TITLE: Proof for Complex Analysis Inequality
QUESTION [6 upvotes]: This is a homework assignment that will be graded; so I'm not specifically asking for an answer.  But I could use a hint, as it's been a few days and I'm still not sure if how I've proved it would really be sufficient.
$$
|e^{z^2}| \le e^{|z|^2}
$$
As far as my attempt goes, I simplified $ |e^{z^2}| $ to $ |e^{x^2-y^2}| $ which I then stated was less than or equal to $ e^{z^2} $ to get rid of the modulus.
Not knowing what to do there, I decided to try to simplify the right term somehow.  All I managed was to square the z, but I couldn't think of a useful way to simplify it further. $ e^{|x^2-y^2+2ixy}| $
So I tried to prove that $ x^2-y^2 \le |x^2-y^2+2ixy| $, but I'm not sure if that would be considered proper.  After that, I figured I would show that, by definition, $|x^2-y^2+2ixy|$ would be $ x^2-y^2 \le ((x^2-y^2)^2+(2xy)^2)^{(1/2)} $.  Squaring both sides, it would end up as $ x^4 - 2x^2y^2 + y^2 \le x^4 + 2x^2y^2 + y^2 $, which would be always true, since x and y are real numbers and they're all being raised to even powers.
I'm not sure if my method is flawed, but if it's wrong or I've made a mistake somewhere, I would appreciate if someone informed me.  Thank you!

REPLY [3 votes]: First, remember that $e^t > 0$ for all real $t$, so $|e^{x^2 - y^2}| = e^{x^2 - y^2}$. Now for a simpler approach: Recall that $|z|^2 = x^2 + y^2$ and work with this directly; you need to prove that (spoiler; mouse over to see)

 $$e^{x^2 - y^2} \le e^{x^2 + y^2}$$

Does this seem a bit more manageable?