# Verifying an Inequality Involving the Complex Exponential Function

1. Nov 3, 2014

### Bashyboy

Demonstrate that $|e^{z^2}| \le e^{|z|^2}$

We have at our disposal the theorem which states $Re(z) \le |z|$. Here is my work:

$e^{|z|^2} \ge e^{(Re(z))^2} \iff$ By the theorem stated above.

$e^{|z|^2} \ge e^x$

We note that $y^2 \ge 0$, and that multiplying by $-1$ will give us $- y^2 \le 0$; adding $x^2$ to both sides gives us $x^2 - y^2 \le x^2$. Substituting this in gives us

$e^{|z|^2} \ge e^{x^2 - y^2}$. I calculated $|e^{z^2}|$ and found that it was $e^{x^2-y^2}$. Therefore,

$e^{|z|^2} \ge |e^{z^2}|$

________________________________________________________________________

Here is the one issue I see with the proof, but I may have resolved this issue: was the first step justly done? I believe so, and here is why:

$e^{f(z)} \ge e^{g(z)} \iff$

$\ln e^{f(z)} \ge \ln e^{g(z)} \iff$

$f(z) \ge g(z)$.

So, one exponential function is greater than the other when its argument function is greater than the other for all $z$.

Does this seem correct?

2. Nov 3, 2014

### RUber

That seems reasonable. Just remember that you can only compare real numbers, so if f(z) and g(z) are purely real, you are allowed to use the inequalities.
You could also get to the result directly by defining $z=x+iy$, then $|z| = \sqrt{x^2+y^2}$ and $z^2 = x^2 -2ixy - y^2$.