solving for z in $|e^z| = 2$

Question

How would I solve for z in the following case:
$|e^z| = 2$, now I know that $|e^z| = e^a$ if we let $z = a+bi$ so then equating moduli we get $a = \ln{2}$

But what about $b$? $2 = 2e^{(0+2\pi n)i}$ so what is the value of b? Can be just be an element of the reals as we place no expectation on $b$?

Answer 1

Let $z=x+iy$, then $e^z = e^x e^{iy}$, and so $|e^z| = e^x$.

Hence $|e^z|=2$ iff $e^x = 2$ iff $x = \log 2$.

(And so $y$ is arbitrary.)