Question

Consider a complex number $z$ on the argand plane satisfying $\arg \left(z^2-\omega^2\right)=\frac{\pi}{2}+\arg \left(z^2-\omega\right)$ (where $\omega= e ^{\frac{ i 2 \pi}{3}}$ ). If minimum value of $| z -2-2 i || z +2+2 i |$ is $\left(\frac{\sqrt{ a }-\sqrt{ b }}{2}\right)( a , b \in N )$ then find the value of $\left(\frac{ a + b }{52}\right)$.

Answer 1

$\arg \left(\frac{ z ^2-\omega^2}{ z ^2-\omega}\right)=\frac{\pi}{2}$
$z ^2$ lies on a circle whose centre is $\left(\frac{-1}{2}, 0\right)$ and radius is equal to $\frac{\sqrt{3}}{2}$ units.
$| z -2(1+ i )|| z +2(1+ i )|=\left| z ^2-4(2 i )\right|=\left| z ^2-8 i \right|$
Minimum value of $\left| z ^2-8 i \right|$ is equal to
$\sqrt{\frac{1}{4}+64}-\frac{\sqrt{3}}{2}=\frac{\sqrt{257}-\sqrt{3}}{2} \equiv \frac{\sqrt{a}-\sqrt{b}}{2} $
$\therefore \frac{a+b}{\sqrt{2}}=\frac{257+3}{52}=5$