Question
All the points in the set $$S = \left\{ {\frac{{\alpha + i}}{{\alpha - i}}:\alpha \in R} \right\}\left( {i = \sqrt { - 1} } \right)$$ lie on a :
A.
straight line whose slope is 1.
B.
circle whose radius is 1.
C.
circle whose radius is $$\sqrt 2 .$$
D.
straight line whose slope is $$- 1.$$
Answer :
circle whose radius is 1.
Solution :
$${\text{Let }}z \in S{\text{ then }}z = \frac{{\alpha + i}}{{\alpha - i}}$$
Since, $$z$$ is a complex number and let $$z = x + iy$$
$$\eqalign{
& {\text{Then, }}x + iy = \frac{{{{\left( {\alpha + 1} \right)}^2}}}{{{\alpha ^2} + 1}}\left( {{\text{by rationalisation}}} \right) \cr
& \Rightarrow \,\,x + iy = \frac{{\left( {{\alpha ^2} - 1} \right)}}{{{\alpha ^2} + 1}} + \frac{{i\left( {2\alpha } \right)}}{{{\alpha ^2} + 1}} \cr} $$
Then compare both sides
$$\eqalign{
& x = \frac{{{\alpha ^2} - 1}}{{{\alpha ^2} + 1}}\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& y = \frac{{2\alpha }}{{{\alpha ^2} + 1}}\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Now squaring and adding equations (1) and (2)
$$ \Rightarrow \,\,{x^2} + {y^2} = \frac{{{{\left( {{\alpha ^2} - 1} \right)}^2}}}{{{{\left( {{\alpha ^2} + 1} \right)}^2}}} + \frac{{4{\alpha ^2}}}{{{{\left( {{\alpha ^2} + 1} \right)}^2}}} = 1$$