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Geometric transformation

  • Thread starter Robin04
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  • #1
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Homework Statement


Let's call the axis of the ##z## complex plain ##x## and ##y##, so a general point can be written as ##z=x+iy##. Reflect the points of the complex plain so that the mirror line of the transformation is a line parallel to the vector ##v## and it passes trough the point ##u##. Give the function that transforms the point ##z=x+iy## to ##z'=x'+iy'## and show that the points on the mirror line are fix points of this transformation. (Sorry for the bad english.)

Homework Equations


##u = 3+5i##
##v=8-5i##

The Attempt at a Solution


First, I constructed the mirror line: ##5x+8y=55##. I checked it with a plotting program and it looks correct.
Then, the way we learnt reflection is the following:
- If the mirror line does not intersect the origin, we have to translate it so that it does. This can be done with any point of the line, let's call this point ##c##.
- Then, if the line does not lie on the x axis, we have to rotate it by an angle ##\beta##.
- Then we do the reflection with the complex conjugate.
-Then we do the inverse of the rotation and the translation.
##z_1 = z -c##
##z_2=z_1 e^{-i \beta}##
##z_3=z_2^{*}##
##z_4=z_3 e^{i \beta}##
##z_5=z_4+c##
So all together: ##z_5=z'=(z-c)^* e^{2i \beta} +c##
I substituted ##z=x+iy##, ##z'=x'+iy'##, ##c=a+bi## and made two functions for the real and the imaginary parts.
##x'=x \cdot cos{2 \beta} +y \cdot sin{2 \beta} -a \cdot cos{2 \beta} -b \cdot sin{2 \beta} +a##
##y'=x \cdot sin{2 \beta} -y \cdot cos{2 \beta} -a \cdot sin{2 \beta} +b \cdot cos{2 \beta} +b##

Now my problem is the last part which is about proving that the fix points are on the mirror line.
I guess I should substitute ##x'=x## and ##y'=y##, add the two equations together and express it as ##y=f(x)## and see if it matches my line equation from the beginning. But they are full of sines and cosines which won't have an exact value if I calculate ##\beta##, so I don't see how to prove this analytically. Numerically, it doesn't work yet, I'm still trying to figure out where did I mess it up.
 

Answers and Replies

  • #2
When you say z' = x' + iy', is the z' supposed to be the image of the point z wrt the mirror line passing through u in the direction of v?
 
  • #3
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When you say z' = x' + iy', is the z' supposed to be the image of the point z wrt the mirror line passing through u in the direction of v?
Yes
 
  • #4
When you are finding the image. Try this instead-
1. Define the line line equation - that is,
z0 - u = λv (u, v are the vectors you mentioned)
2. Consider a point z0 on the line you defined.
3. Find the position vector of z wrt this z0
That is z - z0
4. Find the foot of perpendicular of z on the line (nothing but the point z0) . You do this by taking dot product of perpendicular vectors as 0.
(z-z0).(v) = 0.
Also you know that z0 = u + λv.
Evaluate the value of λ and substitute it in z0 to find it.
5. Using the relation z' - z0 = (z - z0)e(iπ) find the point z'.
 
  • #5
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382
If the mirror line does not intersect the origin, we have to translate it so that it does. This can be done with any point of the line, let's call this point ccc.
c? Why introduce another unknown? You said you can use any point on the line. Why not use the one you are given?
 
  • #6
248
16
When you are finding the image. Try this instead-
1. Define the line line equation - that is,
z0 - u = λv (u, v are the vectors you mentioned)
2. Consider a point z0 on the line you defined.
3. Find the position vector of z wrt this z0
That is z - z0
4. Find the foot of perpendicular of z on the line (nothing but the point z0) . You do this by taking dot product of perpendicular vectors as 0.
(z-z0).(v) = 0.
Also you know that z0 = u + λv.
Evaluate the value of λ and substitute it in z0 to find it.
5. Using the relation z' - z0 = (z - z0)e(iπ) find the point z'.
This seems like a different method. I'll check this out too, but first I would like to realize what's wrong with my calculations.
 
