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I have the following: What are the solutions to:

$$e^{z^{2}}=-1$$

in the circle around $z=0$ of radius $R=2$

Apllying the complex $LOG$

I derived: $$z^2 = \pi*i(1+2k)$$

What is the simplest way to continue from here? I tried to write $z=x+iy$ and then derived that $x=+-y$ and then: $2x^2=\pi*i(1+2k)$ but im getting mixed with positive and negetive values and the values of $k$ are unclear...how can I solve this one easily? Thanks

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    $\begingroup$ $e^z*e^z = e^{2z}$ and not $e^{z^2}$ $\endgroup$ – benji Feb 4 '15 at 0:08
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    $\begingroup$ Also one can respect convention by dropping the $*$ symbol to denote multiplication; we are not discussing programming languages. $\endgroup$ – P Vanchinathan Feb 4 '15 at 0:42
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One way would be to put things in polar form: write $z=r e^{i \theta}$, note that $i=e^{i \pi/2}$, and figure out what r and $\theta$ are.

EDIT: Didn't actually read your whole question. This would be how to solve $z^2 = (2 k + 1) \pi i$, but it's overkill for your actual question, which is how to solve $e^{2z} = -1$.

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  • $\begingroup$ If im going for polar cords, im getting: $R^{2}e^{2\theta i}=e^{\pi /2i+2\pi m}(\pi+2k\pi)$ but now im getting just one root inside the circle radius 2 for $k=0$ and $m=0$ and I know the answer should be 4 roots $\endgroup$ – user3921 Feb 4 '15 at 7:22
  • $\begingroup$ Well, as it's been pointed out above, you're not looking for solutions to $z^2 = (2k+1)\pi i$. $\endgroup$ – Daniel McLaury Feb 4 '15 at 20:21

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