Task B-2.1. (20 points) For complex numbers $z, w$ such that $|z|=|w|=|z-w|$, calculate $\left(\frac{z}{w}\right)^{99}$.
First solution. Let $u=\frac{z}{w}=x+y i$. Then $|u|=1,|u-1|=1$.

Then $x^{2}+y^{2}=1,(x-1)^{2}+y^{2}=1$.

By subtracting these equations, we get $x=\frac{1}{2}, y= \pm \frac{\sqrt{3}}{2}$, so $u=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$.

Now,

$$
\left(\frac{z}{w}\right)^{99}=u^{99}=\left[\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right)^{3}\right]^{33}=\left(\frac{1}{8} \pm \frac{3 \sqrt{3}}{8} i-\frac{9}{8} \mp \frac{3 \sqrt{3}}{8} i\right)^{33}=(-1)^{33}=-1
$$