Old exciting formula!

We have $\frac{1-i}{1+i} = -i = e^{\pi i \left(-\frac{1}{2} + 2n\right)}, \text{ for any integer } n$

Therefore, $2i \ln\left(\frac{1-i}{1+i}\right) = 2i \cdot \left(\pi i \left(-\frac{1}{2} + 2n\right)\right) = \left(1 - 4n\right)\pi$

We select the branch given by $n=0$ to get an answer of $\pi$ (also note that none of the other answers is an integer multiple of $\pi$ , so they cannot be correct).

NOTE AND WARNING: The complex logarithm is multi-valued, so it only makes sense to say that it equals a particular value on a given branch, not in general. In particular, unless you are equipped to deal with a multi-valued logarithm, most of the standard properties of the real logarithm do not hold for the complex logarithm.

$\begin{aligned} z & = 2i\ln \left(\frac {1-i}{1+i} \right) \\ & = 2i\ln \left(\frac {(1-i)^2}{(1+i)(1-i)} \right) \\ & = 2i \ln \left(\frac {-2i}{2} \right) \\ & = 2i \ln ({\color{#3D99F6}-i}) & \small \color{#3D99F6} \text{Consider only the principal value and} \\ & = 2i \ln \left({\color{#3D99F6}e^{-\frac \pi 2 i}} \right) & \small \color{#3D99F6} \text{by Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ & = 2i \left(-\frac \pi 2 i \right) \\ & = \boxed{\pi} \end{aligned}$

we have: so: