How to evaluate a complex integral?

Question

Show that |$\int_{C} \frac{e^z}{\bar z + 1} dz$| ≤ $2\pi e^2$ where C is the circle |z-1| = 1.

I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$

with 0 ≤ t ≤ $2 \pi$

How do i have to continue this proof?

Answer 1

By the ML inequality, we have that

$$\Bigg|\int_\Gamma f(z)\mathrm{d}z\Bigg| \leq \max_{z \in \Gamma}f(z) \cdot l(\Gamma)$$

where $l(\Gamma)$ is the arc length of $\Gamma$.

Thus, for your given integral, it is :

$$\Bigg|\int_C \frac{e^z}{\bar{z}+1}\mathrm{d}z\Bigg| \leq \max_{z \in \Gamma}\Bigg\{\frac{e^z}{\bar{z}+1}\Bigg\}\cdot l(\Gamma)$$

But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(\Gamma) = 2\pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?