Finding the set $\{z: e^z=-1\}$ 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.
 A: 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: $z\in\{0, \pm i\pi, \pm3i\pi, \pm5i\pi,\dots\}$.
A: 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.
A: 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 $\}$.
