Question

For $z$ being complex number prove that $|e^z|\le e^{|z|}$.

Answer 1

For $x,y$ being real numbers we have,

$|x|\le \sqrt{x^2+y^2}$.

We always have, $x\le |x|$.

Since $e^x$ is an increasing function then we get,

$e^x\le e^{|x|}\le e^{\sqrt{x^2+y^2}}$.

or, $|e^z|\le e^{|z|}$. [ Since $|e^z|=e^x$].