Question

If the area of the triangle on the complex plane formed by the complex numbers z, $z,\omega z,z+\omega z$,is $\frac{\sqrt{3}}{100}$ square units then $|z+\omega z|$ equals

• A. 5
• B. $\frac{1}{5}$
• C. $\left|z\right|$
• D. $\left|\omega z\right|$

Answer 1

The correct option is B $\frac{1}{5}$

The triangle formed is an equilateral triangle.
Hence, the area is
$\frac{\sqrt{3}{a}^2}{4}$
$=\frac{\sqrt{3}}{100}$
Therefore
$\frac{a^2}{4}=\frac{1}{100}$
$a^2=\frac{1}{25}$
$|a|=\frac{1}{5}$
Now length of side of the triangle
$=|z-(z+wz)|$
$=|-\left(wz\right)|$
$=|a|$
$=\frac{1}{5}$
Now $|z+wz|$
$=|z||1+w|$
$=|z||-w^2|$
We know that
$|w^2|=|w|=1$
Hence
$|z||-w^2|$
$=|z||w|$
$=|zw|$
$=\frac{1}{5}$
$\Rightarrow |z+wz|=\frac{1}{5}$
Hence, option 'B' is correct.