Question

The locus of z which satisfies the condition, $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$, is

• A. A straight line
• B. A part of the circle
• C. A parabola
• D. An ellipse

Answer 1

The correct option is B A part of the circle

Let $z=x+iy$ and $\alpha =\frac{z-1}{z+1}$

Therefore we get
$\alpha =\frac{(x-1)+iy}{(x+1)+iy}$

$=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{\left(\right(x+1)-iy)}{\left(\right(x+1)-iy)}$
$=\frac{(x^2+y^2-1)+i(xy+y-xy+y)}{(x+1{)}^2+y^2}$
$=\frac{(x^2+y^2-1)+i2y}{(x+1{)}^2+y^2}$
Also, $|\alpha {|}^2={\left(\frac{|z-1|}{|z+1|}\right)}^2$
$=\frac{(x-1{)}^2+y^2}{(x+1{)}^2+y^2}$
Now, $arg\left(\alpha \right)=\frac{\pi }{3}$

$\therefore {\tan }^{-1}\left(\frac{\frac{2y}{(x+1{)}^2+y^2}}{\frac{x^2+y^2-1}{(x+1{)}^2+y^2}}\right)=\frac{\pi }{3}$

$\Rightarrow \left(\frac{2y}{x^2+y^2-1}\right)=\sqrt{3}$

$\Rightarrow 3\left[\right(x^2+y^2-1{)}^2]=(2y{)}^2$

$\Rightarrow x^2+y^2-\frac{2\sqrt{3}y}{3}-1=0$
Hence, it represents a part of a circle.