Question

If area of a triangle whose vertices in argand plane are z, iz, z+iz is 16 square units, find |z|

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $4$
• D. $4\sqrt{2}$

Answer 1

The correct option is D $4\sqrt{2}$

As z and iz are mutually perpendicular and z+iz is resultant of z & iz
$\therefore$ Triangle in argand plane is OAC where OA = z, AC = iz
$\therefore Area=\frac{1}{2}|z||iz|=16$
$\Rightarrow |z{|}^2=32\Rightarrow |z|=4\sqrt{2}$