Let $p$  and $q$  be two complex numbers such that $q\ne 0.$  If the roots of the equation $x^2+px+q^2=0$  have the same absolute value, then $p/q$  is

Let $x_1,x_2$  be the roots of $x^2+px+q^2=0$  and $\left|x_1\right|=\left|x_2\right|=r.$  Note that $x_1{x}_2=q^2\ne 0.$  Now, $\frac{p^2}{q^2}=\frac{{(x_1+x_2)}^2}{x_1{x}_2}=\frac{x_1}{x_2}+\frac{x_2}{x_1}+2$ 

                                         $=\frac{x_1\overline{x_2}}{|x_2{|}^2}+\frac{\overline{x_1}{x}_2}{|x_1{|}^2}+2=\frac{1}{r^2}(x_1\overline{x_2}+\overline{x_1}{x}_2)+2$

                                         $=\frac{2}{r^2}\mathrm{Re}\left(x_1{\overline{x}}_2\right)+2$

$\Rightarrow$  $\frac{p^2}{q^2}$  is a real number.

Next, as $-\mathrm{Re}\left(x_1{\overline{x}}_2\right)\le \left|x_1{\overline{x}}_2\right|=\left|x_1{x}_2\right|=\left|x_1\right|\left|x_2\right|=r^2,$

we get

 $\Rightarrow \frac{p^2}{q^2}\ge -\frac{2r^2}{r^2}+2=-2+2=0$

Therefore, $p/q$  is a real number.