Question

$Z_1=a+ib,Z_2=c+id$ are complex numbers such that $|Z_1|=|Z_2|=1$ and $Re(Z_1.{\overline{Z}}_2)=0$ then the pair of complex numbers $K_1=a+ic,K_2=b+id$ then

• A. $|K_1|=4$
• B. $|K_2|=3$
• C. $Re\left(K_1{K}_2\right)=0$
• D. $Re\left(K_1{K}_2\right)=1$

Answer 1

Answer: (D)

$Re\left(K_1{K}_2\right)=1$

$Re(Z_1.\overline{Z_2})=0$

$Re(Z_1.\overline{Z_2})=Re\left(\right(a+ib\left)\right(c-id\left)\right)=ac+bd=0$

$ac=-bd$ $⟹\frac{a}{b}=-\frac{d}{c}=x$(say)

$\therefore a=bx$ and $c=-\frac{d}{x}$

Given $|Z_1|=1$ $\therefore a^2+b^2=1$

$b^2{x}^2+b^2=1⟹b^2(x^2+1)=1$----------------(1)

Given $|Z_2|=1$ $\therefore c^2+d^2=1$

$\frac{d^2}{x^2}+d^2=1⟹d^2(1+x^2)=x^2$----------------(2)

Dividing (2) by (1) , $\frac{d^2}{b^2}=x^2⟹d^2=b^2{x}^2=a^2$ -----------(3)

Now $a^2+b^2=c^2+d^2$

$=c^2+a^2$

$\therefore b^2=c^2$ ----------(4)

$|K_1|=\sqrt{(a^2+c^2)}$

$a^2+c^2=a^2+b^2=1$ (From equation (4))

$|K_2|=\sqrt{(b^2+d^2)}$

$b^2+d^2=b^2+a^2=1$ (from equation (3))

$\therefore |K_1|=\pm 1$ and $|K_2|=\pm 1$

$Re\left(K_1{K}_2\right)=Re\left(\right(a+ic\left)\right(b+id\left)\right)=\left(ab\right)-\left(cd\right)=ab-(c.(-cx\left)\right)=ab+c^{2}x=ab+b^{2}x$ (from equation (4))

$=ab+b.\left(bx\right)=ab+ba=2ab$

Now we know that $a^2+b^2=1,c^2+d^2=1,a^2+c^2=1$ and $b^2+d^2=1$.

For this, $a^2=b^2=c^2=d^2=\frac{1}{2}$

The ratio between $a$ and $b$ is $x$. So either both $a$ and $b$ has to be positive or both has to be negative. Hence their product is positive.

So $ab=\frac{1}{2}$

$\therefore Re(K_1,K_2)=2ab=2\times \frac{1}{2}=1$