Let z and w be two complex numbers such that | vedclass

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Question

$\omega=z \bar{z}-2 z+2$

$\left|\frac{z+i}{z-3 i}\right|=1$

$\Rightarrow \quad|z+i|=|z-3 i|$

$\Rightarrow \quad z=x+i, \quad x \in R$

$\om

Vedclass · Accepted Answer

$4$

Vedclass · Answer

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

List-$I$	List-$II$
($P$) $\|z\|^2$ is equal to	($1$) $12$
($Q$) $\|z-\bar{z}\|^2$ is equal to	($2$) $4$
($R$) $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	($3$) $8$
($S$) $\|z+1\|^2$ is equal to	($4$) $10$
	($5$) $7$

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 $\operatorname{Re}(w)$ has minimum value. Then, the minimum value of $n \in N$ for which $w ^{ n }$ is real, is equal to..........

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