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

Question

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

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \frac{2+k^2 z}{k+\bar{z}}=k z \
& \Rightarrow 2+k^2 z=k^2 z+k \bar{z} \
& \Rightarrow 2+k|z|^2 \quad\left(z \bar{z}=|z|^2,|z|=1\right) \
& \Rightarrow 2=k \
& \therefore k+k^2 i=2+4 i
\end{aligned}\
\&\therefore \quad \text { The maximum distance is }\
\&\begin{aligned}
& =\sqrt{(4-2)+(2-1)^2}+\text { radius } \
& =\sqrt{(2)^2+(1)^2}+1 \
& =\sqrt{5}+1
\end{aligned}
\end{aligned}$$

Solutions · 1