## Complex Numbers, Fractional Linear Transforms

### Fractional Linear Transforms

Consider the function f(z) = cz, where c is a complex constant.  You may recognize this as a linear transform on the xy plane, since c(a+b) = ca+cb, and c(rz) = rcz for any real number r.  However, not all transforms are accessible.  Apply Demoivre's theorem and you'll see what I mean.  The vector 1 is mapped to c, and i is mapped to ci.  The two unit vectors are always mapped to perpendicular vectors of equal length.  The plane is rotated and magnified; that's it.  This isn't a surprise really.  If all transforms of the real vector space xy were accessible we would need four numbers, building a 2×2 matrix.  The coefficient c only has two numbers, a+bi, and that restricts our options.

Let f(z)= cz+b to shift the plane after the fact, i.e. a translation.

A fractional linear transform is the quotient of two shifted linear transforms.  In other words, f(z) has the form (az+b)/(cz+d).  Remember that z is complex, and all coefficients are complex.  Assume the numerator and denominator are nonzero, and z appears somewhere in the fraction.  Mapping the entire plane to 3/7 isn't very interesting.

Composing two fractional linear transforms gives a third.  In fact, these transforms form a group under function composition.  To invert f(z), set f(z) = w and solve for z.  This gives z = (dw-b)/(a-cw).  All this is discussed in detail for arbitrary rings and fields.

The function f(z) is 1-1 and onto, except when cz+d = 0, or when a-cw = 0.  Create a mythical point at infinity and let f(-d/c) = ∞.  At the same time, map ∞ to a/c, a point that is otherwise inaccessible.  Now f maps the augmented complex plane onto itself, and is invertible.

As z approaches infinity, f(z) approaches a/c.  We know f is continuous everywhere else, hence f is continuous on the augmented plane, where the plane, plus the point at infinity, has the topology of the sphere.  Similarly, f approaches infinity as z approaches -d/c, hence the inverse function is also continuous, and f is bicontinuous.

Fix any three points in the augmented plane, and each transform maps these points to three distinct points in the augmented plane.  Furthermore, the image points determine the transform.  See integral domains for a general proof.

### Lines And Circles

Fractional linear transforms map circles and lines to circles and lines.  If you think of a line as a circle that passes through ∞, the transform maps circles to circles.

Shifting the plane preserves all geometric structures, so we don't have to worry about z+b.

Consider az, where a is a complex constant.  Use Demoivre's formula to multiply by a.  The plane is rotated through some angle, and scaled uniformly.  Geometric shapes grow larger or smaller, but are always similar to the original.  That takes care of az+b.  If there is no z downstairs, we are done.

If the transform is (az+b)/(cz+d), divide numerator and denominator by c and rewrite as (az+b)/(z+d).  Then turn it into a mixed fraction a + (b-ad)/(z+d).  Adding and multiplying by constants preserves shapes, so we only need look at 1/(z+d).  This is 1/z composed with z+d, and the latter preserves shapes, so look at 1/z.

apply the transform, which replaces θ with -θ and r with 1/r.  The former is merely a reflection through the real axis, and preserves shapes.  The latter is symmetric about the origin, so we can rotate the plane, if that is convenient.  Spin the plane so that the line is horizontal.  If it passes through the origin the image is another line passing through the origin, with 0 mapped to ∞ and ∞ mapped to 0.

If the line has the equation y = d, write it in polar form as r = d/sin(θ).  Replace r with 1/r and get r = sin(θ)/d.  Let c = 1/2d for notational convenience.  Thus r = 2c×sin(θ).  Switch back to rectangular coordinates and get this.

sqrt(x2+y2) = 2cy/sqrt(x2+y2)

x2+y2 = 2cy

x2 + (y-c)2 = c2

The shape is a circle of radius c, passing through the origin.  Reverse this algebra to show that circles passing through the origin are mapped to lines.

Now consider a circle that does not pass through the origin.  Spin the plane so that the circle has its center on the positive x axis.  Replacing θ with -θ reflects the circle through the x axis, and doesn't change a thing.  We only need replace radial distance with its reciprocal.  In fact, we can magnify the plane by m, take the reciprocal of distance, and magnify by m again, which effectively maps r onto 1/r.  Magnification takes circles to circles, so we can scale the plane by any factor that is convenient.  Scale the plane so that the circle has center at x and radius 1.  Draw a radius, having length 1, and draw the segment from the end of the radius to the origin.  Bring in the segment from 0 to x and we have a triangle with angle θ.  Use the law of cosines to solve for r.

x2 + r2 - 2xr×cos(θ) = 1

Replace r with 1/r and multiply through by r2 to get the following.  Remember, r cannot be 0; we've already handled the case where the circle passes through the origin.

x2r2 + 1 - 2xr×cos(θ) = r2

(x2-1)r2 - 2xr×cos(θ) = -1

Remember this equation, and start down a new path.  If the image is really a circle, its left and right points are at 1/(x+1) and 1/(x-1).  Let c = x2-1 for notational convenience.  The center is now x/c and the radius is 1/c.  Write the equation for this circle, in polar form, using the law of cosines, as we did above.

x2/c2 + r2 - 2xr/c×cos(θ) = 1/c2

Multiply through by c and move the first term to the right.

cr2 - 2xr×cos(θ) = 1/c - x2/c = (1-x2)/c = -1

This is the same equation we arrived at earlier, by applying the transform.  That completes the proof.