Formula for rotation

\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.

(edited 5 years ago)

Reply 1

DFranklin

Do you understand how a rotation of alpha degrees rotates L onto the real axis?

Reply 2

NotNotBatman

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

Reply 3

DFranklin

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

Reply 4

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.

Reply 5

ghostwalker

Original post by NotNotBatman

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.

Reply 6

DFranklin

Original post by NotNotBatman

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?

Reply 7

NotNotBatman

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.

Original post by DFranklin

What you've written sounds ok (except when you say 'first reflection' you mean 'first rotation' :wink:

. 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'.

Reply 8

DFranklin

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.