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I want to find the set of $z$'s such that $\{z: e^z=-1\}$. Then this just mean that I have to solve $\cos(-iz)+i\sin(-iz)=-1$ which is equivalent to having $\cos(-iz)=-1$ and $\sin(-iz)=0$ Then I find that the set of solutions is such that $-iz=\pi + 2k\pi$ or in other words, $z=(1+2k)\pi i$

Should I also consider the possibility that $\sin(-iz)=i$ and $\cos(-iz)=0$ or is it irrelevant to take this possibility into account? I am not sure. Thx.

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    Why do you think $\cos(-iz)+isin(-iz)=-1$ is equivalent to having $\cos(-iz)$ and $\sin(-iz)=0$?2011-07-30
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    $\cos(-iz)+i\sin(-iz)=-1$ is equivalent to saying that $\cos(-iz)=0$ and $\sin(-iz)=i$. Indeed if both conditions hold then we get: $\cos(-iz)+i\sin(-iz)= 0 + i\sin(-iz) = 0 + i^2=-1$2011-07-30
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    Opps, I mean why do you say in the first paragraph that $\cos(-iz)+i\sin(-iz)=-1$ is equivalent to having $\cos(-iz)=-1$ and $\sin(-iz)=0$. You are trying to solve the equation "case by case". But things may become complicated in this way. For example, $\cos(-iz)=k$,$\sin(-iz)=(k+1)i$ where $k$ is a real number can also satisfy the equation.2011-07-30
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    @Jack: yes thank you that's what I was asking about.2011-07-30

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