Proving $e^z=-1\iff z=i\pi$
Solution 1:
They're wrong and you're right.
One is not allowed to say that “$e^z=-1$ if and only if $z=\log(-1)$ where one (particular) branch of the logarithm is taken”. For the simple reason that the statement is false.
It would be similar to say that $z^2=1$ if and only if $z=1$, because we take a particular branch of the square root.
What one can say is that if $z=\log(-1)$ (with the above meaning for the logarithm), then $e^z=-1$, essentially by definition. But the reverse implication is clearly false, because the exponential function is periodic.
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