Question

Find the image of the unit circle |z| = 2 under the linear fractional transformation (z + 2)
T(z)= (z−1).

Answer 1

$\displaystyle T(z) = \frac{z+2}{z - 1}\\ |z| = 2 \implies x^2 + y^2 = 4\\ \begin{aligned} T(z) &= \frac{(x + 2) + jy}{(x - 1) + jy} \\&= \frac{((x+ 2)+ jy))((x - 1) - jy)}{(x - 1)^2 + y^2} \\&= \frac{x^2 + x - 2 +jy(x - 1 - x - 2) + y^2}{x^2 + y^2 - 2x + 1} \\&= \frac{x^2 + x - 2 - 3jy + y^2}{x^2 + y^2 - 2x + 1} \\&= \frac{4 + x - 2 - 3jy}{4 + 1 - 2x} = \frac{2 + x - 3jy}{5 - 2x} \end{aligned}\\ \begin{aligned} u &= \frac{2 + x}{5 - 2x}\\ 5u - 2ux &= 2 + x \\ x(2u + 1) &= 5u - 2\\ \therefore x &= \frac{5u - 2}{2u + 1} \end{aligned}\\ v = \frac{-3y}{5 - 2x}\\ \textsf{Dividing}\, v\, \textsf{by}\, u\\ \begin{aligned} \frac{v}{u} &= \frac{-3y}{2 + x}\\ \frac{v}{u} &= \frac{-3y}{2 + \frac{5u - 2}{2u + 1}}\\ \frac{v}{u(2u + 1)} &= \frac{-3y}{2(2u + 1) + 5u - 2}\\ \frac{v}{u(2u + 1)} &= \frac{-3y}{4u + 2 + 5u - 2}\\ \frac{v}{u(2u + 1)} &= \frac{-3y}{9u}\\ \frac{v}{(2u + 1)} &= \frac{-y}{3}\\ y &=\frac{-3v}{(2u + 1)} \end{aligned}\\ \textsf{Recall that} \, x^2 + y^2 = 4\\ \therefore\left(\frac{5u - 2}{2u + 1}\right)^2 + \left(\frac{-3v}{2u + 1}\right)^2 = 4\\ (5u - 2)^2 + (3v)^2 = 4(2u + 1)^2\\ 25u^2 - 20u + 4 + 9v^2 = 4(4u^2 + 4u + 1)\\ 25u^2 - 20u + 4 + 9v^2 = 16u^2 + 16u + 4\\ 9u^2 - 36u+ 9v^2 = 0\\ u^2 - 4u+ v^2 = 0\\ (u - 2)^2 + (v - 0)^2 = 4\\ \therefore \textsf{The image of the unit circle}\\ |z| = 2 \, \textsf{under the linear fractional}\\ \textsf{transformation is a circle of centre}\, (2, 0) \\ \textsf{and a radius}\, 2\, \textsf{units.}$