Find the equation of the circle

[jimmycricket]

Homework Statement

The line $$\{z: y=t+x\}$$ is mapped to a circle by the function $$f(z)=\frac{z-1}{1-zi}$$ Find the equation of this circle.

The Attempt at a Solution

One method is to find mappings of three points on the line. These points will be mapped to the circles boundary. Then find the intersection of the perpendicular bisectors of the two chords that join the 3 points to give the center of the circle. This turns out to be far too tedious and messy.

I've found another method that utilizes the "preservation of symmetric points under inversion" property but i only understand it when I'm dealing with a circle not a line as in this case.
For the circle case it goes something like this:
The center of the circle and infinity are inverse points with respect to the circle. Then the images $$f(center)\:\:\:and\:\:\:f(\infty)$$ are inverse with repect to the circle under inversion so $$f(\infty)$$ gives the center.
I'm not sure how to adjust this for the case with a line in the beginning.

 

[mfb]

I don't understand the circle definition. Are (x,y) supposed to represent complex numbers?

One method is to find mappings of three points on the line. These points will be mapped to the circles boundary. Then find the intersection of the perpendicular bisectors of the two chords that join the 3 points to give the center of the circle. This turns out to be far too tedious and messy.

Maybe there are special points that make it easier to guess the center? You can prove it with other methods then.

 

[Ray Vickson]

Homework Statement

The line $$\{z: y=t+x\}$$ is mapped to a circle by the function $$f(z)=\frac{z-1}{1-zi}$$ Find the equation of this circle.

The Attempt at a Solution

One method is to find mappings of three points on the line. These points will be mapped to the circles boundary. Then find the intersection of the perpendicular bisectors of the two chords that join the 3 points to give the center of the circle. This turns out to be far too tedious and messy.

I've found another method that utilizes the "preservation of symmetric points under inversion" property but i only understand it when I'm dealing with a circle not a line as in this case.
For the circle case it goes something like this:
The center of the circle and infinity are inverse points with respect to the circle. Then the images $$f(center)\:\:\:and\:\:\:f(\infty)$$ are inverse with repect to the circle under inversion so $$f(\infty)$$ gives the center.
I'm not sure how to adjust this for the case with a line in the beginning.

You can substitute $z = x + i(t+x)$ into $Z \equiv f(z)$, then take the real and imaginary parts (assuming that $t$ is a real, fixed parameter). This will give you something of the form
$$Z = A(x,t) + i B(x,t)$$
with some known functions $A,B$. As $x$ varies over $(-\infty, \infty)$ the point $Z$ will trace out a curve in $\mathbb{C}$ (again, assuming $t$ is just a fixed real parameter). You can find the maxima and minima of the functions $A(x,t), B(x,t)$ by solving univariate optimization problems and/or looking at limits as $x \to \pm \infty$. Those will give you the vertical and horizontal lines that bound your curve $Z(x)$, and if $Z(x)$ is truly a circle, that will allow you to figure out the radius and center.

 

[jimmycricket]

So $$f(x+i(t+x))=\frac{x-1+i(t+x)}{1+t}=A(x,t)+B(x,t)=\frac{x-1}{1+t}+i\frac{x+t}{1+t}$$

The limits of A and B as $x\rightarrow\pm\infty$ do not exist. This implies the line is mapped to another line and not a circle. Is this true? I was certain it was mapped to a circle.

 

[Ray Vickson]

So $$f(x+i(t+x))=\frac{x-1+i(t+x)}{1+t}=A(x,t)+B(x,t)=\frac{x-1}{1+t}+i\frac{x+t}{1+t}$$

The limits of A and B as $x\rightarrow\pm\infty$ do not exist. This implies the line is mapped to another line and not a circle. Is this true? I was certain it was mapped to a circle.

$$f(x+i(x+t)) = \frac{x-1 +i(x+t)}{1-i[x+i(x+t)]} = A + i B, \text{ where}\\ A = -\frac{1+t}{1+2x+2t+2x^2+2xt+t^2}, \;B = \text{for you to fill in}$$