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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!

Hans Lundmark
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Daniel
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1 Answers1

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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?

  • Ohhh, thank you so much! I had a feeling I was overcomplicating things. This is certainly much simpler and easier to work with. Thank you again for such a prompt and informative answer. – Daniel Apr 24 '14 at 04:56
  • @user119125 You're very welcome. –  Apr 24 '14 at 04:57