Question

Let $z$ and $w$ be two complex numbers such that$w=z \bar{z}-2 z+2,\left|\frac{z+i}{z-3 i}\right|=1 $ and $\quad \operatorname{Re}(w)$ has minimum value. Then, the minimum value of $n \in N$ for which $w ^{ n }$ is real, is equal to ________.

Answer 1

$\omega=z \bar{z}-2 z+2$
$\left|\frac{ z + i }{ z -3 i }\right|=1$
$\Rightarrow |z+i|=|z-3 i|$
$\Rightarrow \quad z = x + i , \quad x \in R$
$\omega=( x + i )( x - i )-2( x + i )+2$
$=x^{2}+1-2 x-2 i+2$
$\operatorname{Re}(\omega)=x^{2}-2 x+3$
For min $(\operatorname{Re}(\omega)), x=1$
$\Rightarrow \omega=2-2 i =2(1- i )=2 \sqrt{2} e ^{- i \frac{\pi}{4}}$
$\omega^{n}=(2 \sqrt{2})^{n} e^{-1 \frac{\pi \pi}{4}}$
For real & minimum value of $n$, $n =4$