First slide

Geometry of complex numbers

Question

If $z = x + i y$ and $w = \frac{1 - i z}{z - i},$ then $| w | = 1$ implies, that, in the complex plane

Moderate

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

Solution

$| w | = 1 \Rightarrow |\frac{1 - i z}{z - i}| = 1$
$\begin{array}{lr} \Rightarrow |- i^{2} - i z| = | z - i | \\ \Rightarrow | z + i | = | z - i | \end{array} \Rightarrow | (- i) (z + i) | = | z - i |$
$\Rightarrow z$ lies on the perpendicular bisector of the segment join-
$i n g i$ and $- i$
$\Rightarrow z$ lies on the real axis.

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