  • #7
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c? Why introduce another unknown? You said you can use any point on the line. Why not use the one you are given?
It can be ##u## too of course, I was just presenting the general idea that I learnt on my lecture.
 
  • #8
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382
- Then, if the line does not lie on the x axis, we have to rotate it by an angle ββ\beta.
Again, beta is not an unknown. Once you translate the given vector v using the given point u the translated v defines beta. In fact it is easier to do the transform with linear algebra rather than creating the angle beta.
 
  • #9
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Again, beta is not an unknown. Once you translate the given vector v using the given point u the translated v defines beta. In fact it is easier to do the transform with linear algebra rather than creating the angle beta.
Yes, that's what I want to avoid: calculating beta. I don't see how can I do it algebraically because ##\beta## doesn't seem to cancel out.
 
  • #10
898
382
It can be ##u## too of course, I was just presenting the general idea that I learnt on my lecture.
No. That doesn’t make any sense. You are trying to construct a particular transform using particular given information. The given information defines the transform. Transform to the given point and vector not some unknown point and vector.
 
  • #11
248
16
No. That doesn’t make any sense. You are trying to construct a particular transform using particular given information. The given information defines the transform. Transform to the given point and vector not some unknown point and vector.
Sure, but I can understand the problem more deeply if I start from the general idea and leave the substitution of the exact numbers to the end. And according to the general idea, the c point can be anywhere on the mirror line. It doesn't change the solution if I don't choose ##c = u##, but it's a good practice of course because I have less work to do.

And I realized what was the problem. First, I made quite some mistakes mostly with the positive and negative signes, and the bigger issue was that I thought I have to calculate ##\beta##, but I realized that I can express the sines and cosines with the components of ##v##. I substituted at the end and I got back the line equation I was looking for.
 
  • #12
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Let's call the axis of the ##z## complex plain ##x## and ##y##...
This is the complex plane (not the z complex plain), and those are axes (plural of axis).
 
  • #13
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This is the complex plane (not the z complex plain), and those are axes (plural of axis).
Thank you! :) I'm still struggling with English.
 
  • #14
248
16

Homework Statement


In my assignement I have to combine three different geometric transformations and determine what kind of transformation is the result and where are its fix points.

The transformations are (in order):
- Reflection where the equation of the mirror line is ##y=-\frac{5}{8}x + \frac{55}{8}##
- Translation with a vector ##u=(3,5)##
- Rotation by ##+90^\circ## where the center point is at ##v=(8,-5)##

Homework Equations


We used complex numbers to work with transformations during the classes.

The Attempt at a Solution


I figured out the function that gives the image of every point in the 2d plane.
##x'=-x\cdot sin{2\beta}+y\cdot cos{2\beta} +u_1\cdot sin{2\beta}-u_2(cos{2\beta} +2) + v_1 + v_2)##
##y'=x\cdot cos{2\beta}+y\cdot sin{2\beta} -u_1( cos{2\beta}-2)-u_2\cdot sin{2\beta} - v_1 + v_2##
where ##\beta## is the angle between the ##x## axis and the mirror line of the reflection, ##u_1,u_2,v_1,v_2## are the components of the ##u,v## vectors.

I calculated the image of 3 points and they look like a rotation to me but I don't know how to prove this.
The general equations for a rotation are:
##x'=x\cdot cos{\alpha} -y\cdot sin{\alpha}-a\cdot cos{\alpha}+b\cdot sin{\alpha}+a##
##y'=x\cdot sin{\alpha} +y\cdot cos{\alpha}-a\cdot sin{\alpha}-b\cdot cos{\alpha}+b##
where ##\alpha## is the angle of the rotation, and ##a,b## are the coordinates of the center.


So I guess I should find the relationship between these, but I'm not sure how to do that.

K_pkiv_g_s.png
 

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  • #15
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I substituted ##x' = x## and ##y'=y## into the equations and tried to solve it but even online linear system of equation solvers say that there's no solution which means there's no fix point. So this should be a translation, but it cannot be because the lines joining the points and their images are not parallel.
 
