Find all complex numbers z that satisfy the conditions…

Find all complex numbers $z$ that satisfy the conditions: $$ 1^{\circ}\ \ \bar z=z^2; \qquad 2^{\circ}\ \ \bar z=z^3. $$

Explanation of Method

Taking the modulus of $\bar z = z^n$ gives $|z|=|z|^n$, which forces $|z|=0$ or $|z|=1$. The case $|z|=0$ always gives the trivial solution $z=0$. For $|z|=1$, the key simplification is $\bar z = 1/z$ on the unit circle, so $\bar z=z^n$ becomes $1/z=z^n$, i.e. $z^{n+1}=1$ — the nonzero solutions are exactly the $(n+1)$-th roots of unity.

Part 1° · $\bar z = z^2$

Step 1 · Take moduli

$$ |\bar z|=|z^2| \quad\Longrightarrow\quad |z|=|z|^2 \quad\Longrightarrow\quad |z|(|z|-1)=0 $$

so $|z|=0$ or $|z|=1$.

Step 2 · Case $|z|=0$

$z=0$: check $\bar0=0=0^2$. ✓ A solution.

Step 3 · Case $|z|=1$

Write $z=e^{i\theta}$, so $\bar z=e^{-i\theta}$ and $z^2=e^{i2\theta}$. The equation becomes:

$$ e^{-i\theta}=e^{i2\theta} \quad\Longrightarrow\quad -\theta \equiv 2\theta \pmod{2\pi} \quad\Longrightarrow\quad 3\theta = 2\pi k,\ k\in\mathbb Z $$

For $\theta\in[0,2\pi)$, this gives three distinct values: $\theta = 0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}$.

Step 4 · List the solutions

$$ z=1, \qquad z=e^{i2\pi/3}=-\frac12+i\frac{\sqrt3}{2}, \qquad z=e^{i4\pi/3}=-\frac12-i\frac{\sqrt3}{2} $$

plus $z=0$ from Step 2. These nonzero solutions are exactly the cube roots of unity ($z^3=1$), consistent with the shortcut $1/z=z^2\Rightarrow z^3=1$.

Step 5 · Verify

For $z=-\frac12+i\frac{\sqrt3}{2}$: $\bar z=-\frac12-i\frac{\sqrt3}{2}$, and $z^2=e^{i4\pi/3}=-\frac12-i\frac{\sqrt3}{2}$. ✓

Step 6 · Cross-check via Cartesian form

Let $z=x+iy$. Then $\bar z=z^2$ becomes $x-iy=(x^2-y^2)+2ixy$, giving $x=x^2-y^2$ and $-y=2xy$. The second equation factors as $y(2x+1)=0$: either $y=0$ (giving $x=x^2 \Rightarrow x=0,1$) or $x=-\frac12$ (giving $y^2=\frac34 \Rightarrow y=\pm\frac{\sqrt3}{2}$) — reproducing exactly the same four solutions.

Part 2° · $\bar z = z^3$

Step 1 · Take moduli

$$ |z|=|z|^3 \quad\Longrightarrow\quad |z|(|z|^2-1)=0 \quad\Longrightarrow\quad |z|=0 \text{ or } |z|=1 $$

Step 2 · Case $|z|=1$

With $z=e^{i\theta}$: $$ e^{-i\theta}=e^{i3\theta} \quad\Longrightarrow\quad 4\theta = 2\pi k \quad\Longrightarrow\quad \theta = \frac{\pi k}{2} $$

For $\theta\in[0,2\pi)$: $\theta=0,\ \dfrac\pi2,\ \pi,\ \dfrac{3\pi}{2}$.

Step 3 · List the solutions

$$ z=1,\ i,\ -1,\ -i \qquad \text{plus } z=0 $$

These are exactly the fourth roots of unity ($z^4=1$), matching the shortcut $1/z=z^3\Rightarrow z^4=1$.

Step 4 · Verify each

$z=i$: $\bar z=-i$, $z^3=i^3=-i$. ✓ $z=-1$: $\bar z=-1$, $z^3=-1$. ✓ $z=-i$: $\bar z=i$, $z^3=(-i)^3=i$. ✓

Left: solutions to $\bar z=z^2$ (origin + cube roots of unity). Right: solutions to $\bar z=z^3$ (origin + fourth roots of unity).

Result

$$ \begin{aligned} 1^{\circ}\ \bar z=z^2: &\quad z \in \left\{0,\ 1,\ -\tfrac12+i\tfrac{\sqrt3}{2},\ -\tfrac12-i\tfrac{\sqrt3}{2}\right\} \\ 2^{\circ}\ \bar z=z^3: &\quad z \in \{0,\ 1,\ i,\ -1,\ -i\} \end{aligned} $$

In general, for $\bar z = z^n$ ($n\ge2$), the solutions are $z=0$ together with the $(n+1)$-th roots of unity.