Question

If $\mid z-iRe\left(z\right)\mid =\mid z-Im\left(z\right)\mid ,$ then z lies on

• A. Re(z) = 2
• B. Im(z) = 2
• C. Re(z) + Im(z)= 2
• D. None of these

Answer 1

The correct option is D None of these

Let $z=x+iy,$ then $\mid z-iRe\left(z\right)\mid =\mid z-Im\left(z\right)\mid$
Implies $\mid x+iy-ix\mid =\mid x+iy-y\mid$
$\Rightarrow x^2+(y-x{)}^2=(x-y{)}^2+y^{2}or{x}^2=y^{2}orx=\pm y.$
Thus,z lies on $Re\left(z\right)=\pm Im\left(z\right).$