Paremetrising the Contour

Question

I'm trying to paremetrise the Contour of a unit circle descibed anti clockwise.

This is so I can integrate $$ \int_{|z| = 1} \frac{e^z}{4z^4} dz $$ Now normally $z(t)=e^{it}$ for $t\in [0,2\pi]$ is sufficient but when I use Contour integral to compute this then I end up with a very messy integral to solve.

There must be an easier or better way to solve this. Can we paremetrise the Contour in a different way which would make the integral easier to solve?

Thanks guys for the quick response! My main focus is to find a better parametrisation. This is because I will be using the estimation theorem.

I'm trying to show (Sorry not good at Latex):

$$ \left|\int_{|z| = 1} \frac{e^z}{4z^4} dz\right| \le \frac{e\pi}{2} $$ .

Answer 1

To prove

$$\left|\int_{|z| = 1} \frac{e^z}{4z^4}\, dz \right| \le \frac{e\pi}{2},$$

one does not need to evaluate the integral. On the circle $|z| = 1$,

$$\left|\frac{e^z}{4z^4}\right| = \frac{e^{\operatorname{Re}z}}{4} \le \frac{e^{|z|}}{4} = \frac{e}{4}.$$

Since the arclength of the unit circle is $2\pi$, the ML-inequality yields

$$ \left|\int_{|z| = 1} \frac{e^z}{4z^4}\, dz\right| \le \frac{e}{4}\cdot 2\pi = \frac{\pi e}{2}.$$

Answer 2

Your integral is of the form $I=\displaystyle\frac{1}{4}\oint_\gamma\displaystyle\frac{f(z)}{(z-0)^{n+1}}dz$, where $n=3$ and $f(z)=e^z$ is analytic within the disk $D$ centered at $0$ and of radius 1. The boundary of this disk is the given circle $\gamma$. By Cauchy formula for the derivatives of $f$ at points from $D$, we know that $f^{(n)}(0)=\displaystyle\frac{n!}{2\pi i} \oint_\gamma\displaystyle\frac{f(z)}{(z-0)^{n+1}}dz$. Hence your integral is $I=\displaystyle\frac{2\pi i}{4n!} f^{(3)}(0)$. The derivatives of $f(z)=e^z$ are calculated as in the real case.