Fractional Linear Transformation of the Image of the Line $y=4-x$

Question

I am trying to find the image of the line $y=4-x$ under the fractional linear transformation $$w=\frac{8}{z-2-2i}.$$

My method is as follows:

Rearranging yields $$z=\frac{8}{w}+2+2i.$$ Now, \begin{align} z=x+iy&=\frac{8}{u+iv}+2+2i \\ &=\frac{8+(u+iv)(2+2i)}{u+iv}\times\frac{u-iv}{u-iv} \\ &=\frac{8u+2u^2+2v^2}{u^2+v^2}+i\frac{2u^2-8v+2v^2}{u^2+v^2} \end{align} Equating real and imaginary components yields $$x=\frac{8u+2u^2+2v^2}{u^2+v^2} \ \ \text{and} \ \ \ y=\frac{2u^2-8v+2v^2}{u^2+v^2}.$$ Hence, $$\frac{2u^2-8v+2v^2}{u^2+v^2}=4-\frac{8u+2u^2+2v^2}{u^2+v^2}\implies v=u.$$

Is my method valid?

Answer 1

Your calculation is correct.

Here is a shorter way using the properties of the given mapping:

• Note that your mapping is a so called Möbius transformation. So, it maps generalized circles (circles and lines) onto generalized circles.
• To find the image of the line $y= 4-x$ which is the same as $z=x + (4-x)i$ you need only the image of $3$ points:

$$x = 0 \Rightarrow z = 4i \mapsto w = -2(1+i)$$ $$x = 4 \Rightarrow z = 4 \mapsto w = 2(1+i)$$ $$\mbox{Finally } z = \infty \mapsto w = \lim_{z \to \infty}\frac{8}{z-2-2i} = 0$$ So, the image is the line passing through $-2(1+i), 2(1+i), 0$. Hence, you get $$\boxed{y = x}$$