  • #16
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How did you find the formula ? Did you tried uaing matrices or smthing else ? If ypu didnt maybe you can try ?

Note: I didnt solve the problem
 
  • #17
mathwonk
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well one clue is to use the concept of orientation. the orientation preserving transformations form a subgroup of all isometries, and consist of precisely the translations and rotations. In addition we have the orientation reversing transformations which are reflections and "glides" which are a combination of a translation and a reflection in the axis of the translation. Since you composed three isometries only one of which is orientation reversing, your composition is itself orientation reversing. Thus it is either a reflection or a glide. In particular it is certainly not a rotation. Glides and reflections can be distinguished by fixed points since a reflection fixes a line and a glide has no fixed points. in particular all you have to do is decide whether or not there are fixed points.

hope this is all correct, it's been a while since I taught this course. i attach some notes from an old course (1996), summarizing the classification of all isometries.
 

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  • #18
LCKurtz
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I had a bit of time for fiddle with Maple this weekend and I got this matrix equation for your transformation:$$
\begin{bmatrix} x' \\ y' \end{bmatrix} =
\begin{bmatrix}
\frac{80}{89} & \frac {39}{89}\\
\frac {39}{89} & - \frac{80}{89}
\end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -\frac{1058}{89} \\ -\frac {340}{89} \end{bmatrix}
$$It seems to agree with the first point you plotted. I got this by composing the matrices that do your three transformations, so I'm guessing I didn't make any mistakes and it must agree with your equations. It looks like a linear transformation plus translation. Can you analyze the matrix to see what kind it is?
 
  • #19
LCKurtz
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  • #20
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  • #21
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16
How did you find the formula ? Did you tried uaing matrices or smthing else ? If ypu didnt maybe you can try ?

Note: I didnt solve the problem
We'll learn matrices in 2 or 3 weeks maybe. Now we only used complex numbers to express the transformations. Translation is described by addition, rotation is described by multiplication and reflection is described by the conjugate of complex numbers. The teacher said however that with matrices it'll be much easier.
 
  • #22
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16
well one clue is to use the concept of orientation. the orientation preserving transformations form a subgroup of all isometries, and consist of precisely the translations and rotations. In addition we have the orientation reversing transformations which are reflections and "glides" which are a combination of a translation and a reflection in the axis of the translation. Since you composed three isometries only one of which is orientation reversing, your composition is itself orientation reversing. Thus it is either a reflection or a glide. In particular it is certainly not a rotation. Glides and reflections can be distinguished by fixed points since a reflection fixes a line and a glide has no fixed points. in particular all you have to do is decide whether or not there are fixed points.

hope this is all correct, it's been a while since I taught this course. i attach some notes from an old course (1996), summarizing the classification of all isometries.
Thank you very much, this helped a lot. I asked my teacher and he said that it's a glide, because there are no fix points, but we didn't learn it.
 
  • #23
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16
I had a bit of time for fiddle with Maple this weekend and I got this matrix equation for your transformation:$$
\begin{bmatrix} x' \\ y' \end{bmatrix} =
\begin{bmatrix}
\frac{80}{89} & \frac {39}{89}\\
\frac {39}{89} & - \frac{80}{89}
\end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -\frac{1058}{89} \\ -\frac {340}{89} \end{bmatrix}
$$It seems to agree with the first point you plotted. I got this by composing the matrices that do your three transformations, so I'm guessing I didn't make any mistakes and it must agree with your equations. It looks like a linear transformation plus translation. Can you analyze the matrix to see what kind it is?
We'll learn matrices in 2 or 3 weeks, I don't really know how to use them to represent transformations yet. A linear transformation and a translation is the same thing as a glide reflection?
 
  • #24
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  • #25
LCKurtz
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A linear transformation and a translation is the same thing as a glide reflection?
Not "linear transformation and translation" but "reflection and translation". Of course, a reflection is a linear transformation, but not all linear transformations are reflections.
 

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