$x^2+y^2\leq4, |z|^2=x^2+y^2\Rightarrow|z|^2\leq4\Rightarrow|z|\leq2$
so,$|z||e^z|=\sqrt{x^2+y^2}e^x$since $x\geq0,y\geq0$
If go through x-axis then y=0 and |f(z)|=$2e^x(0\leq x\leq2)$
then, x=$r\cos t\Rightarrow|f(z)|=2e^{2\cos t},0\leq t\leq \frac{\pi}{2}$around of circle, and then $x=0, |f(z)|=y,0\leq y\leq2$
On the x-axis, the max of |f|is $2e^2$, on the circle is t=0$\Rightarrow|f(z)=2e^2|$, on the y-axis is 2
so, Max${(2e^2,2e^2,2)}=2e^2 $
