Let $z$ be a complex number such that $|z|=1$. If $\frac{2+{\mathrm{k}}^{2}\mathrm{z}}{\mathrm{k}+\overset{¯}{\mathrm{z}}}=\mathrm{kz},\mathrm{k}\in R$, then the maximum distance of $\mathrm{k}+{\mathrm{ik}}^{2}$ from the circle $|\mathrm{z}-(1+2\mathrm{i})|=1$ is: [JEE Main 2025, 2 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) $\sqrt{5}+1$. We are given the equation $\frac{2+{k}^{2}z}{k+\overset{ˉ}{z}}=kz$ and the condition $\mathrm{∣}z\mathrm{∣}=1$. Substitute $\overset{ˉ}{z}=\frac{1}{z}$ into the given equation, we get $\frac{2+{k}^{2}z}{k+\frac{1}{z}}=kz$ $z(2+{k}^{2}z)=kz(kz+1)$ $2z+{k}^{2}{z}^{2}={k}^{2}{z}^{2}+kz$ $2z=kz$ $2=k\text{ }⟹\text{ }k=2$ Substituting $k=2$ into $k+i{k}^{2}$, we get, $2+i({2}^{2})=2+4i$. This corresponds to the point $P(2,4)$. The given circle is $\mathrm{∣}z−(1+2i)\mathrm{∣}=1$. This is a circle with: Center ($C$): $1+2i$, which is the point $(1,2)$. Radius ($r$) : $1$. The distance $d$ between the point $P(2,4)$ and the center of the circle $C(1,2)$ is: $d=\sqrt{(2−1{)}^{2}+(4−2{)}^{2}}=\sqrt{{1}^{2}+{2}^{2}}=\sqrt{1+4}=\sqrt{5}$ Maximum distance $=d+r=\sqrt{5}+1$

Let \(z\) be a complex number such that \(|z|=1\). If \(\frac{2+{\mathrm{k}}^{2}\mathrm{z}}{\mathrm{k}+\overset{¯}{\math…

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