$\log z = 6i$
I am working on a problem very similar.
What I am seeing
$\log z = \ln|z| + i(\theta + 2\pi n)$ for $n\in\mathbb{Z}$
What I am curious about, as if seen obvious to me that
$ \log z = 6i$
Can not exist as $6\geq\pi$
I have looked at polar coordinate such that $6i$ is equivalent to $e^{(6i)} =\cos(6)+ i\sin(6)$
Any thoughts on my approach?
If $\log z = 6i$ where $\log$ is the principal branch of the logarithm, your remark is correct and no solution exists. If $\log$ can be any suitable branch however, then we can find $z$ simply by computing $$z = \exp(6i) \approx 0.960170286650366 - 0.279415498198926i$$ There exist branches of the logarithm, such that $\log \exp 6i = 6i$. The principle branch will evaluate to $\log z = (6-2\pi)i$ nevertheless.