my question is how to solve the equation $z^2 - 2 \bar{z} = 3$, where $z \in \mathbb{C}$?
I know that polar form of any complex number is $z = r \cos(\varphi) + i \cdot r \sin (\varphi)$. And similarly $\bar{z} = r \cos(\varphi) - i \cdot r \sin(\varphi)$. But now, when I plug in into equation, I'm getting a little confused.
So, $(r \cos(\varphi) + i \cdot r \sin(\varphi))^{2} - 2(r \cos(\varphi) - i \cdot r \sin(\varphi)) = 3$, but now I don't know how to solve this equation.
We have $$z^2-2\overline z=3=\overline z^2-2z$$ Therefore \begin{align} &z^2-\overline z^2+2z-2\overline z=0\\ \implies &(z-\overline z)(z+\overline z+2)=0\\ \implies &\overline z = z\qquad\text{or}\qquad\overline z=-2-z \end{align} Putting these one by one in original equation, we get, $\boxed{z=-1,3}$ are the only solutions.
In a problem like this, I prefer Cartesian coordinates, where $z = x + iy.$
$z^2 - 2 \bar{z} = 3$
Therefore $(x + iy)^2 - 2(x - iy) = 3 \implies $
$(x^2 - y^2 - 2x - 3) + i(2xy + 2y) = 0 + i(0)$.
Therefore, $(2y)(x + 1) = 0$.
Therefore, either $y = 0$ or $x = -1$.
$y = 0 \implies (x^2 - 2x - 3) = 0 \implies (x - 3)(x + 1) = 0 \implies x \in \{3, -1\}.$
$x = -1 \implies 3 - y^2 - 3 = 0 \implies y = 0.$
Therefore, the possible answers are $(-1 + i[0], 3 + i[0]).$
