Isometries of the Complex Plane: Fixed Lines
I am trying to find the fixed line in isometries of the plane, and I keep running into contradictions. I have checked both my assumptions and workings 3 times and still get contradictions. I'll start from the beginning, stating all my assumptions, because I do not know where I am going wrong:
I know there are 4 types of isometries: rotations, translations, reflections and glide reflections. (A glide reflection is a reflection, followed by a translation in a direction parallel to the line of reflection).
All isometries can be written as f(z) = az+b or aZ+b, with a,b in the set of complex numbers, lal=1. (I am using capital Z for z conjugate. If you know how to type z conjugate please tell me)
Isometries in the form f(z) = aZ+b are either reflections or glide reflections.
If f(b) = 0, f is a reflection.
Otherwise f is a glide reflection.
I have an isometry that gives a glide reflection, and I am trying to find the line of reflection. I am assuming that the line of reflection in a glide reflection isometry is the only fixed line.
To find the fixed line in a glide reflection, I am using this formula:
f(f(a))-2f(a)+a = 0
All values of a which satisfy this equation should lie on the fixed line. I am aware that this only works for glide reflections, not general functions.
The function I have is f(z)=iZ-2.
f(-2) = -2i -2 =/= 0, so f is a glide reflection, so using the formula above should give me the equation of the fixed line.
f(f(a)) - 2f(a) + a = 0
f(iA-2) - 2(iA-2) + a = 0
i(-ia-2) - 2 - 2(iA-2) + a = 0
a - 2i - 2 - 2iA + 4 + a = 0
a - i - iA + 2 = 0
Let a = x + iy
x + iy - i - ix - y + 2 = 0
x - y + 2 = ix - iy
x - y + 2 = i(x - y)
x and y are real numbers, so this equation has no solution, meaning that the function f(z) = iZ - 2 does not have a fixed line. This is a contradiction with my assumption that a glide reflection always has 1 and only 1 fixed line.
Where am I going wrong?