Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$

Question

Question:

Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.

My try: $$\begin{align}\\ &x^2+y^2-4x=0\\ &\implies (x-2)^2+y^2=4\\ &\implies |z-2|=2\\ \end{align}\\ $$ Now, $w=\frac{2z+3}{z-4}$ $$\begin{align}\\ &\frac w2=\frac{2z+3}{2z-8}\\ &\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\ &\implies\frac{w}{w-2}=\frac{2z+3}{11}\\ &\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\ &\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\ &\implies z=\frac{4w+3}{w-2}\\ &\implies z-2=\frac{2w+7}{w-2}\\ \end{align}\\ $$ $$\therefore\left|\frac{2w+7}{w-2}\right|=2\\ \implies 2w+7=2w-4 $$ Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways?

Answer 1

You can't replace modulus like that in the last line ( because complex numbers with equal magnitude may not be actually equal. For e.g., $2i$ and $-2i$ have magnitude $2$ but aren't equal.)

Use $w = u+iv$

So, $|2(u+iv)+7| = 2|(u+iv)-2|$

Evaluating and squaring both sides,
$\Rightarrow (2u+7)^2 + 4v^2 = 4(u-2)^2 + 4v^2 $

$\Rightarrow 4u^2 + 28u +49 = 4u^2 - 16u + 16 $

$\Rightarrow 44u + 33 = 0 $

$\Rightarrow 4u +3 =0 $

Answer 2

Ak19 answered what I would have answered before I got to it, so here's another possible way of how to do it. These transformations map generalized circles to generalized circles; a generalized circle is either a circle or a line. This particular transformation maps the point $z=0$ on the given circle to $w=-\frac34$, the point $z=2+2i$ on the given circle to $-\frac34-\frac{11}4i$, and the point $z=2-2i$ on the given circle to $-\frac34+\frac{11}4i$, so we can see that it maps the circle to the line $u=-\frac34$.

Answer 3

This is equivalent to showing that $4w+3$ is pure imaginary i.e. $\arg(4w+3)=\pm\frac{\pi}{2}$.

Since $0$ and $4$ are end points of the circle diameter and $4w+3=11\frac{z-0}{z-4}$, then $\arg(4w+3)=\pm\frac{\pi}{2}$.