log(-1)?

[mollwollfumble]

What is wrong with this reasoning?

log(-1) = log(1/-1) = log(1) - log(-1)
So
2 log(-1) = log(1) = 0
log(-1) = 0.

[gmalivuk]

The problem with that reasoning is that it's only valid as the second part of the proposition, "If log(-1) is a real number, then...".

Similarly, we could have
sqrt(-1) = sqrt(1/-1) = sqrt(1)/sqrt(-1)
(sqrt(-1))^2 = sqrt(1) = 1
sqrt(-1) = +/- 1

[tomtom2357]

The problem is 1) you have to check whether it actually works out the other way around e⁰ is not equal to -1, it is equal to 1 and 2) log(x) actually has multiple values, we just normally write the real valued one, but since log(1) also equals 2i*pi your reasoning should look like this:
2log(-1)=log(1)=n*2i*pi
log(-1)=n*i*pi, but since e^2i*pi=1, then:
log(-1)=n*i*pi only for odd n