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**TweedyTL** 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 + 1 = 0

Let a = x + iy

x + iy - i - ix - y + 1 = 0

x - y + 1 = ix - iy + i

x - y + 1 = i(x - y + 1)

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?