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

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 $ $x^{2}+y^{2}-3x+2=y\left(\right.where,y>0\left.\right)$
$\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 $ $a+b=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$