Question

Show that the points representing the complex numbers (3+3i), (-3-3i) and $(-3\sqrt{3}+3\sqrt{3}i)$ on the Argand plane are the vertices of an equilateral triangle.

Answer 1

Let the given numbers be represented on the Argand plane by the point A, B and C respectively. Then,
$AB=|(3+3i)-(-3-3i)|=|6+6i|=\sqrt{(6{)}^2+(6{)}^2}=\sqrt{72}=6\sqrt{2}(d=|z_1|-z_2)$
$BC=|(-3-3i)-(-3\sqrt{3}+3\sqrt{3}i)|=|3\left(\sqrt{3-1}\right)-3(1+\sqrt{3})i|=\sqrt{9(\sqrt{3}-1{)}^2+9(1+\sqrt{3}{)}^2}=\sqrt{9\left[\right(\sqrt{3}-1{)}^2+(\sqrt{3}+1{)}^2]}=\sqrt{9(3+1-2\sqrt{3}+3+1+2\sqrt{3})}$
$=\sqrt{9\times 8}=\sqrt{72}=6\sqrt{2}$
$AC=|(3+3i)-(-3\sqrt{3}+3\sqrt{3}i)|=|3(1+\sqrt{3})+3(1-\sqrt{3})i|=\sqrt{9(1+\sqrt{3}{)}^2+9(1-\sqrt{3}{)}^2}=\sqrt{9(1+\sqrt{3}{)}^2+(1-\sqrt{3}{)}^2}=\sqrt{9\times 8}=\sqrt{72}=6\sqrt{2}$
Thus, AB =BC =AC. Hence, the given points on the Argand plane are the vertices of an equilateral triangle.