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

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(A) Given $\frac{2+k^2z}{k+\overline{z}}=kz$. Since $|z|=1$,we have $\overline{z} = \frac{1}{z}$.Substituting this,we get $\frac{2+k^2z}{k+\frac{1}{z}

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$\sqrt{5}+1$

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(A) Given $\frac{2+k^2z}{k+\overline{z}}=kz$. Since $|z|=1$,we have $\overline{z} = \frac{1}{z}$.Substituting this,we get $\frac{2+k^2z}{k+\frac{1}{z}} = kz$.$\frac{2+k^2z}{\frac{kz+1}{z}} = kz \implies \frac{z(2+k^2z)}{kz+1} = kz$.Assuming $z 
eq 0$,we have $2+k^2z = k(kz+1) = k^2z+k$.This simplifies to $2=k$,so $k=2$.The point is $k+ik^2 = 2+i(2)^2 = 2+4i$.The circle is $|z-(1+2i)|=1$,which has center $C(1,2)$ and radius $r=1$.The distance from the point $P(2,4)$ to the center $C(1,2)$ is $d = \sqrt{(2-1)^2+(4-2)^2} = \sqrt{1^2+2^2} = \sqrt{5}$.The maximum distance from the point to the circle is $d+r = \sqrt{5}+1$.

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

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