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 $z_2$ and $z_3$ are equal to

• A. $-2,1-i\sqrt{3}$
• B. $2,1-i\sqrt{3}$
• C. $-2,1+i\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $-2,1-i\sqrt{3}$

$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$ lie on $\left|z\right|=2$ whose centre is at $O(0,0)$ and radius $2$
$z_1=1+i\sqrt{3}$ Hence $\left|z\right|=2$ and $Arg\left(z_1\right)=\frac{\pi }{3}$
In turn $|z_2|=|z_3|=2$ and $Arg\left(z_2\right)=Arg\left(z_1\right)+{120}^o={180}^o$
$\therefore z_2=-2$
Further, $Arg\left(z_3\right)=Arg\left(z_2\right)+{120}^o={300}^o$
Hence, $z_3=2[\cos (2\pi -\frac{\pi }{3})+i\sin (2\pi -\frac{\pi }{3})]$
$=2[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]=2(\frac{1}{2}-\frac{i\sqrt{3}}{2})$
$=1-\sqrt{3}i$
Thus, $z_2=-2$ and $z_3=1-i\sqrt{3}$