Question

If $\alpha $ and $\beta $ are different complex numbers with $\left| \beta \right|=1$, then the value of $\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|$ is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Hint: We start solving the problem by assuming the term ${{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}$ and make use of the property of complex numbers ${{\left| z \right|}^{2}}=z.\overline{z}$. We then make use of the properties of complex numbers $\overline{\left( \dfrac{a}{b} \right)}=\dfrac{\overline{a}}{\overline{b}}$, $\overline{\left( a-b \right)}=\overline{a}-\overline{b}$, $\overline{ab}=\overline{a}.\overline{b}$, $\overline{1}=1$ and $\overline{\overline{a}}=a$ to proceed through the problem. We then make use of the result $\left| \beta \right|=1$ and perform necessary calculations to get the value ${{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}$ which later gives the value of $\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|$.

Complete step-by-step answer:
According to the problem, we are given that $\alpha $ and $\beta $ are different complex numbers with $\left| \beta \right|=1$. We need to find the value of $\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|$.
Let us assume ${{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}$. We know that ${{\left| z \right|}^{2}}=z.\overline{z}$.
So, we get \[{{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right).\overline{\left( \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right)}\] ---(1).
We know that $\overline{\left( \dfrac{a}{b} \right)}=\dfrac{\overline{a}}{\overline{b}}$. We muse this result in equation (1).
$\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right).\left( \dfrac{\overline{\left( \beta -\alpha \right)}}{\overline{\left( 1-\overline{\alpha }\beta \right)}} \right)$ ---(2).
We know that $\overline{\left( a-b \right)}=\overline{a}-\overline{b}$. We muse this result in equation (2)
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\beta -\alpha }{1-\overline{\alpha }\beta } \right).\left( \dfrac{\overline{\beta }-\overline{\alpha }}{\overline{1}-\overline{\overline{\alpha }\beta }} \right)\] ---(3).
We know that $\overline{ab}=\overline{a}.\overline{b}$ and $\overline{1}=1$. We use this result in equation (3)
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\beta -\alpha }{1-\overline{\alpha }\beta } \right).\left( \dfrac{\overline{\beta }-\overline{\alpha }}{1-\overline{\overline{\alpha }}\overline{\beta }} \right)\] ---(4).
We know that $\overline{\overline{a}}=a$. We muse this result in equation (4)
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\beta -\alpha }{1-\overline{\alpha }\beta } \right).\left( \dfrac{\overline{\beta }-\overline{\alpha }}{1-\alpha \overline{\beta }} \right)\].
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{\beta \overline{\beta }-\beta \overline{\alpha }-\alpha \overline{\beta }+\alpha \overline{\alpha }}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+\overline{\alpha }\beta \alpha \overline{\beta }} \right)\] ---(5).
We know that ${{\left| z \right|}^{2}}=z.\overline{z}$. We muse this result in equation (5)
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{{{\left| \beta \right|}^{2}}-\beta \overline{\alpha }-\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+\overline{\alpha }\alpha \beta \overline{\beta }} \right)\].
From the problem, we are given that $\left| \beta \right|=1$.
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{{{1}^{2}}-\beta \overline{\alpha }-\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}{{\left| \beta \right|}^{2}}} \right)\].
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+\left( {{\left| \alpha \right|}^{2}}\times {{1}^{2}} \right)} \right)\].
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=\left( \dfrac{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}}{1-\overline{\alpha }\beta -\alpha \overline{\beta }+{{\left| \alpha \right|}^{2}}} \right)\].
\[\Rightarrow {{\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|}^{2}}=1\].
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=1\].
So, we have found the value of $\left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|$ as 1.

So, the correct answer is “Option (b)”.

Note: We can see that the given problem contains good amount of calculation so, we need to perform each step carefully. We can also solve this problem as shown below:
We are given $\left| \beta \right|=1$ so, we get ${{\left| \beta \right|}^{2}}=1$. We know that ${{\left| z \right|}^{2}}=z.\overline{z}$.
So, we get $\beta .\overline{\beta }=1\Leftrightarrow \overline{\beta }=\dfrac{1}{\beta }$ ---(6).
We know that \[\left| \dfrac{a}{b} \right|=\dfrac{\left| a \right|}{\left| b \right|}\].
$\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \beta -\alpha \right|}{\left| 1-\overline{\alpha }\beta \right|}$.
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \left( \beta \right)\left( 1-\dfrac{\alpha }{\beta } \right) \right|}{\left| 1-\overline{\alpha }\beta \right|}\] ---(7).
Let us substitute equation (6) in equation (7).
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \left( \beta \right)\left( 1-\alpha \overline{\beta } \right) \right|}{\left| 1-\overline{\alpha }\beta \right|}\].
We know that $\left| a.b \right|=\left| a \right|\left| b \right|$.
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \beta \right|\left| 1-\alpha \overline{\beta } \right|}{\left| 1-\overline{\alpha }\beta \right|}\].
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{1\times \left| 1-\alpha \overline{\beta } \right|}{\left| 1-\overline{\alpha }\beta \right|}\].
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \overline{1-\overline{\alpha }\beta } \right|}{\left| 1-\overline{\alpha }\beta \right|}\].
Let us assume $z=1-\overline{\alpha }\beta $.
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=\dfrac{\left| \overline{z} \right|}{\left| z \right|}\].
We know that $\left| z \right|=\left| \overline{z} \right|$.
\[\Rightarrow \left| \dfrac{\left( \beta -\alpha \right)}{\left( 1-\overline{\alpha }\beta \right)} \right|=1\].