Question

The area of the triangle in the Argand diagram formed by the Complex number $\mathrm{z}$, $\mathrm{iz}$ and $\mathrm{z}+\mathrm{iz}$ is

• A. ${\left|z\right|}^2$
• B. $\left(\sqrt{\frac{3}{2}}\right){\left|z\right|}^2$
• C. $\left(\frac{1}{2}\right){\left|z\right|}^2$
• D. $\left(\frac{3}{2}\right){\left|z\right|}^2$

Answer 1

The correct option is C

$\left(\frac{1}{2}\right){\left|z\right|}^2$

Explanation for the correct answer:

Let,

$$\begin{gathered} \mathrm{A}=\mathrm{z} \\ \mathrm{B}=\mathrm{iz} \\ \mathrm{C}=\mathrm{z}+\mathrm{iz}. \end{gathered}$$

$\mathrm{z}=\mathrm{x}+\mathrm{i}\mathrm{y}$

Then the values of $|\mathrm{A}-\mathrm{B}|,|\mathrm{B}-\mathrm{C}|,|\mathrm{C}-\mathrm{A}|$form an isosceles triangle with equal sides, $\mathrm{AC}=\mathrm{BC}$.

Property:

If $ABC$ is isosceles, then if we join$C$ and midpoints (let the mid point be $P$ of $A$ and $B$, it will be perpendicular to AB.

Since, $P$ is the midpoint so,

$$\begin{gathered} \mathrm{P}=\frac{\mathrm{A}+\mathrm{B}}{2} \\ =\frac{\mathrm{z}+\mathrm{iz}}{2} \end{gathered}$$

Applying the property in the area of the triangle equation,

$\text{Area}=\frac{1}{2}\times \mathrm{AB}\times PC$

Finding the area.

$\begin{array}{rcl}A& =& \frac{1}{2}\times \frac{|\mathrm{z}+\mathrm{iz}|}{2}\times |\mathrm{z}-\mathrm{iz}\mid \\ & =& \frac{\left|{\mathrm{z}}^2+{\mathrm{z}}^2\right|}{4}\\ & =& \frac{|\mathrm{z}{|}^2}{2}\end{array}$

Hence, option (C) is the correct answer