Using complex numbers to find reflections

Question

if the equation of the curve of the reflection of ellipse $ \frac{(x-4)^2}{16} + \frac{ (y-3)^2}{9} = 1$ about the line $x-y-2=0$ is $16x^2 + 9y^2 + k_1 x -36 y+k_2 =0 $ , then $ \frac{k_1 +k_2}{33}$ =?

So, I thought of this method using complex numbers to find reflection point (z'), so say I want to reflect a point $ z=(x+iy)$ about a line, then I do this sequence of transformation.

$$ z' = \overline{z} e^{ 2i \arctan(m) } $$

where 'm' is slope of line,

So, I first do a coordinate transform for the ellipse

$ x= 4 + 4 \cos \theta$

$ y = 3 + 3 \sin \theta$

And thne,

I put $ z= (4 + 4 \cos \theta) + i( 3+ 3 \sin \theta)$

$\overline{z} = (4 +4 \cos \theta) - i (3 + 3 \sin \theta)$

so, $ m= \frac{\pi}{4}$

I get,

$ z' = \overline{z} (i)$\

but this doesn't give me the right answer for some reason...

P.s: the motivation for the reflection point formula is that, first I make the line my x axis byrotating whole plane by negative of slope of line i.e : \overline{ z e^{i \arctan(m)} , then to find reflection I took reflection of this point about 'x' axis by conjugating it then I multiply it by $ e^{ i \arctan(m)} $ to find the point in original coordinate system

Where exactly am I going wrong?

If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ...

I saw this, but I want to do this using complex numbers reference for my method:

1. I make my line of refleciton my axis
2. I conjugate my point
3. I rotate my point back by how much I rotated it originally

Part that I am confused with : Why do we shift the line such that intercepts line up with origin? I know we either shift x intercept to origin or y intercept to origin coz x intercept and y intercept related. But why do we start with this?

Answer 1

The diagram is on the right track, but you will also have to do a translation given the fact that $y=x-2$.

The steps as I would envision are:

1. Make the following substitutions: $$ x = \frac{z+ \overline{z}}{2}$$ $$ y = \frac{z- \overline{z}}{2i}$$

2. Apply the rotational coordinate transformation: $$ z' = ze^{-i\theta}$$ $$\theta = artcan(m)$$

3. Apply the translation coordinate transformation: $$ z'' = z' + ai $$ $$ a = \sqrt{2} $$

4. Perform the reflection: $$ z'' = \overline{z''}$$

5. 'Undo' the translation coordinate transformation: $$ z' = z'' - ai $$

6. 'Undo' the rotational coordinate transformation: $$ z = z'e^{i\theta}$$

Diagram of translation first approach (Ellipse position is very approximate):

Answer 2

I've got a result which is analogous to reflection of a point w.r.t. a line in Cartesian coordinates.

After following the step 1 in the answer of @gigo318, you get the complex equation of the line as $$a\overline z+\overline az+b=0$$ Now, the reflection of the point $z_1$ about this line can be given as

$$\begin{align*}\displaystyle\frac{(z+\overline z) -(z_1+\overline{z_1})}{a+\overline a}=\frac{(z-\overline z)-(z_1-\overline{z_1})}{a-\overline a}&=\frac{-(a\overline{z_1}+\overline az_1+b)}{|a|^2}\end{align*}$$

Answer 3

Conceptually, what you are doing in the complex plane is to rotate the line of reflection to the horizontal, take the conjugate, and rotate it back by the same angle. So, say you have a line from $z_1$ to $z_2$, at an angle $\alpha=\tan^{-1}(m)$ and you want the reflection of point $z$, say $z'$.

Then

$$ \begin{align*} z'=\big((z-z_1)e^{-i\alpha}\big)^*e^{i\alpha}+z_1\ &=(z^*-z_1^*)e^{i2\alpha}+z_1 \end{align*}$$