show that the transformation $w=\frac{(2z+3)}{(z−4)}$ maps the circle $x^2+y^2=4x$ on the straight line $4u+3=0$

Question

show that under the transformation $w=\frac{(2z+3)}{(z−4)}$,the circle $x^2+y^2=4x$ is transformed into the straight line $4u+3=0$ in the w plane

Answer 1

The given circle is $\;(x-2)^2+y^2=4\;$ , so take any element $\;z=(x,y)\sim x+iy\;$ in it and apply to it $\;w\;$ :

$$w(z):=\frac{2z+3}{z-4}=\frac{2x+3+2iy}{x-4+iy}\cdot\frac{x-4-iy}{x-4-iy}=\frac{(2x^2-5x-12+2y^2)-11yi}{(x-4)^2+y^2}$$

Let us check whether the above belongs to the line $\;4u+3=0\;$ . Observe also that the denominator above equals $\;(x-4)^2+4-(x-2)^2=-4x+16=-4(x-4)\;$ , and like wise the numerator is

$$2x^2-5x-12+2y^2-11yi=2x^2-5x-12+8-2(x-2)^2-11yi=3(x-4)-11yi$$

so that :

$$4\frac{3(x-4)-11yi}{-4(x-4)}+3=-3+\frac{11y}{x-4}i+3=\frac{11y}{x-4}i$$

Perhaps there's some mistake in the above...or in the question,. of course.