Question regarding logarithms 2

Question

What is $\ln(-1)$? And would there a taylor series for $$\ln\frac{1+x^m}{1-x^m}$$?

Answer 1

Use Euler's Identity:

$e^{i\pi}+1=0\implies-1=e^{i\pi}\implies\ln(-1)=\ln(e^{i\pi})=i\pi$

Answer 2

The most beautiful formula in mathematics is $$\Large e^{i \pi}+1=0$$ so the result as given by barak manos while I was typing.

With regard to the second question, consider $$\log(1+y)=\sum_{i=1}^{\infty}\frac{(-1)^{i-1}y^i}{i}$$ $$\log(1-y)=-\sum_{i=1}^{\infty}\frac{y^i}{i}$$ and now subtract to get $$\log\frac{1+y}{1-y}=2\sum_{i=1}^{\infty}\frac{y^{2i-1}}{2i-1}$$ Now, you can replace $y$ by $x^m$.

Answer 3

For the first question, I'd like to make some complement to other answers: $$\ln(-1)=\ln(e^{(1+2k)i\pi})=(1+2k)i\pi$$