For what values of $z$ is $e^z=-\frac{i}{2}$. Question: For what values of $z$ is $e^z=-\frac{i}{2}$.
Thoughts: I've done a couple of problems like this, and I've found it best to either begin by writing $re^{i\theta}=r\cos\theta+ir\sin\theta$, or writing $e^z=e^x(\cos\theta+i\sin\theta)$, just depending on which is easier to go from.  For the $re^{i\theta}$ way of writing it, I find that $r=\frac{1}{4}$, and so I would like for $\sin\theta=-2$ and for $\cos\theta=0$, so since $\cos\theta=0$, we have that $\theta=\frac{\pi}{2}+\pi k$, for $k\in\mathbb{Z}$, but $\sin\theta$ certainly doesn't equal $-2$ at any of those values.
So, we try and play with the other form.  I would like for $e^x=1$ and for $\sin\theta=-\frac{1}{2}$ and $\cos\theta=0$, but there is no such $\theta$.  I suppose I could try a different combination, but since I need $Re(z)=0$, I am not quite sure how.  I'm sure, by this point, I am over looking something very simple.  Any help is greatly appreciated!  Thank you.
 A: I think your mistake was saying $r=1/4$. $-i/2$ has modulus $1/2$, and so $\exp(z)=\exp(s+it)=\exp(s)\exp(it)=-i/2$ shows you $\exp(s)$ must be $1/2$ (and therefore $s=\ln(1/2)$), where I take $s,t$ to be real, since $\exp(it)$ always lies on the unit circle - modulus $1$.
$\exp(it)=-i$: immediately one knows $t=-\pi/2$, since $-i$ is a quarter turn down from the positive real line: $-\pi/2$ is its argument. You can verify this using Euler’s formula and trigonometry if you wish.
Altogether now: the principal value $z$ s.t. $\exp(z)=-i/2$ is $\ln(1/2)-i\pi/2$. Add an additional term of $+2\pi ik,\,k\in\Bbb{Z}$, and you get all solutions.
A: An alternative approach is:
Since the real part of $\frac{-i}{2} = 0, ~\theta ~$ must be chosen such that 
$\cos(\theta) = 0 \implies \sin(\theta) = \pm 1.$ 
Next, since the value has form $i \times k$, where $k < 0$, you must have that $\sin(\theta) = -1.$
The problem is then completed following the analysis of the answer of FShrike.
That is, $\theta$ must have form $(-\pi/2) + 2n\pi ~: ~n \in \Bbb{Z}.$
Further, $r > 0$ must be chosen such that 
$r \times -1 = (-1/2) \implies r = (1/2).$
Then, noticing that $e^{\ln(1/2)} = (1/2)$ 
you conclude that $z = \ln(1/2) + i[(-\pi/2) + 2n\pi] ~: ~n \in \Bbb{Z}.$

The original problem is actually a special case of for how to compute $z$ such that $e^z = (a + ib) ~: ~(a + ib) \neq (0 + i[0]).$
As has been indicated, the very first step is to compute $r > 0$ such that $r = \sqrt{a^2 + b^2}$.  Then, the real part of $z$ will be $\ln(r).$
Then, within a modulus of $2\pi$, you identify the unique value $\theta$ such that $\cos(\theta) = (a/r), \sin(\theta) = (b/r).$
Note that as defined, you will have that

*

*$\cos^2(\theta) + \sin^2(\theta) = 1.$


*The identification of the unique value of $\theta$, within a modulus of $2\pi$, will always be possible.
Thus, you will have that $(a + bi) = re^{i\theta} = e^{\ln(r) + i(\theta + 2n\pi)} ~: ~n \in \Bbb{Z}.$
