Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions

Question

I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that $$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$

has only real solutions

This is what I got till now.

$$\left(\frac{1+i(a+ib)}{1-i(a+ib)}\right)^n = a+ib$$ $$\left(\frac{(1-b)+ia}{(1+b)-ia}\right)^n = a+ib$$

$$\frac{(1-b)+ia}{(1+b)-ia}.\frac{(1+b)+ia}{(1+b)+ia} = \frac{1-b^2+2ia-a^2}{1+2b+b^2+a^2}$$

then

$$\left(\frac{1-b^2-a^2}{1+2b+b^2+a^2}+\frac{2ia}{1+2b+b^2+a^2}\right)^n = a+ib$$ where $$a=0 \text{; & } 1+2b+b^2\neq0$$

Answer 1

Note that $$ |1-iz|^2 - |1+iz|^2 = -2i(z -\bar z) = 4 \operatorname{Im}(z) $$ so that $$ z \in \Bbb R \iff \left | \frac{1+iz}{1-iz}\right|= 1 \, . $$ It follows that $$ \left(\frac{1+iz}{1-iz}\right)^n = a+ib $$ has only real solutions $z$ if and only if $|a+ib|=1$, i.e. if $a^2+b^2=1$.

Answer 2

that is an awesome answer given by Martin R. alternatively, let $$ a+ib = \sqrt{a^2+b^2} e^{i \theta} $$ $c$ be the real solution for satisfied $a,b$. One would obtain $$ \label{eq} \sqrt[n]{a^2+b^2} e^{ \frac{i (\theta + 2k\pi )}{n} } = \frac{1-c^2 +i 2c}{1+c^2} \tag1 $$ note that $$ ( \frac{1-c^2 }{1+c^2} ) ^2 + ( \frac{2c}{1+c^2} ) ^2 = 1 $$ which means $\sqrt[n]{a^2+b^2} =1$ equivalently $$a^2+b^2 =1 $$ And for each $e^{ \frac{i (\theta + 2k\pi )}{n} } $ there exists only one $c \in \mathbb{R}$ satisfy $\eqref{eq}$

Answer 3

Set $$ w=\frac{1+iz}{1-iz} $$ so you can solve for $z$, getting $$ z=i\frac{1-w}{1+w} $$ This is real if and only if $$ i\frac{1-w}{1+w}=-i\frac{1-\bar{w}}{1+\bar{w}} $$ that becomes $$ 1+\bar{w}-w-w\bar{w}=-1+\bar{w}-w+w\bar{w} $$ that is, $w\bar{w}=1$.

Therefore $|a+bi|=1$ is a necessary condition. Now if we write $a+bi$, the equation becomes $$ \frac{1+iz}{1-iz}=u $$ where $u$ is any $n$-th root of $a+bi$ and $|u|=1$. By the same argument as before, the solution is real.