1. **Stating the problem:** We want to find the form of a complex number $z$ if $z$ is equal to the multiplicative inverse of its conjugate $\overline{z}$. That is, $$z = \frac{1}{\overline{z}}.$$ 2. **Recall complex number and conjugate:** Let $z = a + bi$ where $a,b \in \mathbb{R}$ and $i^2 = -1$. Its conjugate is $\overline{z} = a - bi$. 3. **Express the equation:** Given $$z = \frac{1}{\overline{z}},$$ substitute $z$ and $\overline{z}$: $$a + bi = \frac{1}{a - bi}.$$ 4. **Find the inverse of the conjugate:** The reciprocal of $a - bi$ can be found by multiplying numerator and denominator by the conjugate: $$\frac{1}{a - bi} = \frac{a + bi}{(a - bi)(a + bi)} = \frac{a + bi}{a^2 + b^2}.$$ 5. **Set up the equality:** $$a + bi = \frac{a + bi}{a^2 + b^2}.$$ 6. **Multiply both sides by $a^2 + b^2$ to clear denominator:** $$(a + bi)(a^2 + b^2) = a + bi.$$ 7. **Rewrite:** $$ (a^2 + b^2)(a + bi) - (a + bi) = 0$$ $$ (a + bi)((a^2 + b^2) - 1) = 0.$$ 8. **Analyze product:** For the product to be zero, either: - $a + bi = 0$ (i.e., $z=0$), or - $(a^2 + b^2) - 1 = 0$ Since $z=0$ does not have a multiplicative inverse, discard the first case. 9. **Solve the norm equation:** $$a^2 + b^2 = 1.$$ 10. **Interpretation:** This means $z$ lies on the unit circle in the complex plane, i.e. $|z| = 1$. **Final answer:** The complex number $z$ must satisfy $|z|=1$, or equivalently, it lies on the unit circle.