Question
If $${\text{amp}}\frac{{z - 1}}{{z + 1}} = \frac{\pi }{3}$$ then $$z$$ represents a point on
A.
a straight line
B.
a circle
C.
a pair of lines
D.
None of these
Answer :
a circle
Solution :
$$\eqalign{
& {\text{If }}\left| {\frac{{z - 1}}{{z + 1}}} \right| = r\,{\text{then }}\frac{{z - 1}}{{z + 1}} = r\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right) = r\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right) \cr
& {\text{or, }}\frac{{\left( {x - 1} \right) + iy}}{{\left( {x + 1} \right) + iy}} = \frac{r}{2} + i\frac{{r\sqrt 3 }}{2} \cr
& {\text{or, }}\left( {x - 1} \right) + iy = \frac{r}{2}\left( {x + 1} \right) - \frac{{yr\sqrt 3 }}{2} + i\left\{ {\frac{{ry}}{2} + \frac{{r\sqrt 3 }}{2}\left( {x + 1} \right)} \right\} \cr} $$
\[\left. \begin{array}{l}
\Rightarrow \,\,x - 1 = \frac{r}{2}\left( {x + 1} \right) - \frac{{yr\sqrt 3 }}{2}\\
\,\,\,\,\,\,\,y = \frac{{ry}}{2} + \frac{{r\sqrt 3 }}{2}\left( {x + 1} \right)
\end{array} \right\}\]
$$ \Rightarrow \,\,\frac{{x - 1}}{y} = \frac{{x + 1 - y\sqrt 3 }}{{y + \sqrt 3 \left( {x + 1} \right)}}.$$
On simplification, $$\sqrt 3 \left( {{x^2} + {y^2}} \right) - 2y - \sqrt 3 = 0,$$ which is a circle.