\gamma}\frac{cos(z)}{z}dz=2\pi{}i cos(0)=2\pi{}i $ thus $|\int_{\gamma}\frac{cos(z)}{z}dz|=2\pi{}<2\pi{}e^2$.