Solve $(1+iz)^n=m(1-iz)^n$

Question

So, the problem is like this : We have $n\in\mathbb{N}$. " The set of values of $m\in\mathbb{C}$ for which the equation $(1+iz)^n=m(1-iz)^n$ has all the roots real is :

a)$m\in\mathbb{R}$

b) $ m\in[-1,1]$

c) $ m\in ${-1,1}$

d) $m\in ${-1,1,-i,i}$

e) $m\in ${$z\in\mathbb{C}$;|z|=1} $

f) $m\in ${$z\in\mathbb{C} ;|z|\leq 1$}$

I have no idea how to start . Any hints will be helpful . I know i should post my attempt as well but i don't know where to start .

Answer 1

The entire equation can be rearranged as:

$\dfrac{1+iz}{1-iz}=\sqrt[n]{m}$

The numerator and denominator are conjugates i.e their product is a positive real number. If you check they have the same absolute value.

Using polar co-ordinates to work out the division:

$\dfrac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}$ where $\theta=\tan^{-1} z$

This means that if $m$ also has the form $re^{i\theta}$, then $r=1$ since $\sqrt[n]{r}$ has to match that on the other side which is $1$.

Therefore the set of $m$ that satisfy this restriction have $|m|=1$. Thus in my justification I'd go for (e)

Answer 2

Hint: The equation rearranges to $$\Big(\frac{2i}{i+z}-1\Big)^n=m,$$ so the solutions are $$z=i\,\frac{1-\zeta^k m^{1/n}}{1+\zeta^k m^{1/n}}$$ for $k=1,\dots n$, where $\zeta$ is a primitive $n$th root of unity. Can you study this expression for the different cases?

Answer 3

For real $z,$ let $z=\tan t$ where $t$ is real

$$\dfrac{1+iz}{1-iz}=\dfrac{1+i\tan t}{1-i\tan t}=\cos2t+i\sin2t$$

$$m=(\cos2t+i\sin2t)^n=\cos2nt+i\sin2nt$$

$$|m|=?$$

Answer 4

Hint: a real root also satisfies $(1-iz)^n=m^\ast(1+iz)^n$ by conjugation, so $(1-|m|^2)(1+iz)^n=0$. See if you can now prove $|m|=1$ is necessary and sufficient.