Formula for rotation Watch

NotNotBatman
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 \alpha \in [0,\pi/2]

Find a formula for the reflection of each z (in the complex set) across the straight line L making an angle alpha with the positive real axis.



The solution says We first rotate L into the real axis by  z \to e^{-i\alpha}z . Then we reflect across the real axis by z \to \overline{z} .Finally, we rotate back via  z \to e^{i\alpha} z.

I have no idea what this is saying. Could someone explain please, because I couldn't find an explanation anywhere.
Last edited by NotNotBatman; 4 weeks ago
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DFranklin
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Do you understand how a rotation of alpha degrees rotates L onto the real axis?
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NotNotBatman
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(Original post by DFranklin)
Do you understand how a rotation of alpha degrees rotates L onto the real axis?
Yes, if you mean clockwise so  -\alpha ?
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DFranklin
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So, imagine that you've drawn z and L on a diagram, and then rotated by -alpha.

Then z \to z e^{-i\alpha} (*) and the line you want to reflect in is now the x-axis.

So now the reflection is simply taking the conjugate.

And then you need to rotate back.

(*) Note that it *is* -alpha and not alpha; I assumed you'd just made a typo, but it's possible this is your cause of confusion
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NotNotBatman
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If z is on L, I understand because z goes back on L, as each z on L would be invariant.

But if arg(z) = theta < alpha, then the first reflection gets you to (theta - alpha) , reflecting in x then gets you to (alpha - theta), then rotating by alpha gets you to 2alpha - theta. I feel like I'm misinterpreting the question.

Yes, the alpha was just a typo.
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ghostwalker
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(Original post by NotNotBatman)
If z is on L, I understand because z goes back on L, as each z on L would be invariant.

But if arg(z) = theta < alpha, then the first reflection gets you to (theta - alpha) , reflecting in x then gets you to (alpha - theta), then rotating by alpha gets you to 2alpha - theta.
Yes.

Here's a diagram. I've used a,t, rather than alpha, theta, as I can't do Greek in Paint. A is the original point, and A' the reflected point.

Arg(A)= t

Arg(A' ) = t + (a-t) + (a-t) = 2a-t.

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DFranklin
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(Original post by NotNotBatman)
If z is on L, I understand because z goes back on L, as each z on L would be invariant.

But if arg(z) = theta < alpha, then the first reflection gets you to (theta - alpha) , reflecting in x then gets you to (alpha - theta), then rotating by alpha gets you to 2alpha - theta. I feel like I'm misinterpreting the question.

Yes, the alpha was just a typo.
What you've written sounds ok (except when you say 'first reflection' you mean 'first rotation'). I'm not sure why you think there's a problem?
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NotNotBatman
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(Original post by ghostwalker)
Yes.

Here's a diagram. I've used a,t, rather than alpha, theta, as I can't do Greek in Paint. A is the original point, and A' the reflected point.

Arg(A)= t

Arg(A' ) = t + (a-t) + (a-t) = 2a-t.

Name:  Untitled.jpg
Views: 4
Size:  13.0 KB
(Original post by DFranklin)
What you've written sounds ok (except when you say 'first reflection' you mean 'first rotation'. I'm not sure why you think there's a problem?
Thanks, both, I'm not even sure why I was confused, seems really obvious now. And yes, I did mean 'first rotation'.
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DFranklin
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(Original post by NotNotBatman)
Thanks, both, I'm not even sure why I was confused, seems really obvious now. And yes, I did mean 'first rotation'.
One point I'd make: what you've written is fine *if* you know what arg z is. But if you had z in the form x+iy it would all get rather messy - there's no nice way of finding arg z when z is in this form.

I rather think this is the form you're supposed to be dealing with - the rotate by -alpha so reflection isijust complex conjugation only reallyrmakes sense if you're thinking of z as x+iy.
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