Question

Suppose $z_1,z_2,z_3$ are the vertices of an equilateral triangle inscribed in the circle | z | = 2 . If $z_1=1+i\sqrt{3}$, then $z_2=....z_3=...$.

Answer 1

Given $|z|=2$

$z_1=1+i\sqrt{3}$

It can be written as $z_1=2(\frac{a}{2}+i{\sqrt{3}}_{/2})$

$=2e^{\pi i_{/3}}$

Equilateral ${\Delta }^4$ means rotating $2{\pi }_{/3}$ about centre.

$z_2=z_1{e}^{2\pi i_{/3}}=2e^{\pi i}=2(\cos \pi +i\sin \pi )=-2$

$z_3=z_1{e}^{4\pi i_{/3}}$ $=2e^{5i{\pi }_{/3}}=$ $2(\cos 5{\pi }_{/3}+i\sin 5{\pi }_{/3})$

$=2(\frac{1}{2}-i{\sqrt{3}}_{/2})$

$=1-i\sqrt{3}$.

$\therefore z_2=-2,z_3=1-i\sqrt{3}$