Question

If the arguments of $\left(1-i\right)\left(\sqrt{3}+i\right)\left(1+\sqrt{3}i\right)$ and $\left(Z-2\right)\left(\bar{Z}-1\right)$ are equal, then the locus of $Z$ is part of a circle with centre $\left(a,b\right)$ . The value of $a+b$ is

Answer 1

Answer: 2

Let, $Z=x+iy$
Now,
$\left(Z-2\right)\left(\bar{Z}-1\right)=Z\bar{Z}-2\bar{Z}-Z+2$
$=x^2+y^2-3x+2+iy$
argument of $\left(1-i\right)\left(\sqrt{3}+i\right)\left(1+\sqrt{3}i\right)=-\frac{\pi }{4}+\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{4}$
Now, since both arguments are equal
$\Rightarrow$ $x^2+y^2-3x+2=y(where,y>0)$
$\Rightarrow$ $x^2+y^2-3x-y+2=0$
Hence, the centre of the circle is $\left(\frac{3}{2},\frac{1}{2}\right)=\left(a,b\right)$
$\Rightarrow$ $a+b=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$