# Verifying an Inequality Involving the Complex Exponential Function

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}|##

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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?