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?