## Task A-2.1.

Determine all complex numbers $z$ for which $z^{2}=\frac{1}{z}+\frac{1}{\bar{z}}$.
## First Solution.

Let $x$ and $y$ be the real and imaginary parts of the number $z$, respectively, i.e., $z = x + i y$. The equation from the problem can then be written as

$$
x^{2} - y^{2} + 2 x y i = \frac{1}{x + i y} + \frac{1}{x - i y} = \frac{x - i y}{x^{2} + y^{2}} + \frac{x + i y}{x^{2} + y^{2}} = \frac{2 x}{x^{2} + y^{2}}
$$

Clearly, the right-hand side is a real number, so the left-hand side must also have an imaginary part equal to zero. This implies $2 x y = 0$, i.e., $x = 0$ or $y = 0$.

If we substitute $x = 0$ into the above expression, the equation becomes $-y^{2} = 0$, so $z = 0 + 0 i = 0$, and for this $z$, the original equation makes no sense.

If $y = 0$, the equation reduces to $x^{2} = \frac{2}{x}$, or $x^{3} = 2$. The only real solution to this equation is $x = \sqrt[3]{2}$.

The solution to the original equation is $z = x + i y = \sqrt[3]{2}$.