Contradiction in Euler’s Identity? $2i\pi = 0$

Question

Many of us know Euler’s famous identity:

$e^{i\pi} + 1 = 0$.

But if we add -1 (subtract 1) to both sides we get:

$e^{i\pi} = -1$

then natural log of both sides and:

$i\pi = \ln(-1)$

Next we multiply both sides by 2:

$2i\pi = 2\ln(-1)$

Which by basic logarithm rules is equal to:

$2i\pi = \ln({-1}^2) = \ln(1) = 0$

so

$2i\pi = 0$

which can’t be true as either 2, i or $\pi$ would have to equal 0. My best guess is that logarithms just plainly aren’t defined for negative values, I just assumed this was just true logarithms bounded under the real numbers, though I might be wrong. If someone could help it would be great.

Thanks.

Answer 1

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= \log w + 2\pi i n$$ where $n$ is an arbitrary integer, $n=\dots, -2,-1,0,1,2,\dots$.

In your case, you will get $$ 2\pi i = 0 + 2\pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).