Question

ABCD is a rhombus in the Argand plane. If the affixes of the vertices be $z_1,z_2,z_3,z_4$ and taken in anti-clockwise sense and $\angle CBA=\pi /3$, Find $2z_2=$

• A. $z_1(1-i\sqrt{2})+z_3(1+i\sqrt{2})$
• B. $z_1(1+i\sqrt{3})+z_3(1-i\sqrt{3})$
• C. $z_1(1+i\sqrt{2})+z_3(1-i\sqrt{2})$
• D. $z_1(1-i\sqrt{3})+z_3(1+i\sqrt{3})$

Answer 1

The correct option is B $z_1(1+i\sqrt{3})+z_3(1-i\sqrt{3})$

We know that ABCD is a rhombus in Argand plane.

The affixes of the vertices are $z_1,z_2,z_3$ and $z_3$

$\angle CBA=\frac{\pi }{3},$ $2z_2=?$

$\therefore ABC$ is an equilateral triangle.

$\Rightarrow |t_3-t_2|=|t_2-t_1|=|t_3-t_1|$

By using rotation of complex number.

$\Rightarrow t_2-t_1=|t_3-t_1|\frac{(t_3-t_1)}{|t_3-t_1|}$

$\Rightarrow t_2=t_1=\frac{(1+i\sqrt{3})}{2}+t_3\frac{(1-i\sqrt{3})}{2}$

$\therefore 2t_2=t_1(1+i\sqrt{3})+t_3(1-i\sqrt{3})$

$\Rightarrow t=z$

$\therefore 2z_2=z_1(1+i\sqrt{3})+z_3(1-i\sqrt{3})$

Hence, the answer is $z_1(1+i\sqrt{3})+z_3(1-i\sqrt{3}).$