Question

If the points in the set S are represented by \[S=\left\{ \dfrac{\alpha +i}{\alpha -i}:\alpha \in R \right\}\], $\left( i=\sqrt{-1} \right)$, then find the locus on which all these points?
(a) A circle whose radius is 1
(b) A straight line whose slope is 1
(c) A straight line whose slope is –1.
(d) A circle whose radius is $\sqrt{2}$.

Answer 1

Hint: We start solving the problem by assigning a complex variable to the given general representation of points. We do required calculations and equate real and imaginary parts on both sides. Now, we add the squares of the real and imaginary parts that we have just obtained. We make necessary calculations and check the equation to get the desired locus.

Complete step-by-step answer:
Given that the points in the set S are represented as \[S=\left\{ \dfrac{\alpha +i}{\alpha -i}:\alpha \in R \right\}\], $\left( i=\sqrt{-1} \right)$. We need to find locus of all these points in the set S.
Let us assume $z=\dfrac{\alpha +i}{\alpha -i}$, where $z=x+iy$ is a complex number.
So, we have $x+iy=\dfrac{\alpha +i}{\alpha -i}$.
$\Rightarrow x+iy=\dfrac{\alpha +i}{\alpha -i}\times \dfrac{\alpha +i}{\alpha +i}$.
$\Rightarrow x+iy=\dfrac{\left( \alpha +i \right)\times \left( \alpha +i \right)}{\left( \alpha -i \right)\times \left( \alpha +i \right)}$.
$\Rightarrow x+iy=\dfrac{\left( \alpha .\alpha \right)+\left( \alpha .i \right)+\left( i.\alpha \right)+\left( i.i \right)}{\left( \alpha .\alpha \right)+\left( \alpha .i \right)-\left( i.\alpha \right)-\left( i.i \right)}$.
\[\Rightarrow x+iy=\dfrac{{{\alpha }^{2}}+i\alpha +i\alpha +{{i}^{2}}}{{{\alpha }^{2}}+i\alpha -i\alpha -{{i}^{2}}}\].
\[\Rightarrow x+iy=\dfrac{{{\alpha }^{2}}+2i\alpha +{{\left( \sqrt{-1} \right)}^{2}}}{{{\alpha }^{2}}-\left( {{\left( \sqrt{-1} \right)}^{2}} \right)}\].
\[\Rightarrow x+iy=\dfrac{{{\alpha }^{2}}+2i\alpha +\left( -1 \right)}{{{\alpha }^{2}}-\left( -1 \right)}\].
\[\Rightarrow x+iy=\dfrac{{{\alpha }^{2}}+2i\alpha -1}{{{\alpha }^{2}}+1}\].
\[\Rightarrow x+iy=\left( \dfrac{{{\alpha }^{2}}-1}{{{\alpha }^{2}}+1} \right)+i\left( \dfrac{2\alpha }{{{\alpha }^{2}}+1} \right)\].
We compare real and imaginary parts on both sides and we get \[x=\dfrac{{{\alpha }^{2}}-1}{{{\alpha }^{2}}+1}\] and \[y=\dfrac{2\alpha }{{{\alpha }^{2}}+1}\].
Let us find the value of ${{x}^{2}}+{{y}^{2}}$.
So, we have ${{x}^{2}}+{{y}^{2}}={{\left( \dfrac{{{\alpha }^{2}}-1}{{{\alpha }^{2}}+1} \right)}^{2}}+{{\left( \dfrac{2\alpha }{{{\alpha }^{2}}+1} \right)}^{2}}$.
$\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{\left( {{\alpha }^{2}}-1 \right)}^{2}}}{{{\left( {{\alpha }^{2}}+1 \right)}^{2}}}+\dfrac{{{\left( 2\alpha \right)}^{2}}}{{{\left( {{\alpha }^{2}}+1 \right)}^{2}}}$.
$\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{\alpha }^{4}}-2{{\alpha }^{2}}+1}{{{\alpha }^{4}}+2{{\alpha }^{2}}+1}+\dfrac{4{{\alpha }^{2}}}{{{\alpha }^{4}}+2{{\alpha }^{2}}+1}$.
$\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{\alpha }^{4}}-2{{\alpha }^{2}}+1+4{{\alpha }^{2}}}{{{\alpha }^{4}}+2{{\alpha }^{2}}+1}$.
$\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{\alpha }^{4}}+2{{\alpha }^{2}}+1}{{{\alpha }^{4}}+2{{\alpha }^{2}}+1}$.
$\Rightarrow {{x}^{2}}+{{y}^{2}}=1$.
We know that the equation ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ represents a circle with its center at $\left( 0,0 \right)$ and radius a.
So, we have found all the points in the set \[S=\left\{ \dfrac{\alpha +i}{\alpha -i}:\alpha \in R \right\}\], $\left( i=\sqrt{-1} \right)$ lies on a circle of radius 1.
∴ The locus of all the points in the set \[S=\left\{ \dfrac{\alpha +i}{\alpha -i}:\alpha \in R \right\}\], $\left( i=\sqrt{-1} \right)$ lies on a circle of radius 1.
The correct option for the given problem is (a).

Note: We can also find the magnitude of the complex number z to prove this result. Because $\sqrt{{{x}^{2}}+{{y}^{2}}}$ is the magnitude of the complex number z. We should not have negative values for radius as we can never find a circle with negative radius. Whenever we get this type of problem, we solve by assigning a complex variable to the given representation.