Complex Numbers Ques 23
If $z=x+i y$ and $w=(1-i z) /(z-i)$, then $|w|=1$ implies that, in the complex plane
(1983, 1M)
(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
Show Answer
Answer:
Correct Answer: 23.(b)
Solution:
Formula:
- Since, $|w|=1 \Rightarrow |\frac{1-i z}{z-i}|=1$
$ \begin{array}{lll} \Rightarrow & |z-i|=|1-i z| \\ \Rightarrow & |z-i|=|z+i| \quad[\because|1-i z|=|-i||z+i|=|z+i|] \end{array} $
$\therefore$ It is a perpendicular bisector of $(0,1)$ and $(0,-1)$ i.e. $X$-axis.
Thus, $z$ lies on the real axis.
BETA
