Question

Let $z$ be a complex number satisfying $|z-5i|\le 1$ such that amp $z$ is minimum. Prove that
$z=\frac{2\sqrt{6}}{5}+\frac{24i}{5}.$

Answer 1

$|z-5i|\le 1$ represent all points lying inside and on

the circle centred at $(0,5)$ and of radius $1$. Clearly

the point $A$ has minimum amplitude $\theta =\angle

AOX=$\angle ACO$

${OA}^2=25-1=24$ i.e., $AO=2\sqrt{6}$.

$\therefore \cos \theta =\frac{1}{5}\Rightarrow \sin \theta =\frac{2\sqrt{6}}{5}$

$x=OA\cos \theta =2\sqrt{6}.\frac{1}{5},$

$y=OA\sin \theta =2\sqrt{6}.\frac{2\sqrt{6}}{5},$

$\therefore A$ is $z=x+iy=\frac{2\sqrt{6}}{5}$$(1+2\sqrt{6}i)$