Task A-2.4. (4 points)

Determine all complex numbers $z$ for which

$$
|z|=|z+1|=\left|\frac{1}{z}\right|
$$
## First solution.

From $|z|=\left|\frac{1}{z}\right|$, we get $|z|^{2}=1$, and thus $|z|=1$.

(1 point)

If we set $z=a+bi$, we get $a^{2}+b^{2}=1$.

On the other hand, since $|z+1|=|z|=1$, we have $|a+bi+1|=1$,

or, $(a+1)^{2}+b^{2}=1$, which, after rearranging, gives $a^{2}+2a+b^{2}=0$.

(1 point)

Since $a^{2}+b^{2}=1$, we get $2a=-1$, or $a=-\frac{1}{2}$,

and then it follows that $b= \pm \frac{\sqrt{3}}{2}$. The desired numbers are $\quad z_{1,2}=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$.

(1 point)