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I need to find the Image of the region $\{z\in\mathbb{C}:\Re(z)>\Im(z)>0\}$ under the map $z\to e^{z^2}$

$z=x+iy\Rightarrow z^2=x^2-y^2+2ixy \Rightarrow e^{z^2}=e^{x^2-y^2}[\cos(2xy)+i\sin(2xy)]$

What I know is $x>y\Rightarrow x^2-y^2>0\Rightarrow e^{x^2-y^2}>0$ and infact as $x^2=y^2$ so $e^{x^2-y^2}>1$ and the modulas of any complex number in the image plane is always $=e^{x^2-y^2}>1$ so can I conclude the image set as $\{w\in\mathbb{C}:|w|>1\}$?

Please tell If I am wrong in my logic anywhere. Thank you for discussion.

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This is incorrect. Note that $x > y$ does not imply that $x^2 > y^2$. (Take $x = 0$, $y=-1$.) Also, you haven't argued why the image is all of $\{ |w| > 1 \}$, just that it's a subset thereof.

The easiest way to do this is to map $\Omega = \{ x > y \}$ first using $z \mapsto z^2$ and then using $z \mapsto e^z$. To do this, observe that $\Omega$ is bounded by the rays $re^{i\pi/4}$ and $re^{-3i\pi/4}$. Hence (how does $z \mapsto z^2$ map rays from the origin?), by mapping $\Omega$ with $z \mapsto z^2$, we get the entire complex plane with the execption of the positive imaginary axis.

Finally $z \mapsto e^z$ maps the complex plane onto $\mathbb{C} \setminus{0}$, and since $e^z$ is periodic with period $2\pi i$, removing the positive imaginary axis doesn't change the image. In conclusion, $z \mapsto e^{z^2}$ maps $\Omega$ onto $\mathbb{C} \setminus{0}$.

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I have edited my question! Thank you for your explanation, could you just tell me if my logic is now correct? –  Une Femme Douce Aug 8 '13 at 9:46
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It's hardly fair to edit your question in a way that makes my answer wrong. I already pointed out the logic flaws in your argument. You haven't showed that all points in $\{ |w| > 1 \}$ are included in the image. –  mrf Aug 8 '13 at 10:10
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@TaxiDriver You can, however, adapt my argument above to cover the new situation. –  mrf Aug 8 '13 at 10:11

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