Complex Numbers 2 Question 24
25. If $z _1=a+i b$ and $z _2=c+i d$ are complex numbers such that $\left|z _1\right|=\left|z _2\right|=1$ and $\operatorname{Re}\left(z _1 \bar{z} _2\right)=0$, then the pair of complex numbers $w _1=a+i c$ and $w _2=b+i d$ satisfies
(a) $\left|w _1\right|=1$
(b) $\left|w _2\right|=1$
(c) $\operatorname{Re}\left(w _1 \bar{w} _2\right)=0$
(d) None of these
$(1985,2 M)$
Passage Based Problems
Read the following passages and answer the questions that follow.
Passage I
Let $A, B, C$ be three sets of complex number as defined below
$$ \begin{aligned} & A={z: \operatorname{lm}(z) \geq 1} \\ & B={z:|z-2-i|=3} \\ & C={z: \operatorname{Re}((1-i) z)=\sqrt{2}} \end{aligned} $$
(2008, 12M)
Show Answer
Answer:
Correct Answer: 25. (b)
Solution:
- Since, $z _1=a+i b$ and $z _2=c+i d$
$\Rightarrow\left|z _1\right|^{2}=a^{2}+b^{2}=1$ and $\left|z _2\right|^{2}=c^{2}+d^{2}=1$
$$ \left[\because\left|z _1\right|=\left|z _2\right|=1\right] $$
Also, $\operatorname{Re}\left(z _1 \bar{z} _2\right)=0 \quad \Rightarrow \quad a c+b d=0$
$$ \Rightarrow \quad \frac{a}{b}=-\frac{d}{c}=\lambda $$
From Eqs. (i) and (ii), $b^{2} \lambda^{2}+b^{2}=c^{2}+\lambda^{2} c^{2}$
$\Rightarrow \quad b^{2}=c^{2}$ and $a^{2}=d^{2}$
Also, given $w _1=a+i c$ and $w _2=b+i d$
Now, $\quad\left|w _1\right|=\sqrt{a^{2}+c^{2}}=\sqrt{a^{2}+b^{2}}=1$
$$ \left|w _2\right|=\sqrt{b^{2}+d^{2}}=\sqrt{a^{2}+b^{2}}=1 $$
and $\operatorname{Re}\left(w _1 \overline{w _2}\right)=a b+c d=(b \lambda) b+c(-\lambda c) \quad$ [from Eq. (i)]
$$ =\lambda\left(b^{2}-c^{2}\right)=0 $$
