Complex Numbers Ques 91
- If $z$ and $w$ are two complex numbers such that $|z w|=1$ and $\arg (z)-\arg (w)=\frac{\pi}{2}$, then
(2019 Main, 10 April II)
(a) $\bar{z} w=-i$
(b) $z \bar{w}=\frac{1-i}{\sqrt{2}}$
(c) $\bar{z} w=i$
(d) $z \bar{w}=\frac{-1+i}{\sqrt{2}}$
Show Answer
Answer:
Correct Answer: 91.(a)
Solution:
Formula:
- It is given that, there are two complex numbers $z$ and $w$, such that $|z w|=1$ and $\arg (z)-\arg (w)=\pi / 2$ $\therefore \quad|z||w|=1$ $\left[\because\left|z _1 z _2\right|=\left|z _1\right|\left|z _2\right|\right]$ and $\arg (z)=\frac{\pi}{2}+\arg (w)$
Let $|z|=r$, then $|w|=\frac{1}{r}$
and let $\arg (w)=\theta$, then $\arg (z)=\frac{\pi}{2}+\theta$
So, we can assume
$$ z=r e^{i(\pi / 2+\theta)} $$
$[\because$ if $z=x+i y$ is a complex number, then it can be written as $z=r e^{i \theta}$ where, $\left.r=|z| \operatorname{and} \theta=\arg (z)\right]$
$$ \text { and } \quad w=\frac{1}{r} e^{i \theta} $$
Now, $\quad \bar{z} \cdot w=r e^{-i(\pi / 2+\theta)} \cdot \frac{1}{r} e^{i \theta}$
$=e^{i(-\pi / 2-\theta+\theta)}=e^{-i(\pi / 2)}=-i \quad\left[\because e^{-i \theta}=\cos \theta-i \sin \theta\right]$
and
$$ \begin{aligned} & z \bar{w}=r e^{i(\pi / 2+\theta)} \cdot \frac{1}{r} e^{-i \theta} \\ & \quad=e^{i(\pi / 2+\theta-\theta)}=e^{i(\pi / 2)}=i \end{aligned} $$
BETA
