Solution to complex equation with conjugate squared

Question

Say we have the equation

$$(\overline{z})^2=1$$

How do we solve this for $z$

Answer 1

Since $(\overline{z})^2=\overline{(z^2)}$, it is easy to see that

$$(\overline{z})^2=1$$ $$\iff\overline{(z^2)}=1$$ $$\iff z^2 = 1$$ $$\iff z=\pm 1$$

It is true in general that if $P$ is a polynomial with real coefficients, then $\overline{P(z)}=P(\overline{z})$ for all $z$. This is false if $P$ has a nonreal coefficient.

Answer 2

let $$z=x+iy$$ then we have $$(x-iy)^2=1$$ or $$x^2-y^2-1-2xyi=0$$ so we get $$x^2-y^2=1$$ and $$2xy=0$$ can you solve this?

Answer 3

Hint. The given equation is equivalent to $$(\overline{z})^2-1=(\overline{z}-1)(\overline{z}+1)=\overline{(z-1)(z+1)}=0.$$ Can you take it from here?