Question

If $P$ and $Q$ are respectively by the complex numbers $z_1$ and $z_2$ such that $|\frac{1}{z_1}+\frac{1}{z_2}|=|\frac{1}{z_1}-\frac{1}{z_2}|$, then the circumcentre of $△OPQ$ (where O is the origin) is

• A. $\frac{z_1-z_2}{2}$
• B. $\frac{z_1+z_2}{2}$
• C. $\frac{z_1+z_2}{3}$
• D. $z_1+z_2$

Answer 1

The correct option is B $\frac{z_1+z_2}{2}$

$z_1=(a,b)z_2=(c,d)$

$|\frac{1}{z_1}+\frac{1}{z_2}|=|\frac{1}{z_1}-\frac{1}{z_2}|$

$\Rightarrow |z_2+z_1{|}^2=|z^2-z_1{|}^2$

$\Rightarrow (c+a{)}^2+(b+d{)}^2=(c-a{)}^2+(d-b{)}^2$

$\Rightarrow a^2+2ac+b^2+2bd+c^2+d^2=a^2-2ac+b^2-2bd+c^2+d^2$

$4ac+4bd=0$

$ac+bd=0,OPQ$ is a right $△$ because

In right $△$ circumcentre is the mid pt of the hypotenuse is $PQ$,

mid pt is $(z_1+z_2)/2$