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(C) Given $\left| \frac{z+i}{z-3i} \right| = 1$, we have $|z+i| = |z-3i|$.This represents the perpendicular bisector of the segment joining $-i$ and $

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$4$

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(C) Given $\left| \frac{z+i}{z-3i} ight| = 1$, we have $|z+i| = |z-3i|$.This represents the perpendicular bisector of the segment joining $-i$ and $3i$ on the complex plane, which is the line $\operatorname{Im}(z) = 1$.Let $z = x + i$, where $x \in \mathbb{R}$.Given $w = z \bar{z} - 2z + 2$, substituting $z = x + i$:$w = (x+i)(x-i) - 2(x+i) + 2$$w = (x^2 + 1) - 2x - 2i + 2$$w = (x^2 - 2x + 3) - 2i$.Thus, $\operatorname{Re}(w) = x^2 - 2x + 3$.To find the minimum value of $\operatorname{Re}(w)$, we differentiate with respect to $x$ or use the vertex formula $x = -b/(2a) = 2/2 = 1$.At $x = 1$, $\operatorname{Re}(w) = 1 - 2 + 3 = 2$.Then $w = 2 - 2i = 2(1 - i) = 2\sqrt{2} e^{-i\pi/4}$.For $w^n$ to be real, the argument of $w^n$ must be a multiple of $\pi$.$\operatorname{arg}(w^n) = n 	imes (-\pi/4) = -n\pi/4$.For this to be $k\pi$, we need $n/4$ to be an integer.The minimum $n \in N$ is $4$.

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

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