If one vertex of a triangle having maximum area that can be inscribed in the circle $|z - i| = 5$ is $3 - 3 i$ then the other vertices of a triangle are

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$- \frac{1}{2} \pm \frac{\sqrt{3} i}{2}$

$- \frac{1}{2} \pm \frac{i}{2}$

None

answer is D.

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Detailed Solution

$\frac{z - i}{3 - 3 i - i} = \frac{| z - i |}{| 3 - 3 i - i |} e^{i \frac{2 π}{3}}$

$= \frac{R}{R} e^{\frac{i 2 π}{3}}$

$= \cos \frac{2 π}{3} + i \sin \frac{2 π}{3}$

$z - i = (3 - 4 i) (- \frac{1}{2} \pm i \frac{\sqrt{3}}{2})$

$z = i + (3 - 4 i) (- \frac{1}{2} \pm i \frac{\sqrt{3}}{2})$

$\begin{aligned} Z = i + (3 - 4 i) (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) \\ = i - \frac{3}{2} + i \frac{3 \sqrt{3}}{2} + 2 i + 2 \sqrt{3} \\ Z = - \frac{3}{2} + 2 \sqrt{3} + i (3 + \frac{3 \sqrt{3}}{2}) \end{aligned}$

$\begin{aligned} z = i + (3 - 4 i) (- \frac{1}{2} - i \frac{\sqrt{3}}{2}) \\ = i - \frac{3}{2} - i \frac{3 \sqrt{3}}{2} + 2 i - 2 \sqrt{3} \\ = - \frac{3}{2} - 2 \sqrt{3} + i (3 - \frac{3 \sqrt{3}}{2}) \end{aligned}$