Why does Natural Log exhibits such behaviour with negative numbers?

Question

Out of curiosity I tried negative numbers in the function Ln(x), and following were the outputs:

And sorry for awkward writing, I don't know LaTeX that much, any edit will be appreciated!

$ \ln(-1) = \pi \ i$
$\ln(-2) = \pi \ i$ + $\ln(2)$
And it goes so on!

With a bit of change:

$\ln(-2 \times i) = \ln(-i \times 2)$
Here why does it stays the same?

And the last one:
$\ln(i)= \frac{i\pi}{2}$

Can anyone give a thorough explaination on this! And also why do the natural log function exhibits such behaviour. And also can this be related to the exponential function somehow? Thanks!

Answer 1

Take the complex number $z=x+iy$ which has length $|z|=\sqrt{x^2+y^2}$. We can also represent this using Euler's identity, $$z=|z|(\cos\theta+i\sin\theta)=|z|(\cos(\theta+2\pi k)+i\sin(\theta+2\pi k))=|z|e^{i\theta+2i\pi k}$$ where $\theta\in[-\pi,\pi)$ and $k\in\mathbb{Z}$. Now take logarithms, $$\log z=\log(|z|e^{i\theta+2\pi k})=\log|z|+i(\theta+2\pi k)$$ if you choose a $k$ once and for all, then you have chosen a particular "branch."

Answer 2

Euler's identity: $e^{i \pi} = -1$

Take the logarithm of both sides to get:

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

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

If $x > 0$, then $\ln(-x) = ln(-1 \times x) = \ln(-1) + \ln(x)$

From above, $\ln(-1) = i \pi$, so $\ln(-x) = i \pi + \ln(x)$

For the last one, $\ln(i) = \frac{i\pi}{2}$, take the logarithm of both sides of $e^{i\pi/2} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$