Find all complex numbers z satisfying z^(n-1)=bar z , for a…
Find all complex numbers $z$ satisfying $\ z^{n-1}=\bar z$, for a fixed integer $n\ge2$.
Explanation of Method
Multiply both sides by $z$ to turn $\bar z$ into $|z|^2$, take moduli to pin down $|z|$, and then use $\bar z=1/z$ on the unit circle to reduce the equation to $z^n=1$.
Step 1 · The trivial solution
$z=0$ clearly satisfies $0^{n-1}=\bar0=0$.
Step 2 · Determine $|z|$ for nonzero solutions
Taking the modulus of both sides of $z^{n-1}=\bar z$:
$$ |z|^{n-1}=|\bar z|=|z| $$
For $z\neq0$, divide by $|z|$: $\ |z|^{n-2}=1$, so $|z|=1$ (the unique positive real solution).
Step 3 · Reduce to a roots-of-unity equation
Since $|z|=1$, we have $\bar z=\dfrac1z$. Substituting into $z^{n-1}=\bar z$:
$$ z^{n-1}=\frac1z \quad\Longrightarrow\quad z^n=1 $$
Step 4 · Conclusion
Every nonzero solution is an $n$-th root of unity, and conversely every $n$-th root of unity satisfies $|z|=1$ (so $\bar z=1/z$) and $z^n=1$ (so $z^{n-1}=1/z=\bar z$). Hence:
$$ \boxed{z=0 \quad\text{or}\quad z=e^{2k\pi i/n},\ \ k=0,1,\ldots,n-1} $$
a total of $n+1$ solutions.
Verification, $n=4$
| $k$ | $z$ | $z^{3}$ | $\bar z$ |
|---|---|---|---|
| 0 | $1$ | $1$ | $1$ |
| 1 | $i$ | $-i$ | $-i$ |
| 2 | $-1$ | $-1$ | $-1$ |
| 3 | $-i$ | $i$ | $i$ |
All four checks agree to full numerical precision, confirming the general derivation.
Result
The equation's only nontrivial constraint is that $z$ be an $n$-th root of unity — the modulus condition $|z|=1$ makes conjugation and inversion interchangeable.