(Algebra) Isometries on the complex plane

[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.

 

[micromass]

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??

 

[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.

 

[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 $\tan\frac\theta 2 = -\frac ab$.

 

[blue_raver22]

Hello! I understand that mathematics can be challenging and frustrating at times, but don't worry, I am here to help you.

First, let's define some terms. Isometries are transformations that preserve distances and angles. In the context of the complex plane, isometries are transformations that preserve the distance between points and the angle between lines.

Now, let's break down the problem. We are given that we can identify complex numbers with vectors in ℝ2 and form their dot product. The dot product of two vectors is defined as the product of their magnitudes and the cosine of the angle between them. In this case, the dot product of (a,b) and (c,d) is ac+bd.

Next, we are asked to prove that ζ\bar{c} + c = 0 iff c is orthogonal to \sqrt{ζ}. Let's start by understanding what this means. The conjugate of a complex number is a complex number with the same real part but an opposite imaginary part. So, if c = a+bi, then \bar{c} = a-bi. Now, let's look at the term \sqrt{ζ}. In the context of the complex plane, the square root of a complex number is a complex number that, when squared, gives the original complex number. So, if ζ = cosθ + i sinθ, then \sqrt{ζ} = cos(θ/2) + i sin(θ/2).

Now, let's look at the left side of the equation: ζ\bar{c} + c. Substituting in the definitions we just discussed, we get (cosθ + i sinθ)(a-bi) + (a+bi). Expanding this, we get (acosθ + bsinθ) + i(asinθ - bcosθ) + a + bi. Simplifying this, we get (a + acosθ - bsinθ) + i(b + asinθ + bcosθ). Now, let's look at the right side of the equation: 0 iff c is orthogonal to \sqrt{ζ}. This means that the dot product of c and \sqrt{ζ} is equal to 0. Using the dot product formula we discussed earlier, we get acos(θ/2) + bsin(θ/2) = 0.

Now,