Question

Find the bilinear transformation which maps the points ${{z}_{1}}=2$ , ${{z}_{2}}=2i$ , and ${{z}_{3}}=-2$ into the points ${{\omega }_{1}}=1$ , ${{\omega }_{2}}=i$ , and ${{\omega }_{3}}=-1$ .

Answer 1

Hint:First considering the bilinear transformation of the form $\omega =f\left( z \right)=\dfrac{az+b}{cz+d}$ then putting the values of $\omega $ and $z$ one by one in the above bilinear transformation, we will get four equations in terms of a, b , c ,d and then solving these four equations for these variables we get the values of a, b , c, d and hence we get the required bilinear transformation.

Complete step by step answer:
Let us assume $\omega =f\left( z \right)=\dfrac{az+b}{cz+d}$ ………(i) be the required bilinear transformation, then
Putting z = 2 and $\omega =1$ in above equation (1), we have
$\begin{align}
  & f\left( 2 \right)=1 \\
 & \Rightarrow \dfrac{a(2)+b}{c(2)+d}=1 \\
\end{align}$
Simplifying it further, we have
$2a+b=2c+d$ ……………..(ii)
Again, putting z = 2i and $\omega =i$ in equation (i), we have
$\begin{align}
  & f\left( 2i \right)=i \\
 & \Rightarrow \dfrac{a(2i)+b}{c(2i)+d}=i \\
\end{align}$
Simplying it further, we get
$\begin{align}
  & 2ia+b=i(2ic+d) \\
 & \Rightarrow 2ia+b=2{{i}^{2}}\text{c+}i\text{d} \\
 & \Rightarrow (b-2{{i}^{2}}\text{c)+i}(2a-d)=0 \\
 & \Rightarrow (b+2c)+i(2a-d)=0 \\
\end{align}$
Now equating real parts and imaginary parts of this equation with 0, we have
(b + 2c) = 0 …………(iii) and (2a - d) = 0 ……..(iv)
Again, putting z = -2 and $\omega =-1$ in equation (i), we have
$\begin{align}
  & f\left( -2 \right)=-1 \\
 & \Rightarrow \dfrac{a(-2)+b}{c(-2)+d}=-1 \\
\end{align}$
Simplifying it further, we get
$\begin{align}
  & -2a+b=-1(-2c+d) \\
 & \Rightarrow b-2a=2c-d\text{ }.........\text{(v)} \\
\end{align}$
From equation (iii), and (iv), we get
$b=-2c\text{ and }d=2a$ ……..(vi)
Now substituting d = 2a in equation (i), we get
$b=2c$ but $b=-2c$ from (iii)
$\begin{align}
  & \therefore b=-b \\
 & \Rightarrow b+b=0 \\
 & \Rightarrow 2b=0 \\
 & \Rightarrow b=0 \\
\end{align}$
$\begin{align}
  & \because b=2c\text{ and b=0} \\
 & \Rightarrow c=0 \\
\end{align}$
Now,
 $\begin{align}
  & \omega =f\left( z \right)=\dfrac{az+b}{cz+d} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{az+0}{(0)z+d} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{az}{2a}\,\,\left[ \text{since d = 2a from equation (vi)} \right] \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{z}{2} \\
\end{align}$
Hence, the required bilinear transformation is $\omega =f\left( z \right)=\dfrac{z}{2}$

Note:
Always remember that to get values of n variables, we should have at least n equations. Here also as we have four variables a, b, c, d so before trying to solve for variables must remember first to get 4 equations in terms of a, b, c, d.