Question

If $z_1=a+ib$ and $z_2=c+id$ are complex numbers such tat $\left|z_1\right|=\left|z_2\right|=1$ & $Re\left(z_1\overline{z_2}\right)=0$, then the pair of complex numbers $w_1=a+ic$ and $w_2=b+id$ satisfies -

• A. $\left|w_1\right|=1$
• B. $\left|w_2\right|=1$
• C. $Re\left(w_1\overline{w_2}\right)=0$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\left|w_1\right|=1$
B $\left|w_2\right|=1$
C $Re\left(w_1\overline{w_2}\right)=0$

Let, $z_1=a+ib$ and $z_2=c+id$ such that $|z_1|=1\Rightarrow \sqrt{a^2+b^2}=1\Rightarrow a^2+b^2=1$......(1)

and $|z_2|=1\Rightarrow \sqrt{c^2+d^2}=1\Rightarrow c^2+d^2=1$......(2).

Also, $Re\left(z_1\overline{z_2}\right)=0$

or, $Re\{ac+bd+i(bc-ad\left)\right\}=0$

or, $ac+bd=0$

or, $a=-\frac{bd}{c}$.....(3).

Using (3) in (1) we get,

$\frac{b^2{d}^2}{c^2}$ $+b^2=1$

or, $\frac{b^2(d^2+c^2)}{c^2}=1$

or, $b^2=c^2$ [Using (2)]........(4).

Similarly using (3) in (2) and using (1) we shall have

$a^2=d^2$.........(5).

Now, according to the problem $w_1=a+ic$ and $w_2=b+id$.

$\therefore |w_1|=\sqrt{a^2+c^2}=\sqrt{a^2+b^2}=1$ [Using (4)].

$\therefore |w_2|=\sqrt{b^2+d^2}=\sqrt{b^2+a^2}=1$ [Using (5)].

Now, $w_1\overline{w_2}=(a+ic)(b-id)=(ab+cd)+i(bc-ad)$.

$\therefore Re\left(w_1\overline{w_2}\right)=ab+cd=-\frac{b^{2}d}{c}$ $+cd=\frac{d(c^2-b^2)}{c}$ $=0$ [Using (3) and (4)].

So, option (A), (B) and (C) are true.