Why can a Complex Logarithm have infinitely many values? What does this mean, that "Due to the periodicity of the trigonometric functions, a complex logarithm can have infinitely many values"?
$\ln z=\ln (\cos x +i\sin x)=?$
 A: The complex logarithm is the inverse of the complex exponential. By Euler's formula,
$$e^{iy} = \cos y + i \sin y.$$
Hence for any integer $n$,
$$e^{x + (y + 2\pi n)i} = e^x (\cos y + i \sin (y + 2\pi n)) = e^x (\cos y + i \sin y) = e^{x + iy}.$$
So the complex exponential takes each of its values infinitely often, from which it follows that the complex logarithm has infinitely many values (for this reason, we much choose a branch cut for the complex logarithm).
A: The exponential function $f(x) = e^x$, as a function that takes real numbers to positive real numbers, is a one-to-one and onto function, and therefore has a well-defined inverse function: for each positive $y$, there is just one number (denoted $\log(y)$) such that $e^{\log(y)} = y$. 
But the exponential function $f(z) = e^z = \sum_{n \geq 0} \frac{z^n}{n!}$, considered as a function on the complex numbers, is not one-to-one. There is in particular the wonderful identity due to Euler, 
$$e^{i x} = \cos(x) + i\sin(x),$$ 
which shows for example that there are infinitely many solutions to the equation $e^{z} = 1$, namely $z = 2\pi k$ for $k = \ldots, -3, -2, -1, 0, 1, 2, 3 \ldots$. So the complex exponential function is not one-to-one. 
More fundamentally, while we could just randomly choose a solution $w$ to the equation $e^w = z$ for each $z$, and call those choices collectively "the logarithm of $z$", there is no way to choose so that the result will give a continuous logarithm function. 
A: 
Why can a Complex Logarithm have infinitely many values?

For the same reason that $$e^{n2\pi i}=1$$ is true for infinitely many $n\in\mathbb{Z}$.
Taking the inverse of $y=e^x$, which is $\ln(y)$, then leaves you with this non-uniqueness in the result.
