Question

Consider two complex numbers $\alpha$ and $\beta$ as $\alpha ={{\left[ \dfrac{\left( a+bi \right)}{\left( a-bi \right)} \right]}^{2}}+{{\left[ \dfrac{\left( a-bi \right)}{\left( a+bi \right)} \right]}^{2}}$where $a,b\in \mathbb{R}$ and $\beta =\dfrac{\left( z-1 \right)}{\left( z+1 \right)}$, where $\left| z \right|=1$, then find the correct statement
(a) Both $\alpha$and $\beta$ are purely real
(b) Both $\alpha$and $\beta$ are purely imaginary
(c) $\alpha$ is purely real $\beta$ is purely imaginary
(d) $\beta$ is purely real and $\alpha$ is purely imaginary

Answer 1

In order to find the solution of the given question that is to find the nature of $\alpha$ and $\beta$ where $\alpha ={{\left[ \dfrac{\left( a+bi \right)}{\left( a-bi \right)} \right]}^{2}}+{{\left[ \dfrac{\left( a-bi \right)}{\left( a+bi \right)} \right]}^{2}}$ here $a,b\in \mathbb{R}$ and $\beta =\dfrac{\left( z-1 \right)}{\left( z+1 \right)}$, here $\left| z \right|=1$ by applying the following concepts, $a+bi={{e}^{i\theta }}$ and the componendo dividendo rule, which is a theorem on proportions that allows for a quick way to perform calculations and reduce the amount of expansions needed and it states that If $a,b,c$and $d$ are numbers such that b and d are non-zero and $\dfrac{a}{b}=\dfrac{c}{d}$ then the following holds: Componendo: $\dfrac{a+b}{b}=\dfrac{c+d}{d}$; Dividendo: $\dfrac{a-b}{b}=\dfrac{c-d}{d}$; for $k\ne \dfrac{a}{b},\dfrac{a+kb}{a-kb}=\dfrac{c+kd}{c-kd}$ and for $k\ne \dfrac{-b}{d},\dfrac{a}{b}=\dfrac{a+kc}{b+kd}$.

Complete step by step solution:
According to the question, given is as follows:
$\alpha ={{\left[ \dfrac{\left( a+bi \right)}{\left( a-bi \right)} \right]}^{2}}+{{\left[ \dfrac{\left( a-bi \right)}{\left( a+bi \right)} \right]}^{2}}$ where $a,b\in \mathbb{R}$ and
and $\beta =\dfrac{\left( z-1 \right)}{\left( z+1 \right)}$, where $\left| z \right|=1$
Now simplify $\alpha$, by using the result $a+bi={{e}^{i\theta }}$ and $a-bi={{e}^{-i\theta }}$, we get:
$\Rightarrow \alpha ={{\left[ \dfrac{\left( a+bi \right)}{\left( a-bi \right)} \right]}^{2}}+{{\left[ \dfrac{\left( a-bi \right)}{\left( a+bi \right)} \right]}^{2}}$
$\Rightarrow \alpha ={{\left[ \dfrac{{{e}^{i\theta }}}{{{e}^{-i\theta }}} \right]}^{2}}+{{\left[ \dfrac{{{e}^{-i\theta }}}{{{e}^{i\theta }}} \right]}^{2}}$
Simplify it further with the help of division and multiplication of exponents, we get:
$\Rightarrow \alpha ={{e}^{4i\theta }}+{{e}^{-4i\theta }}$
Apply the formula ${{e}^{i\theta }}=\cos \left( \theta \right)+i\sin \left( \theta \right)$ in the above equation, we get:

& \Rightarrow \alpha =\cos \left( 4\theta \right)+i\sin \left( 4\theta \right)+\cos \left( 4\theta \right)-i\sin \left( 4\theta \right) \\\ & \Rightarrow \alpha =2\cos \left( 4\theta \right) \\\ \end{aligned}$$ Therefore, $$\alpha $$ is purely real. After this simplify for $$\beta $$ by using the componendo dividendo rule, which is a theorem on proportions that allows for a quick way to perform calculations and reduce the amount of expansions needed and it states that If $$a,b,c$$ and $$d$$ are numbers such that b and d are non-zero and $$\dfrac{a}{b}=\dfrac{c}{d}$$ then the following holds: Componendo: $$\dfrac{a+b}{b}=\dfrac{c+d}{d}$$; Dividendo: $$\dfrac{a-b}{b}=\dfrac{c-d}{d}$$; for $$k\ne \dfrac{a}{b},\dfrac{a+kb}{a-kb}=\dfrac{c+kd}{c-kd}$$ and for $$k\ne \dfrac{-b}{d},\dfrac{a}{b}=\dfrac{a+kc}{b+kd}$$, we get: $$\Rightarrow \dfrac{\beta +1}{\beta -1}=\dfrac{z-1+z+1}{z-1-\left(z+1\right)}$$ $$\Rightarrow \dfrac{\beta +1}{\beta -1}=\dfrac{-2z}{2}$$ $$\Rightarrow \dfrac{1+\beta }{1-\beta }=z$$ Let $$\beta =x+iy$$, applying this in the above equation we get: $$\Rightarrow \dfrac{(x+1)+iy}{(1-x)-iy}=z...\left( 1 \right)$$ Now $$\left| z \right|=1$$ $$\Rightarrow \sqrt{{{(1+x)}^{2}}+{{y}^{2}}}=\sqrt{{{(1-x)}^{2}}+{{y}^{2}}}$$ $$\Rightarrow 1+x=1-x$$ $$\Rightarrow x=0$$ Here $$y$$ can have any value, for say $$y=k$$, then equation $$\left( 1 \right)$$ becomes as follows: $$\Rightarrow z=\dfrac{1+ki}{1-ki}$$ Here x=0 so$ \beta = i y$ Therefore, we can clearly see $$\beta $$ is purely imaginary. **Hence, $$\alpha $$ is purely real and $$\beta $$ is purely imaginary. Therefore option (c) is correct.** **Note:** Students make a lot of mistakes while applying the componendo dividendo rule. It is important to remember it’s is a theorem on proportions that allows for a quick way to perform calculations and reduce the amount of expansions needed and it states that If $$a,b,c$$ and $$d$$ are numbers such that b and d are non-zero and $$\dfrac{a}{b}=\dfrac{c}{d}$$ then the following holds: Componendo: $$\dfrac{a+b}{b}=\dfrac{c+d}{d}$$; Dividendo: $$\dfrac{a-b}{b}=\dfrac{c-d}{d}$$; for $$k\ne \dfrac{a}{b},\dfrac{a+kb}{a-kb}=\dfrac{c+kd}{c-kd}$$ and for $$k\ne \dfrac{-b}{d},\dfrac{a}{b}=\dfrac{a+kc}{b+kd}$$.