Question

If $z=x+iy$ is a complex number satisfying $\left|z-i\mathrm{Re}\left(z\right)\right|=\left|z-\mathrm{Im}\left(z\right)\right|$ then $z$ lies on

• A. $y=x$
• B. $y=-x$
• C. $y=x+1$
• D. $y=-x+1$

Answer 1

Answer: (A), (B)

$y=-x$

Explanation for the correct option :

Find the equation of the line in which $z$ lies on.

It is given that $z=x+iy$ is a complex number. The real part of $z$ is $\mathrm{Re}\left(z\right)=x$ and its imaginary part is $\mathrm{Im}\left(z\right)=y$.

Now in the equation $\left|z-i\mathrm{Re}\left(z\right)\right|=\left|z-\mathrm{Im}\left(z\right)\right|$, substitute $x+iy$ for $z$, $x$ for $\mathrm{Re}\left(z\right)$, and $y$ for $\mathrm{Im}\left(z\right)$.

$$\begin{gathered} \left|z-i\mathrm{Re}\left(z\right)\right|=\left|z-\mathrm{Im}\left(z\right)\right| \\ \Rightarrow \left|x+iy-ix\right|=\left|x+iy-y\right| \\ \Rightarrow \left|x+(y-x)i\right|=\left|(x-y)+iy\right| \\ \Rightarrow \sqrt{x^2+{\left(y-x\right)}^2}=\sqrt{{\left(x-y\right)}^2+y^2}\left[\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}\mathbf{}\mathbf{;}\mathbf{}\mathbf{\left|}\mathbf{z}\mathbf{\right|}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\right] \end{gathered}$$

Now square both sides and simplify the equation.

$$\begin{gathered} {\left(\sqrt{x^2+{\left(y-x\right)}^2}\right)}^2={\left(\sqrt{{\left(x-y\right)}^2+y^2}\right)}^2 \\ \Rightarrow x^2+{\left(y-x\right)}^2={\left(x-y\right)}^2+y^2 \\ \Rightarrow x^2+{\left(-\left(x-y\right)\right)}^2={\left(x-y\right)}^2+y^2 \\ \Rightarrow x^2+{\left(x-y\right)}^2={\left(x-y\right)}^2+y^2 \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=\pm y \end{gathered}$$

So the complex number $z$ might lie on the line $y=x$ or the line $y=-x$.

Hence, the correct options are A and B.