Question

For a complex number $Z$ , if the argument of $\left(Z-a\right)\left(\bar{Z}-b\right)$ is $\frac{\pi }{4}$ or $\frac{-3\pi }{4}$ (where $a,$ $b$ are two real numbers), then the value of $ab$ such that the locus of $Z$ represents a circle with centre $\frac{3}{2}+\frac{i}{2}$ is (where, $i^2=-1$ )

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

Let, $Z=x+iy$ , then
$\left(Z-a\right)\left(\bar{Z}-b\right)=Z\bar{Z}-a\bar{Z}-bZ+ab$
$=\left(x^2+y^2\right)-a\left(x-iy\right)-b\left(x+iy\right)+ab$
$=\left(x^2+y^2\right)-\left(a+b\right)x+i\left(a-b\right)y+ab$
If the argument of the above complex number is $\frac{\pi }{4}$ or $-\frac{3\pi }{4}$ , then
$x^2+y^2-\left(a+b\right)x+ab=\left(a-b\right)y$
$\Rightarrow$ Centre of the circle is $\left(\frac{a+b}{2},\frac{a-b}{2}\right)$
$\Rightarrow \frac{a+b}{2}=\frac{3}{2}\&\frac{a-b}{2}=\frac{1}{2}$
$\Rightarrow a=2,b=1\Rightarrow ab=2$