Simplify the expression $ \left(\displaystyle \frac{1+i z}{1-iz}\right)^i, i^2=-1. $

Question

Simplify the expression $$ \left(\frac{1+i z}{1-iz}\right)^i, i^2=-1, z \in \mathbb{R}. $$ Really I have no any good idea to do it.

Answer 1

Converting to polar form, we have

$$1 +iz = \sqrt{1+z^2} e^{i(\arctan z + 2n\pi)}, 1-iz = \sqrt{1+z^2}e^{-i(\arctan z + 2m\pi)}$$

where $n, m$ can be any integers due to the periodicity of the complex exponential.

Since the modulus is the same for both complex numbers, it cancels out in the fraction and we are left with:

\begin{align} \left(\frac{1+iz}{1-iz}\right)^i &= \left(\frac{e^{i\arctan z}e^{2n\pi i}}{e^{-i\arctan z}e^{2m\pi i}}\right)^i\\ &=\left(e^{2i\arctan z}e^{2i\pi k}\right)^i\\ &=e^{-2(\arctan z +\pi k)} \end{align}

where $k = n - m$ is also an integer.

Answer 2

Since I've never dealt with complex powers of complex numbers I'm just testing out what I think should happen following Ananthakrishna's advice, please suggest any changes or help fill in the gaps. Here goes-

Rationalizing said expression $=(\frac{1-z^2+2iz}{1-z^2})^i=(1+i\frac{2z}{1-z^2})^i=r(e^{i\theta})^i=re^{i^2\theta}=re^{-\theta}$

$$where,\ r=\frac{1+z^2}{1-z^2} \ and\ \tan\theta=\frac{2z}{1-z^2} $$