consider a regular \(2n-gon\) show that a horizontal flip followed by a vertical flip is equivalent to 180 degree rotation

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consider a regular \(2n-gon\) show that a horizontal flip followed by a vertical flip is equivalent to 180 degree rotation

Mathematics
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|dw:1440928722800:dw|
what are you thinking of, interrupting it with geometrical coordinates or just prove symmetric on 2n-gon in group way ?
I think I have it using coordinate geometry(transformation rules) : Horizontal flip (x coordinate remains same, y coordinate changes sign) : \((x,y)\longrightarrow (x,-y)\) after that, we do Vertical flip(x coordinate changes sign, y coordinate remains same) : \((x,-y)\longrightarrow (-x,-y)\)

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next recalling that the rule for rotating 180 degrees is \((x,y)\longrightarrow (-x,-y)\) ends the proof
Also consider the fact that the angle between the lines of your two reflections determines the angle of rotation (if you wanted to go that route)
i.e. Two reflections are equivalent to a specific angle rotation. And the angle between the lines determines the rotation about their intersection point.
Oh, and I forgot to mention that it's twice the angle between them. This means that since there is a 90 degree angle between the x- and y-axis, there will be a 180 degree rotation.
what im trying to understand is why u have chosen a regular \(2n-gon\) xD i believe it applies on each n-gon, this is confusing me since im trying to know what is special about it but in general all rotation and fillips are alike for Dihedral groups xD or maybe i'm missing something out. |dw:1440934919407:dw|
@jabberwock thanks! that works perfectly!
@ikram002p when the number of sides is not even, we will not be haveing the element \(R_{180}\) in the dihedral group right ?
well that part is confusing me, since basically rotation in abstract algebra (R_something) mean identity and u need it to take origin form, how ever in geometry u can rotate anything u want as much as u want without such restrict xD if you know what i mean.
|dw:1440936280877:dw|
Yes, i get what you mean. In dihedral groups, the rotations are restricted to a finite number : integer multiples of \(\large \dfrac{360}{n}\)
where \(n\) = number of sides
exactly, but now i got what ur asking for xD
after rotating, the polygon must look exactly same as before (overlap exactly)
that applies to reflections also we only consider the reflections that overlap exactly with the initial polygon
they are precisely the reflections over "symmetry lines" of the polygon
and as you know every \(n-gon\) has \(n\) symmetry lines
it make sense to me now, been long time these stuff need to be reviewed.
You can also prove this from the perspective of complex analysis. Let \(z\) and \(w\) be complex numbers such that \(z^{2n}=w\), and let \(z=r\exp\left(i\theta_0\right)\) so that \(|z|=r\) and \(\arg z=\theta_0\). The \(2n\)th roots of \(w\) are then \[\large w^{\frac{1}{2n}}=z=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\] As you probably know, the \(n\)th roots of any complex number form a regular \(n\)-gon around the origin. A reflection across the real axis (a "horizontal flip", as you put it) is equivalent to the mapping \(\xi\mapsto\bar{\xi}\) for any \(z\) in the plane, where \(\bar{\xi}\) is the complex conjugate of \(\xi\). A reflection across the imaginary axis ("vertical flip") is the same as \(\xi\mapsto-\bar{\xi}\). Putting these transformations together yields \(\xi\mapsto-\xi\). So, any of the \(2n\)th roots \(z\) simply become \(-z\), which can be written as \[\large -z=-r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\] On the other hand, a rotation of \(180^\circ\), or \(\pi\text{ rad}\), can be described by writing \(\arg z\) as \(\arg z^*:=\arg z+i\pi\), so we have \[\large z^*=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\] Now we just have to show that \(-z=z^*\): \[\large\begin{align*} -r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right)\\[2ex] -\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=\exp\left(\frac{\theta_0+(2n+2k)\pi}{2n}i\right)\end{align*}\] Take the logarithm of both sides, using the principal branch, i.e. \(\ln \xi=\ln(se^{it})=\ln s+it\). \[\large\begin{align*} \ln(-1)+\frac{\theta_0+2\pi k}{2n}i&=\ln1+\frac{\theta_0+(2n+2k)\pi}{2n}i\\[2ex] i\pi+\frac{\theta_0+2\pi k}{2n}i&=\frac{\theta_0+(2n+2k)\pi}{2n}i\end{align*}\] and you can easily work out that the LHS=RHS.

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