Question

If the complex numbers $z_1,z_2$ and $z_3$ denote the vertices of an isosceles triangle, right angled at $z_1$, then $(z_1-z_2{)}^2+(z_1-z_3{)}^2$ is equal to

• A. $0$
• B. $(z_2+z_3{)}^2$
• C. $2$
• D. $3$
• E. $(z_2-z_3{)}^2$

Answer 1

The correct option is A $0$

Since, the triangleis isosceles and right angled, we can say that

$(z_3-z_1)=(z_2-z_1)(\cos {90}^o+i\sin {90}^o)$

$\Rightarrow (z_3-z_1)=i(z_2-z_1)$

Squaring bothe sides, we get

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

$\Rightarrow (z_3-z_1{)}^2+(z_2-z_1{)}^2=0$