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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 posibility 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 $\sin(-iz)=0$? –  Jack Jul 30 '11 at 2:27
    
$\cos(-iz)+i\sin(-iz)=-1$ is equivalent to saying that $\cos(-iz)=0$ and $\sin(-iz)=i$. Indeed if both conditions hold then we get: $\cos(-iz)+i\sin(-iz)= 0 + i\sin(-iz) = 0 + i^2=-1$ –  user786 Jul 30 '11 at 2:37
    
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. –  Jack Jul 30 '11 at 2:48
    
@Jack: yes thank you that's what I was asking about. –  user786 Jul 30 '11 at 3:19

4 Answers 4

Since $e^{i\pi}=-1$, we can rewrite the equation $e^z=-1$ as $e^z=e^{i\pi}$, or equivalently $e^{z-i\pi}=1$.

The solutions of $e^w=1$ are $w=i(2n\pi)$, where $n$ ranges over the integers. Thus the solutions of $e^{z-i\pi}=1$ are $\:i\pi(2n+1)$, where $n$ ranges over the integers.

About your Calculation: Although that calculation happens to give the right answer, the logic is not right. In the calculation, $\sin$ and $\cos$ are functions of a complex variable, and take on complex values.

You are treating the complex $\cos$ and $\sin$ functions as if they had the same formal properties as the corresponding real functions. For example, from $\cos(−iz)+i\sin(−iz)=−1$, you conclude that one of the summands is $0$ and the other is $−1$. But (until one proves otherwise) the imaginary parts of $\cos(−iz)$ and $i\sin(−iz)$ could each be non-zero, but cancel. And as you point out, there is the possibility of considering $\sin(-iz)=i$, $\cos(-iz)=0$. Unfortunately, these are by no means the only possibilities to consider.

Detailed analysis of the complex sine and cosine may enable you to push an argument through. But it is certainly not immediate.

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Thank you but I already know what the solutions are: I wrote them down in my question. I am asking about the possibility of having $\sin(-iz)=i$ and $\cos(-iz)=0$ and if it reduces to the first case. –  user786 Jul 30 '11 at 2:42
    
Does this prove that those are the ONLY solutions? –  Michael Hardy Jul 30 '11 at 3:00
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@user786: Although your calculation gives the right answer, the logic is not right. You are treating the complex $\cos$ and $\sin$ functions as if they had the same formal properties as the corresponding real functions. For example, from $\cos(-iz)+i\sin(-iz)=-1$ you cannot conclude that one term is $0$ and the other is $-1$. The imaginary parts of $\cos(-iz)$ and $i\sin(-iz)$ could be non-zero, but cancel. –  André Nicolas Jul 30 '11 at 3:12
    
@André: thank you. That's the answer I needed. –  user786 Jul 30 '11 at 3:18
    
@Michael Hardy: Yes, if we take the solutions of $e^w=1$ as a "standard fact." –  André Nicolas Jul 30 '11 at 3:35

Supposing $z = x + iy$, we have $e^z = e^{x+iy} = e^x e^{iy} = e^x(\cos y + i\sin(y))$.

(That much is true even if $x$ and $y$ are not real!)

Now suppose $x$ and $y$ are real. Then $e^x$ is positive and $\cos y + i\sin y$ is on the unit circle centered at $0$, so that the absolute value $|\cos y + i\sin y|$ is $1$. So the absolute value of $e^x(\cos y + i\sin(y))$ is $e^x$. Since $|-1|=1$, we need $e^x =1$. Since $x$ is real, this means $x=\ln1 = 0$. So we want $\cos y + i\sin y = -1$. Therefore we must have $\cos y = -1$ and $\sin y = 0$. That $\sin y=0$ means $y\in\{0, \pm\pi, \pm 2\pi, \pm3\pi,\dots\}$. But at some of those points the cosine is $+1$ rather than $-1$. The points where $\cos y = -1$ are $\pm\pi, \pm3\pi,\pm5\pi,\dots$.

Bottom line: $z\in\{0, \pm i\pi, \pm3i\pi, \pm5i\pi,\dots\}$.

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Yes indeed this is exactly what I've found, but I look at my equation above, i could very well also have the possibility that $\sin(-iz)=i$. What do I do with this possibility? –  user786 Jul 30 '11 at 2:20
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In bottom line, you shouldn't include 0. –  sdcvvc Jul 30 '11 at 2:28
    
@user786: The possibility that $\sin(-iz) = i$ is not the only possibility. Those possibilities do arise and bother you in a way that the path you're choosing is most probably not the right one to take to prove such a thing. You have plenty of suggestions of how to take another path to solve your problem, so I guess you should just take a look at them. I'm not saying it's impossible to go on with your idea, but it most probably is. Michael Hardy's idea is the closest one to yours though. –  Patrick Da Silva Jul 30 '11 at 3:29

Write $z = x + iy$, with $x$ and $y$ real. Then $e^z = e^{x + iy} = e^x e^{iy}$. Note this writes $e^z$ in polar form, with $r = e^x$ and $\theta = y$.

Next, note that in polar form $-1 = 1*e^{i\pi}$. The polar form of any nonzero complex number is unique, except that any multiple of $2\pi$ may be added to the argument. So the statement that $e^z = -1$ is equivalent to the statement that $e^x = 1$ and $y = \pi + 2k\pi$ for some integer $k$. Equivalently, $x = 0$ and $y$ is an odd multiple of $\pi$.

So the set of all such $z$ are exactly $\{\pi k i: k$ is an odd integer $\}$.

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When you say "finding the set ${z:e^z=-1}$", I think you are talking about "finding the complex roots of $e^z=-1$. Since $\{z\in{\mathbb R}:e^z=-1\}$ and $\{z\in{\mathbb C}:e^z=-1\}$ are totally different. On the other hand, the set is already, so "finding the set" may be somewhat confusing.

For solving $e^z=-1$, you would like to look at complex logarithm.

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