Question

If $z=x+iy$ and $\omega =\frac{1-iz}{z-i}$, then $\left|\omega \right|=1$ implies that, in the complex plane

• A. $z$ lies on the imaginary axis
• B. $z$ lies on the real axis
• C. $z$ lies on the unit circle
• D. None of these

Answer 1

The correct option is B

$z$ lies on the real axis

Explanation for the correct option.

Find the locus of $z$.

In the equation $\left|\omega \right|=1$ substitute the value of $\omega$ as $\omega =\frac{1-iz}{z-i}$ and then substitute $z=x+iy$.

$$\begin{gathered} \left|\frac{1-iz}{z-i}\right|=1 \\ \Rightarrow \frac{\left|1-i\left(x+iy\right)\right|}{\left|x+iy-i\right|}=1 \\ \Rightarrow \left|1-ix-i^{2}y\right|=\left|x+(y-1)i\right| \\ \Rightarrow \left|\left(1+y\right)-ix\right|=\left|x+(y-1)i\right| \\ \Rightarrow \sqrt{{\left(1+y\right)}^2+{\left(-x\right)}^2}=\sqrt{x^2+{(y-1)}^2} \end{gathered}$$

Now square both sides and form the equation.

$$\begin{gathered} {\left(\sqrt{{\left(1+y\right)}^2+{\left(-x\right)}^2}\right)}^2={\left(\sqrt{x^2+{(y-1)}^2}\right)}^2 \\ {\left(1+y\right)}^2+x^2=x^2+{(y-1)}^2 \\ \Rightarrow 1+y^2+2y+x^2=x^2+y^2-2y+1 \\ \Rightarrow 4y=0 \\ \Rightarrow y=0 \end{gathered}$$

So the locus of $z$ is $y=0$ which represents the real axis on the complex plane.

So in the complex plane $z$ lies on the real axis.

Hence, the correct option is B.