What is wrong with this logarithm?

WBahn

Joined Mar 31, 2012
26,398
So what is the mistake?

Is it that (a^x)^2 isn't a^(2x)? Or is it that (-1)^2 isn't 1?

If you are being careful, it should be noted that

\(e^{ix} \; = \; e^{i\left(x+2n\pi\right)}\)

I would say that the problem is here:

\(ln \left( e^{2 \pi i} \right) \; = \; 2 \pi i\)

This should be

\(ln \left( e^{2 \pi i} \right) \; = ln \left( e^{2n \pi i} \right) \; = \; 2n \pi i \;\; \text{(n is an integer)}\)

So now in the next set you have

\(2n \pi i \; = \; 0 \)

which is true if we choose n=0
 
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