The number of complex numbers z, satisfying |z|=1 and |zz+zz|=1, is :

Question

The number of complex numbers $z$, satisfying $|z|=1$ and $|\frac{z}{\vec{z}}+\frac{\vec{z}}{z}|=1$, is :

Accepted Answer

$\begin{aligned} & |z|=1 \ & \left|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}ight|=1 \ & \Rightarrow\left|z^2+(\bar{z})^2ight|=1 \ & 	ext { Let } z=x+i y \ & \Rightarrow\left|(x+i y)^2+(x-i y)^2ight|=1 \ & \Rightarrow\left|2 x^2-2 y^2ight|=1 \ & \Rightarrow\left|x^2-y^2ight|=\frac{1}{2} \ & \Rightarrow x^2-y^2=\frac{ \pm 1}{2} \ & 	ext { and } x^2+y^2=1\end{aligned}$

Case I: $x^2-y^2=\frac{1}{2}$

Case II: $x^2-y^2=-\frac{1}{2}$

Hence, we get 8 complex numbers.

Solutions · 1