Which is the value of this expression

I want to prove the following:

$$\Big|\int_C \frac{e^z}{\overline{z}+1}dz\Big|\leq 2\pi e^2$$where $C$ is the unit circle $|z-1|=1$.

I used the parametrization $\lambda: t \Longrightarrow 1+e^{i2\pi t}$, and I think I am left to show that $$\Big|\int_{0}^1 \frac{e^{i2\pi t + e^{i2\pi t}}}{2+e^{-i2\pi t}}dt\Big|\leq e$$but I don’t know how to follow.

Any help or hints would be appreciated.

Solution 1:

In general for a continous function $f: U \to \Bbb C$ and a differentiable (even for piecewise differentiable; just split it into differentiable paths and sum them up) path $\gamma: [a,b] \to U$

$$\Big\vert \int_\gamma f(z) dz \Big\vert = \Big\vert \int\limits_{a}^b f(\gamma(t)) \dot \gamma(t) dt \Big\vert \leq \int\limits_{a}^b \big\vert f(\gamma(t)) \dot \gamma(t) \big\vert\ dt \leq \|f\|_{\infty,\gamma} \int_{a}^b \vert \dot \gamma(t) \vert dt = \|f\|_{\infty,\gamma}\cdot L(\gamma)$$

where $L(\gamma)$ denotes the lenght of $\gamma$. In your case this means, if you use the parametrization $t \mapsto 1+e^{it},\ t\in [0,2\pi]$:

$$\Big\| \frac{e^{z}}{\bar{z} + 1} \Big\|_{\infty, C} \leq e^2$$ since for all $t \in [0,2\pi]$: $$\vert 1+e^{-it}+1 \vert \geq 1 \Rightarrow \left\vert \frac{1}{1+e^{-it}+1}\right\vert \leq 1$$ $$\big\vert e^{1+e^{it}} \big\vert = e\cdot\big\vert e^{\cos(t) + i\sin(t)} \big\vert = e \big\vert e^{\cos{t}} \big\vert \leq e^2$$

Now you get

$$ \int_C \frac{e^z}{\bar{z}+1} dz \leq 2\pi e^2$$