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

& A)1 \\\ & B)2 \\\ & C)\sqrt{2} \\\ & D)\dfrac{1}{2} \\\ \end{aligned}$$

Answer 1

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

& \Rightarrow a+ib=\dfrac{x+iy-\alpha }{x+iy+\alpha } \\\ & \Rightarrow a+ib=\dfrac{x-\alpha +iy}{x+\alpha +iy} \\\ \end{aligned}$$ Now let us multiply and divide with $$x+\alpha -iy$$ on R.H.S. $$\begin{aligned} & \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{aligned}$$ We know that the value of $${{i}^{2}}$$ is equal to -1. $$\begin{aligned} & \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{aligned}$$ We know that a complex number is said to be purely imaginary number if the real part of the complex number is equal to zero. So, we can write that $$\begin{aligned} & \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{aligned}$$ From the question, it is given that $$\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{aligned} & \Rightarrow \left| z \right|=2 \\\ & \Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=2 \\\ \end{aligned}$$ Now let us square on both sides, then we get $$\Rightarrow {{x}^{2}}+{{y}^{2}}=4.................(4)$$ Now let us substitute equation (4) in equation (3), then we get $$\begin{aligned} & \Rightarrow {{\alpha }^{2}}=4 \\\ & \Rightarrow \alpha =2……….......(5) \\\ \end{aligned}$$ **So, it is clear that option C is correct.** **Note:** Some students may have a misconception that the value of $$\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.