$\int_{|z+1|=2}^{}\frac{e^z}{z-2}dz \leq2\pi e$

Question

This is how I attempted to solve this but I could not get the exact inequality.

$\gamma(t)=1+2e^{it}, t\in [0,2\pi],f(z)=\frac{e^z}{z-2}$

$|e^{1+2it}|=e|e^{2it}|=e.|e^{it}|.|e^{it}|=e$

$|(1+2e^{it})-2|=|2e^{it}-1|\geq ||2e^{it}|-1|=1$

Thus,|$f(\gamma(t))|=\frac{|e^{1+2it}|}{|(1+2e^{it})-2|}\leq e$

Hope someone could hep me out.thanks

Answer 1

This is a continuation of Timbuc's answer.

|$\int_{|z+1|=2}^{}\frac{e^z}{z-2}dz| \leq \int_{|z+1|=2}^{}|\frac{e^z}{z-2}|dz=\int_{[0,2\pi]}^{}|\frac{e^{\gamma (t)}}{\gamma(t)-2}|\gamma'(t) dz \leq e \int_{[0,2\pi]}^{}\gamma'(t) dt$=$$e\int_{[0,2\pi]}^{}\ 2ie^{it} dt=0 \leq2\pi e$$

Answer 2

The function $$ f(z)=\frac{\mathrm{e}^z}{z-2} $$ is holomorphic in the disk $D(-1,3)$, and $\{z:\lvert z+1\rvert=2\}\subset D(-1,3)$.

Hence, by virtue of Cauchy's Theorem $$ \int_{\lvert z+1\rvert=2}\frac{\mathrm{e}^z\,dz}{z-2}=\int_{\lvert z+1\rvert=2}f(z)\,dz=0. $$

Answer 3

The parametrization of $\;|z+1|=2\;$ is

$$\gamma(t)=-1+2e^{it}\;,\;\;t\in[0,2\pi]\;,\;\;\gamma'(t)=2ie^{it}\implies |\gamma'(t)|=2$$

and also

$$e^{-1+2e^{it}}\implies |e^z|=\left|e^{-1+2\cos t+2i\sin t}\right|=e^{2\cos t-1}$$

so that finally

$$\left|\oint\limits_\gamma \frac{e^z}{z-2}dz\right|\le\oint\limits_\gamma\frac{|e^z|}{|z-2|}dz\le \frac{e^{2\cos t-1}}{5}\cdot 4\pi$$

and then we must prove that

$$\frac{e^{2\cos t-1}}{5}\cdot 4\pi\le 2\pi e\iff e^{2\cos t-1}\le\frac52e\;\;(**)$$

But if we denote

$$f(t):=e^{2\cos t-1}\implies f'(t)=-2\sin t\,e^{2\cos t-1}$$

so that the function is monotone decreasing on $\;(0,\pi)\;$ and monotone increasing on $\;(\pi, 2\pi)\;$, and thus the maximal values are obtained when

$$t=0\implies \;\;\;f(0)=e^1=e\\ t=2\pi\implies f(2\pi)=e^1=e$$

and thus we get that anyway (**) is true.