1. **State the problem:** Solve the equation $z^* = z^3$ where $z^*$ denotes the complex conjugate of $z$. 2. **Recall definitions and formulas:** For a complex number $z = x + yi$ where $x,y \in \mathbb{R}$ and $i^2 = -1$, the conjugate is $z^* = x - yi$. 3. **Rewrite the equation:** Substitute $z = x + yi$ and $z^* = x - yi$ into the equation: $$x - yi = (x + yi)^3$$ 4. **Expand the right side:** Use binomial expansion: $$(x + yi)^3 = x^3 + 3x^2(yi) + 3x(yi)^2 + (yi)^3$$ Recall $i^2 = -1$, $i^3 = i^2 \cdot i = -i$. So, $$(x + yi)^3 = x^3 + 3x^2 yi + 3x y^2 i^2 + y^3 i^3 = x^3 + 3x^2 y i - 3x y^2 - y^3 i$$ Group real and imaginary parts: $$= (x^3 - 3x y^2) + i(3x^2 y - y^3)$$ 5. **Equate real and imaginary parts:** Since $x - yi = (x^3 - 3x y^2) + i(3x^2 y - y^3)$, we have Real parts: $$x = x^3 - 3x y^2$$ Imaginary parts: $$-y = 3x^2 y - y^3$$ 6. **Simplify the real part equation:** $$x = x^3 - 3x y^2 \implies 0 = x^3 - 3x y^2 - x = x^3 - x - 3x y^2$$ Factor out $x$: $$x(x^2 - 1 - 3 y^2) = 0$$ 7. **Simplify the imaginary part equation:** $$-y = 3x^2 y - y^3 \implies 0 = 3x^2 y - y^3 + y = y(3x^2 - y^2 + 1)$$ 8. **Analyze cases:** - Case 1: $x = 0$ From the imaginary part: $$y(3(0)^2 - y^2 + 1) = y(- y^2 + 1) = 0$$ So either $y=0$ or $y^2 = 1$ which gives $y = \pm 1$. - Case 2: $x \neq 0$ From real part: $$x^2 - 1 - 3 y^2 = 0 \implies x^2 = 1 + 3 y^2$$ From imaginary part: $$y(3x^2 - y^2 + 1) = 0$$ If $y \neq 0$: $$3x^2 - y^2 + 1 = 0$$ Substitute $x^2$: $$3(1 + 3 y^2) - y^2 + 1 = 0 \implies 3 + 9 y^2 - y^2 + 1 = 0 \implies 4 + 8 y^2 = 0$$ This cannot be true for real $y$ since $8 y^2 \geq 0$ and $4 + 8 y^2 > 0$. So $y=0$. Then from real part: $$x^2 = 1 + 3(0) = 1$$ So $x = \pm 1$. 9. **Summary of solutions:** - From case 1: $z = 0 + 0i = 0$, $z = 0 + i = i$, $z = 0 - i = -i$ - From case 2: $z = 1 + 0i = 1$, $z = -1 + 0i = -1$ 10. **Final answer:** $$\boxed{z \in \{0, \pm 1, \pm i\}}$$