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

Answer: (B)

z lies on the real axis

Given $w=\frac{1-iz}{z-i}$ and

$\left|w\right|=1$

$\Rightarrow \left|\frac{1-iz}{z-i}\right|=1$ ...(1)

Substitute $z=x+iy$ in equation (1)
$\Rightarrow \left|\frac{1-i(x+iy)}{(x+iy)-i}\right|=1$
$\Rightarrow |1+y-ix|=|x+i(y-1)|\Rightarrow {(1+y)}^2+x^2=x^2+{(y-1)}^2\Rightarrow y=0$
Therefore z lies on the real axis.
Ans: B