Let m be a nonzero integer and w∈mathbb C . What condition…
Let $m$ be a nonzero integer and $w\in\mathbb C$. What condition on $w$ makes every root $z$ of $\ \left(\dfrac{1+iz}{1-iz}\right)^m=w\ $ real (allowing $z=\infty$ to count as the single non-real exception)?
Explanation of Method
The map $f(z)=\dfrac{1+iz}{1-iz}$ (a Cayley transform) is a Möbius transformation, hence a bijection of the extended real line $\mathbb R\cup\{\infty\}$ onto the unit circle $|\zeta|=1$. Since $f$ is a bijection, the equation $f(z)^m=w$ has exactly $m$ solutions $z_k=f^{-1}(\zeta_k)$, one for each $m$-th root $\zeta_k$ of $w$. All $m$ solutions are real (finite) exactly when all $m$ roots $\zeta_k$ lie on the unit circle and none of them equals $-1$ (the single point of the circle corresponding to $z=\infty$).
Step 1 · $f$ maps $\mathbb R\cup\{\infty\}$ bijectively onto the unit circle
For real $x$: $\ |f(x)|^2=\dfrac{|1+ix|^2}{|1-ix|^2}=\dfrac{1+x^2}{1+x^2}=1$, so $f(x)$ always lies on the unit circle. Writing $f(x)=\dfrac{(1+ix)^2}{1+x^2}=\dfrac{1-x^2+2ix}{1+x^2}$ and comparing to the double-angle formulas with $x=\tan\varphi$ shows:
$$ f(x)=e^{2i\arctan x} $$
As $x$ ranges over $\mathbb R$, $\arctan x$ ranges over $\left(-\frac\pi2,\frac\pi2\right)$, so $2\arctan x$ ranges over $(-\pi,\pi)$ — every angle except $\pi$. Including $x=\infty$: $\ f(\infty)=\lim_{x\to\infty}\frac{1+ix}{1-ix}=\frac i{-i}=-1$, which fills in exactly the missing point. So $f$ is a bijection $\mathbb R\cup\{\infty\}\to\{|\zeta|=1\}$, with $f(\infty)=-1$.
Step 2 · The equation $f(z)^m=w$ has exactly $m$ roots (for $w\neq0$)
Since $f$ is a bijection of the Riemann sphere, setting $\zeta=f(z)$ turns $f(z)^m=w$ into $\zeta^m=w$, which (for $w\neq0$) has exactly $m$ distinct solutions $\zeta_0,\ldots,\zeta_{m-1}$ (the $m$-th roots of $w$). Each corresponds to exactly one $z_k=f^{-1}(\zeta_k)$.
Step 3 · Translate "all roots real" into a condition on $w$
By Step 1, $z_k=f^{-1}(\zeta_k)$ is a finite real number exactly when $\zeta_k$ lies on the unit circle and $\zeta_k\neq-1$; it equals $\infty$ (not real) exactly when $\zeta_k=-1$; and it is genuinely complex (non-real) when $|\zeta_k|\neq1$.
So all $m$ roots are real if and only if:
- every $m$-th root of $w$ lies on the unit circle — since all $m$-th roots of $w$ share modulus $|w|^{1/m}$, this is equivalent to $\ |w|=1$;
- none of the $m$-th roots of $w$ equals $-1$ — this fails exactly when $w=(-1)^m$ (i.e. $-1$ itself is among the roots).
Step 4 · Conclusion
$$ \boxed{|w|=1 \ \text{ and } \ w\neq(-1)^m} $$
equivalently: $w$ lies on the unit circle, excluding $w=1$ when $m$ is even, or $w=-1$ when $m$ is odd.
Verification
Checked numerically for $m=1,2,3,4$: for every tested $w=e^{i\theta}$ with $w\neq(-1)^m$, all $m$ roots $z_k=f^{-1}(\zeta_k)$ came out real to full floating-point precision. For $w=(-1)^m$ exactly (e.g. $m=2,w=1$; $m=4,w=1$; $m=1,w=-1$; $m=3,w=-1$), exactly one of the $m$ roots was the point at infinity — confirming the exclusion is necessary, not just sufficient. Testing $w$ off the unit circle (e.g. $|w|=2$) always produced at least one genuinely complex root.
Result
Since $f(z)=\frac{1+iz}{1-iz}$ bijects the real line (plus $\infty$) onto the unit circle, the roots of $f(z)^m=w$ are all real exactly when every $m$-th root of $w$ stays on that circle and none of them lands on the single "missing" point $\zeta=-1=f(\infty)$.