${{2\pi i=0}}$? [duplicate]

I was trying to solve for $i^i$. I got distracted and did the following:
$e^{i \theta} = \cos(\theta) + i\sin(\theta)$
$e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation:
$\ln(e^{i2\pi}) = \ln(1)$
$i2\pi = 0$
Clearly this is false. What did I do wrong?

Solution 1:

You cannot conclude that $2i\pi=0$.

If $f$ is an injective function, you can indeed argue that $f(x)=f(y)$ implies $x=y$. However the exponential function is not injective. Indeed, it is periodic with period $2i\pi$: $$ e^{z+2i\pi}=e^z $$ for every complex $z$. The complex logarithm is a wild beast, compared to the standard functions: it cannot be the inverse of the exponential exactly because the exponential is not injective; it's an example of a multivalued function and cannot be used like standard functions.

Solution 2:

You should ask yourself, if you have understood the definitions of all the relevant objects?

It turns out, that defining $\ln$ is not so easy to do for complex numbers, and that a definition of the $\ln$ does not have the nice properties that they have for real numbers.

For example $$ \ln(e^z) = z $$ does not need to be true for a complex number $z$. (this can be regarded as a mistake in your question)

Just because something is true for real numbers, it doesnt mean that the same thing is also true for complex numbers.