Find the minimum value of $|z_1−z_2|$ of $e^{ia}z^2-(i+3e ^{3ia}) z+3ie^{2ia}=0$

Question

Numbers $z_1$ and $z_2$ - solutions of the equation $$e^{ia}z^2-(i+3e ^{3ia}) z+3ie^{2ia}=0$$ with the real parameter a . Find the minimum value of $|z_1−z_2|$

My answe: find $D =(i+3e^{3ia})^2 -4e^{ia}3ie^{2ia}= (3e^{3ia}-i)^2 $

$z_2=(i+3e^{3ia} + 3e^{3ia}-i)/e^{ia}z^2 = 3e^{2ia} $

$z_2=(i+3e^{3ia} - 3e^{3ia}-i)/e^{ia}z^2 =ie^{-ia} $

$|z_1−z_2| = (sin(a)+3sin(2a)+3isin(2a)+cos(a))^{1/2}$

min of $|z_1−z_2|=2$

Answer 1

Note,

$$|z_1−z_2|^2=(z_1−z_2)(\bar{z}_1−\bar{z}_2)$$ $$=\frac{i+3e^{3ia}}{e^{ia}}\frac{-i+3e^{-3ia}}{e^{-ia}}=10-3i(e^{3ia}-e^{-3ia}) =10+6\sin(3a)$$

which has its minimum value at $10-6=4$. Thus, the minimum $|z_1−z_2|$ is 2.