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

Question

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

Accepted Answer

From $2+k^2z=kz(k+\bar z)=k^2z+kz\bar z$, and using $z\bar z=|z|^2=1$:
$$2+k^2z = k^2z+k \implies 2=k.$$

So $k=2$, and the point is $k+ik^2=2+4i$. The circle $|z-(1+2i)|=1$ has centre $(1,2)$ and radius $1$. Distance from $(1,2)$ to $(2,4)$:
$$\sqrt{(2-1)^2+(4-2)^2}=\sqrt{1+4}=\sqrt5.$$

Maximum distance from the point to the circle $=\sqrt5+1$.

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