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

Hint: 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:
\[\begin{align}
  & \Rightarrow \alpha =\cos \left( 4\theta \right)+i\sin \left( 4\theta \right)+\cos \left( 4\theta \right)-i\sin \left( 4\theta \right) \\
 & \Rightarrow \alpha =2\cos \left( 4\theta \right) \\
\end{align}\]
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 \[k\ne \dfrac{-b}{d},\dfrac{a}{b}=\dfrac{a+kc}{b+kd}\].