# Isometries of the Complex Plane: Fixed Lines

• Aug 22nd 2009, 12:36 AM
TweedyTL
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?
• Aug 22nd 2009, 03:27 AM
HallsofIvy
You have f(x+ yi)= x- yi- 2 and a line, in the complex plane, must be of the form y= mx+ c. Saying that a line is fixed means that $f(x_1+ (mx_1+c)i)= x_2+ (mx_2+ c)i$ (That is, that one point on the line is mapped into itself or another point on the line). In fact, it is sufficient to start by using the fact that a specific point on the line, (0, c) is simplest, is mapped into a point (x, mx+ c).

$f(ci)= -ci+ 2= x+ (mx+ c)i$

Setting the real parts equal, we have x= 2 and, setting the imaginary parts equal, $-c= mx+c= 2m+ c$ so that c= -m. That is, the line is of the form y= mx- m. Now look at the next easiest point, (1, 0). f(1)= 1- 2= -1. But (-1, 0) will be on the line y= mx- m only is 0= -m-m= -2m: m= 0.

This glide reflection fixes the real axis. Actually, that should have been obvious since $\overline{z}$ reflects around the real axis and then "- 2" translates parallel to the real axis.

By the way, I got $\overline{z}$ by using LaTex. The code is [ math ]\overline{z}[ /math ] (without the spaces).
• Aug 22nd 2009, 08:43 AM
TweedyTL
Thanks for the help, but I am still having problems, and I think you're making mistakes as well.

Firstly, for f(ic), c $\in$ R, I get
f(ic) = i( $\overline{ic}$) - 2
= i( $\overline{i}{c}$) - 2
= i(-ic) - 2
= c - 2
which is different from your value of f(ic)

Secondly, where do you get the point (1,0) from?
f(1) = i - 2
f(i - 2) = -1 -2i
The points (1,0) (-2,1) and (-1,-2) are not colinear so (1,0) is not on a fixed line. I know that the fixed line is not parallel to either axis.

It would help if someone could tell me where the contradiction arises in my reasoning:

(i) my assumption that the function f(z) = i $\overline{z}$-2 is a glide reflection isometry of the complex plane

(ii) my assumption that a glide reflection always has exactly 1 fixed line, and that this line is the line of reflection prior to translation. (The line stays the same, but loses some of the properties that a line of reflection has). This is because a glide reflection is a reflection in a line, followed by a translation parallel to that line.

(iii) my assumption that solving the equation f(f(a)) - 2f(a) + a = 0 for a will find the values of a which lie on the fixed line, if f is a glide reflection.

(iv) my calcultion to find the values of a which solve the equation when f(z) = i $\overline{z}$-2. I found no solutions, meaning that the mapping of f contains no fixed lines. Contradion!
• Aug 23rd 2009, 01:20 AM
running-gag
Quote:

Originally Posted by 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?

I have marked your mistakes in red
The line is x - y + 1 = 0
• Aug 23rd 2009, 03:14 AM
TweedyTL
Thanks. I did that equation about 5 times and somehow managed to get the wrong answer each time (Headbang)

Anyway I had an epiphany this morning and have worked out a foolproof method to find the fixed line:

(i)Find the line of reflection $L$ that reflects z to a $\overline{z}$
(ii)Find b as the sum of the 2 vectors: b $\sb{1}$ which is parallel $L$, and b $\sb{2}$ which is perpendicular to $L$
(iii)To find the fixed line, translate L by $\frac{1}{2}$ b $\sb{2}$

Problem Solved!