Question

If a expression \[\dfrac{z-\alpha }{z+\alpha }\left( \alpha \in R \right)\] is a purely imaginary number and \[\left| z \right|=2\] then the value of \[\alpha \] is equal to
\[\begin{align}
  & A)1 \\
 & B)2 \\
 & C)\sqrt{2} \\
 & D)\dfrac{1}{2} \\
\end{align}\]

Answer 1

Hint: We know that a complex number is said to be purely imaginary number if the real part of the complex number is equal to zero. We know that the value of \[{{i}^{2}}\] is equal to -1. We know that the value of \[\left| z \right|\], if \[z=a+ib\], is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]. By using these concepts, we can find the value of \[\alpha \].

Complete step-by-step solution:
Let us assume \[z=x+iy\].
Let us consider
\[z=x+iy....(1)\]
From the question, we were given a complex number \[\dfrac{z-\alpha }{z+\alpha }\left( \alpha \in R \right)\]. Let us assume \[\dfrac{z-\alpha }{z+\alpha }\left( \alpha \in R \right)\] is equal to \[a+ib\].
So, let us consider
\[a+ib=\dfrac{z-\alpha }{z+\alpha }....(2)\]
Now let us substitute equation (2) in equation (1), then we get
\[\begin{align}
  & \Rightarrow a+ib=\dfrac{x+iy-\alpha }{x+iy+\alpha } \\
 & \Rightarrow a+ib=\dfrac{x-\alpha +iy}{x+\alpha +iy} \\
\end{align}\]
Now let us multiply and divide with \[x+\alpha -iy\] on R.H.S.
\[\begin{align}
  & \Rightarrow a+ib=\dfrac{x-\alpha +iy}{x+\alpha +iy}\times \dfrac{x+\alpha -iy}{x+\alpha -iy} \\
 & \Rightarrow a+ib=\dfrac{\left( x-\alpha +iy \right)\left( x+\alpha -iy \right)}{\left( x+\alpha +iy \right)\left( x+\alpha -iy \right)} \\
 & \Rightarrow a+ib=\dfrac{{{x}^{2}}-\alpha x+ixy+\alpha x-{{\alpha }^{2}}-{{i}^{2}}{{y}^{2}}}{{{\left( x+\alpha \right)}^{2}}-{{i}^{2}}{{y}^{2}}} \\
\end{align}\]
We know that the value of \[{{i}^{2}}\] is equal to -1.
\[\begin{align}
  & \Rightarrow a+ib=\dfrac{{{x}^{2}}-\alpha x+ixy+\alpha x-{{\alpha }^{2}}-\left( -1 \right){{y}^{2}}}{{{\left( x+\alpha \right)}^{2}}-\left( -1 \right){{y}^{2}}} \\
 & \Rightarrow a+ib=\dfrac{{{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}}+ixy}{{{\left( x+\alpha \right)}^{2}}+{{y}^{2}}} \\
 & \Rightarrow a+ib=\dfrac{{{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}}}{{{\left( x+\alpha \right)}^{2}}+{{y}^{2}}}+i\dfrac{xy}{{{\left( x+\alpha \right)}^{2}}+{{y}^{2}}} \\
\end{align}\]
We know that a complex number is said to be purely imaginary number if the real part of the complex number is equal to zero.
So, we can write that
\[\begin{align}
  & \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}-{{\alpha }^{2}}}{{{\left( x+\alpha \right)}^{2}}+{{y}^{2}}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}={{\alpha }^{2}}....(3) \\
\end{align}\]
From the question, it is given that \[\left| z \right|=2\].
We know that the value of \[\left| z \right|\], if \[z=a+ib\], is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\].
\[\begin{align}
  & \Rightarrow \left| z \right|=2 \\
 & \Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=2 \\
\end{align}\]
Now let us square on both sides, then we get
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=4.................(4)\]
Now let us substitute equation (4) in equation (3), then we get
\[\begin{align}
  & \Rightarrow {{\alpha }^{2}}=4 \\
 & \Rightarrow \alpha =2……….......(5) \\
\end{align}\]
So, it is clear that option C is correct.

Note: Some students may have a misconception that the value of \[\left| z \right|\], if \[z=a+ib\], is equal to \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]. But we know that the value of \[\left| z \right|\], if \[z=a+ib\], is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]. If this misconception is followed, then we cannot get the correct answer. So, this misconception should be avoided by students.