Why is $2i\ln(-i)=\pi$

Question

So I always wanted to make my own approximation for pi. I thought of using trigonometric functions to help me start, so I decided to use arccosine. I put $\cos^{-1}(x) = \pi $ in a calculator to solve for $x$. The calculator has a section called "alternate form" where they change the equation a bit. So they gave me this $$\frac{\pi}{2} + i\ln(\sqrt{1-x^2}+ix) = \pi$$ Substituting $-1$ for $x$:$$\frac{\pi}{2}+i\ln(-i)=\pi$$ Then I combine the first and second terms on the left side to get $2i\ln(-i)$. But I don't know how the first equation even exists. How did the calculator come up with this? Or was that first equation is already well-known?

Answer 1

I think you are referring to the principal complex branch of the logarithm. Then if $z=re^{i\theta}$, we have $$ \ln(z)=\ln r + i Arg(z). $$ In this case, $z=-i$ so $Arg(z)=-\frac{\pi}{2}$.

However, the identity breaks down for different branches.

Answer 2

For a complex number $z = r e^{i \theta}$, $$ \ln \left(z\right) = \ln r + i\left(\theta + 2n\pi\right), n \in {\mathbb Z} $$ See this link for an explanation.

With Euler's formula you should be able to see that $-i = e^{-i \pi/2}$.

For $z = -i = e^{-i \pi/2}$, $r = 1$ and $\theta = -\pi/2$, so $$ \begin{eqnarray} \ln \left(-i\right) &=& \ln \left(1\right) + i\left(-\frac{\pi}{2} + 2n\pi\right) \\ &=& i \pi \left(-\frac{1}{2} + 2n\right) \\ &=& ..., -\frac{9\pi i}{2}, -\frac{5\pi i}{2}, -\frac{\pi i}{2}, \frac{3\pi i}{2}, \frac{7\pi i}{2}, ... \end{eqnarray} $$

To derive your first equation, see this post, equation 29 here, or the Inverse Trigonometric Functions section here.