Question

If the area of triangle formed by by the points z , z + iz and iz on the complex plane is 18, then the value of $\left|z\right|$ is

• A. $6$
• B. $9$
• C. $3\sqrt{2}$
• D. $2\sqrt{3}$

Answer 1

The correct option is A $6$

Given,

$z,iz,z+iz$

$z=a+ib$

$iz=-b+ia$

$z+iz=(a-b)+i(b+a)$

$Area=\frac{1}{2}\left|\begin{array}{cccc}a& -b& a+b& a\\ b& a& b-a& b\end{array}\right|$

$=\frac{1}{2}[a^2+b^2-b^2-ab-a^2+ab+ab-b^2-ab-a^2]$

$\Rightarrow 18=\frac{1}{2}|a^2+b^2|$

$36=|a^2+b^2|$

but, $|z|=\sqrt{a^2+b^2}$

$\Rightarrow \sqrt{36}=\sqrt{a^2+b^2}$

$\therefore |z|=6$