First slide

Algebra of complex numbers

Question

Let $z$ and $w$ be two complex numbers such that $| z | = | w | = 1$ and $| z + i w | = | z - i \bar{w} | = 2 .$ Then $z$ equals

Moderate

A

$1 o r i$
B

$1 o r - i$
C

$1 o r - 1$
D

$i o r - 1$

Solution

We have $| - i w | = | - i | | w | = 1$
and $| i \bar{w} | = | i | \bar{w} ∣= 1$
$\Rightarrow - i w$ and $i \bar{w}$ lie on the circle $| z | = 1 .$
As $| z - (- i w) | = | z - i \bar{w} | = 2$ we get $z$ $- i w$ , as well
as $z$ and $i w$ are the end points of the same diameter, with one end point at $z .$
$∴ - i w = i \bar{w} \Rightarrow w + \bar{w} = 0$
$\Rightarrow$ $w$ is purely imaginary.
Let $w = i k$ where $k \in R$
As $| w | = 1,$ we get $| i k | = 1$
$\begin{aligned} \Rightarrow | k | = 1 \Rightarrow k = \pm 1 . \\ ∴ w = \pm i \Rightarrow - i w = i \bar{w} = \pm 1 \end{aligned}$
When $i \bar{w} = 1,$ then $z = - 1$ and
when $i \bar{w} = - 1$ then $z = 1$

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