Let z be a complex number such that |z|=1.

Question

Let $z$ be a complex number such that $|z|=1$. If $\frac{2+{k}^{2}z}{k+\vec{z}}=kz,k\in R$, then the maximum distance of $k+i{k}^{2}$ from the circle $|z-(1+2i)|=1$ is :

Accepted Answer

$\begin{gathered} \frac{2+{k}^{2}z}{k+\vec{z}}=kz \ \Rightarrow 2+{k}^{2}z={k}^{2}z+k\vec{z} \ \Rightarrow 2+k|z{|}^{2}\, (z\vec{z}=|z{|}^{2},|z|=1) \ \Rightarrow 2=k \ 	herefore k+{k}^{2}i=2+4i \ 	herefore \, \ =\sqrt{(4-2)+(2-1{)}^{2}}+ \ =\sqrt{(2{)}^{2}+(1{)}^{2}}+1 \ =\sqrt{5}+1 \end{gathered}$

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