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
$\left(Z - a\right)\left(\bar{Z} - b\right)=Z\bar{Z}-a\bar{Z}-bZ+ab$
$=\left(x^{2} + y^{2}\right)-a\left(x - i y\right)-b\left(x + i y\right)+ab$
$=\left(x^{2} + y^{2}\right)-\left(a + b\right)x+i\left(a - b\right)y+ab$
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 $\left(\frac{a + b}{2} , \frac{a - b}{2}\right)$
$\Rightarrow \frac{a + b}{2}=\frac{3}{2} \, \& \, \frac{a - b}{2}=\frac{1}{2}$
$\Rightarrow a=2, \, b=1\Rightarrow ab=2$