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4-1.Complex numbers
hard
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..........
A
$5$
B
$2$
C
$4$
D
$6$
(JEE MAIN-2021)
Solution
$\omega=z \bar{z}-2 z+2$
$\left|\frac{z+i}{z-3 i}\right|=1$
$\Rightarrow \quad|z+i|=|z-3 i|$
$\Rightarrow \quad z=x+i, \quad x \in R$
$\omega=(x+i)(x-i)-2(x+i)+2$
$=x^{2}+1-2 x-2 i+2$
$\operatorname{Re}(\omega)=x^{2}-2 x+3$
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 $n$
$n =4$
Standard 11
Mathematics
