# (Algebra) Isometries on the complex plane

1. Feb 19, 2012

### praecox

So this is the problem as written and I'm totally lost. Any help or explanation would be greatly appreciated.

"Viewing ℂ=ℝ2 , we can identify the complex numbers z = a+bi and w=c+di with the vectors (a,b) and (c,d) in R2 , respectively. Then we can form their dot product, (a,b)$\bullet$(c,d)=ac+bd.
Prove that ζ$\bar{c}$ + c = 0 iff c is orthogonal to $\sqrt{ζ}$."

I feel like there are too many things undefined - or maybe I just don't get what things are supposed to be. $\bar{c}$ is supposed to be the conjugate of c, I know that much. And in another problem ζ was defined as cosθ + i sinθ, but I'm not sure how to use this information (or if it even applies to this problem). I know that c and ζ being orthogonal means they're both vectors and their dot product is zero. It's in the chapter on isometries of ℝ and ℂ.

I've been trying to figure this problem out for hours and am frustrated to the point of tears. Please help.

2. Feb 19, 2012

### micromass

Staff Emeritus
c orthogonal to $\sqrt{\xi}$? That doesn't even make sense. Those two things are numbers, what does it even mean that they're orthogonal??

3. Feb 19, 2012

### praecox

You can see my frustration then. I checked to make sure I copied the problem word for word. I'm so confused by it - it's making me crazy.

The only thing I can think is ζ is a rotation (as it had been used before). So it makes sense to me that ζ$\bar{c}$ = -c, where ζ rotates the conjugate -90°. but the c orthogonal to √ζ is what's killing me.

I'm totally lost.

4. Feb 19, 2012

### morphism

Assuming $\zeta = \cos\theta + i \sin\theta$ and $\sqrt \zeta = \cos\frac \theta 2 + i \sin\frac \theta 2$, and writing $c = a+ib$, we see c is orthogonal to $\zeta$ iff $a \cos\frac \theta 2 + b\sin\frac \theta 2 = 0$ iff $\tan\frac\theta 2 = -\frac ab$. Now use the usual formulas expressing $\sin\theta$ and $\cos\theta$ in terms of $\tan\frac\theta 2$ to see that $\zeta \bar c + c = 0$ iff $\tan\frac\theta 2 = -\frac ab$.