Question

If $z_1+z_2+z_3=0$ and $|z_1|=|z_2|=|z_3|=1$, then area of triangle whose vertices are $z_1,z_2$ and $z_3$ is:

• A. $\frac{3\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{4}$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$\frac{3\sqrt{3}}{4}$

${|z_1+z_2|}^2+{|z_1-z_2|}^2=2({|z_1|}^2+{|z_2|}^2)$ ....... $\left(1\right)$

$z_1+z_2+z_3=0$

$z_3=-z_2-z_1$

By taking modulus both sides

$|-z_3|=|z_2+z_1|$ ......... $\left(2\right)$

Using $\left(2\right)$ in $\left(1\right),$ we have

${|-z_3|}^2+|z_1-z_2{|}^2=2(1+1)$

$1+|z_1-z_2{|}^2=2(1+1)$

$|z_1-z_2{|}^2=3$

$|z_1-z_2|=\sqrt{3}$

Similarily $|z_2-z_3|$=$|z_3-z_1|=\sqrt{3}$

Hence sides of triangle formed are $\sqrt{3},$ $\sqrt{3},$ $\sqrt{3}$ and are equal.

So triangle formed is an equilateral triangle

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{s}^2$

where $s$ is side of equilateral triangle

So area of triangle formed by $z_1,z_2,z_3=3\frac{\sqrt{3}}{4}$