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 $\omega =e^{\frac{i2\pi }{3}}$ ). If minimum value of $|z-2-2i||z+2+2i|$ 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

Answer: 5

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