First slide

Geometry of complex numbers

Question

A regular hexagon is drawn with two of its vertices forming a shorter diagonal at $z = - 2$ and $z = 1 - i \sqrt{3}$ . The other four vertices are:

Moderate

A

$\pm 2 \sqrt{3}, \pm i$
B

$\pm \sqrt{3}, \pm i$
C

$\sqrt{3}, \sqrt{3} \pm i, - 1 - i \sqrt{3}$
D

None of these

Solution

$A C = 2 \sqrt{3} = 2 r \sin 60^{0} = 2$
$| O A | = 2 \Rightarrow O$ is origin.
$∴$ $z_{4} = 2$
$\frac{z_{2} - 0}{- 2 - 0} = e^{- π / 3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$
$\Rightarrow$ $z_{2} = - 1 - i \sqrt{3}$
$∴$ $z_{5} = 1 + i \sqrt{3}$
Also $\frac{- 2 - 0}{z_{6} - 0} = e^{i π / 3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$
$\Rightarrow$ $z_{6} = \frac{- 4}{1 + i \sqrt{3}} = \frac{- 4 (1 - i \sqrt{3})}{3} = - 1 + i \sqrt{3}$
$∴$ Remaining vertices are
$2, - i + i \sqrt{3}, 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

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