hey all, i got this problem
find all the solutions z^(i)=1
so i wrote it out as e^(i(ln(1)+i@)= 1
and i end up getting e^(-@)=1
is that right?
This question also appeared in this thread.
To complete the answer I gave there (since Tonio's answer is incorrect): is equivalent to (up to this point, the values of the left-hand sides depend on the determination of the logarithm), which is equivalent to for some , and thus to , . Hence the solutions are the real numbers for every ; it is worth noting that since the answer is real, the choice of the determination of the logarithm is irrelevant (and you can forget about it if you don't know what I mean).
Edit: drumist was quicker than me, but I leave my post because of the remark on the definition of logarithm