A Linear Fractional Transformation
Find the images of the points 0, 1 + i, i, and ∞ under the linear fractional transformation T(z) = (2z + 1)/ (z − i).
For z = 0 and z = 1+i we have:
T(0)={\frac{2(0)+1}{0-i}}={\frac{1}{-i}}=i\quad{\mathrm{and}}\quad T(1+i)={\frac{2(1+i)+1}{(1+i)-i}}={\frac{3+2i}{1}}=3+2i.Identifying a = 2, b = 1, c = 1, and d = −i in (3), we also have:
T(z)=\left\{\begin{array}{l} \frac{az+b}{cz+d} , \quad z\neq -\frac{d}{c},z\neq \infty \\\infty , \quad\quad z=-\frac{d}{c} \\ \frac{a}{c},\quad\quad z=\infty \end{array}\right. (3)
T(i)=T\left(-\frac{d}{c}\right)=\infty\;\;\;\mathrm{and}\;\;\;T(\infty)=\frac{a}{c}=2.