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

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

• Are you sure of the data? I get that circle is mapped into the line $\;4u+3=k\;$ , with $\;0\neq k=$ a complex constant... Commented Nov 17, 2016 at 9:18
• Yes I am sure the data is correct. Can You post the whole answer here. Commented Nov 17, 2016 at 9:38
• It's true that the circle is mapped into the line 4u+3=0. Commented Nov 17, 2016 at 21:20
• yes it is mapped into the line 4u+3=0. Commented Nov 18, 2016 at 5:02

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.

• actually you took wrong equation for circle it is $x^2+Y^2=4x$ Commented Nov 17, 2016 at 10:37
• @PrashantMore No, that's exactly what I took and used: $$x^2+y^2=4x\iff (x-2)^2+y^2=4$$ Commented Nov 17, 2016 at 17:07
• while solving we can take $2$ common from $2x^2$ and $2y^2$ and put $x^2+y^2=4x$ and same in denominator by solving that $(x-4)^2$ we can do the same thing and then sort out real and imaginary part of w. and then put real part value in line equation ie u's value in line equation. Like this I got the answer. Commented Nov 18, 2016 at 5:07