Solving the complex equation $\cos z = 2$ After doing some work, I understand that the correct way to go about this is to start with 
$$
\cos z = \frac{e^{iz} + e^{-iz}}{2} = 2
$$
...and then solving the quadratic equation
$$
(e^{iz})^2 -4(e^{iz}) + 1 = 0
$$
to get $2 \pm \sqrt{3}$, but why do we then take the natural log of this to solve for $z = i\ln(2 \pm \sqrt{3})$?
And why can't we start by using the fact that the roots of $\cos z$ will be 
$(n +\frac{1}{2})π, n \in \Bbb Z$?
Thank you!
 A: So the equation you're trying to solve is
$$(e^{iz})^2 -4(e^{iz}) + 1=0$$
Suppose we make the substitution $a = e^{iz}$ - this will make what comes out clearer. Then this becomes
$$a^2 - 4a + 1=0$$
Solving through the quadratic formula yields
$$a = 2 \pm \sqrt{3}$$
But through our substitution, we also know $a = e^{iz}$ and therefore
$$e^{iz} = 2 \pm \sqrt{3}$$
Therefore, we take the natural logarithm of both sides. But we have to be careful, since the exponential is actually a periodic function: that is,
$$e^{iz} = e^{iz + i2\pi} = e^{iz + i4\pi} = ... = e^{i(z + 2k\pi)}$$
for all integers $k$. So we generalize our result by using this latter result, concluding
$$i(z + 2k\pi) = \ln(2 \pm \sqrt 3)$$
Divide through by $i$ (note that $1/i = -i$) and then subtract the even multiple of $\pi$ from both sides. Thus,
$$z = -i \cdot \ln(2 \pm \sqrt 3) - 2k\pi$$
In particular, taking $k=0$, we get the result from your opening post, $$z = -i \cdot \ln(2 \pm \sqrt 3)$$

So that's my first warning to you: there are multiple solutions to this equation, owing to the periodic nature of the exponential function.
But anyhow, the reason we take the natural logarithm of the solution of the quadratic, is because by design $e^{iz}$ is the root of the quadratic.


And why can't we start by using the fact that the roots of cos z will be 
  $(n +\frac{1}{2})π, n \in \Bbb Z$?

The real question is, why would this be useful? I can't think of why it would be. The roots of $\cos(z)$ in the real line - the values for which $\cos(x) = 0$, where $x\in\Bbb R$ - don't seem to have much usefulness whatsoever here.
