PROBLEM SHEET FOR WEEKS 3 AND 4 Exercise 15. Consider a

PROBLEM SHEET FOR WEEKS 3 AND 4
UNIVERSITY OF TRENTO – DISCRETE FOURIER ANALYSIS (2012/13) – 14 MARCH 2013
To avoid degeneracies, in the problems about polygons you should assume that the
vertices are distinct (otherwise some of the statements may fail). However, excluding
that any three consecutive vertices are collinear is probably superfluous (and so we may
allow angles of π or 0). Do not assume that the polygons are convex, or that their sides
do not cross each other. One may always assume that the centroid is in the origin.
Exercise 15. Consider a polygon z with k vertices.
Prove that if k is odd and Dr z is regular for some r ≥ 1, then z is regular as well.
Prove that if k is arbitrary and Dr z is regular for some r ≥ 2, then Dz is regular as
well.
Give a geometric construction of a quadrilateral z which is not a square, such that Dz
is a square.
Characterize, in terms of their DFT, all polygons with an even number k of vertices
such that Dz is a regular polygon. Can you also give a geometric construction for them
starting from their derived polygon?
Exercise 16. Use the discrete Fourier transform to prove that the derived polygon of
an arbitrary quadrilateral is a parallelogram.
Exercise 17. Under some natural assumptions, show that a polygon z with k vertices
is symmetric with respect to the origin if and only if k is even and ẑ(a) = 0 for every
even a. (In particular, we must have ẑ(0) = 0, and so the centroid must coincide with
the origin, naturally.)
Hint: You should assume that no vertex is in the origin, otherwise the statement becomes
false. (Why?) Under this assumption k is clearly even.
Now assume that the symmetric of z(a) is z(a + k/2) to prove the desired conclusion.
This last assumption is justified provided the central symmetry preserves the cyclic order of
the vertices. What if it reverses the order, is that possible?
Exercise 18. Suppose that the polygon z has the vertex z(0) on the real axis. Show
that z is symmetric with respect to the real axis if and only if ẑ(a) is real for every a.
Hint: Here the symmetric of z(a) is z(−a).
Remark: If an arbitrary symmetry axis passes through at least one vertex one can always
assume that it is the real axis by applying a suitable rotation. The only case left out here is
that of a polygon with an even number of vertices, none of which is on the real axis.
Exercise 19. Prove that every R-linear map of C to itself can be written in the form
z 7→ αz + β z̄ for uniquely determined α, β ∈ C.
Furthermore, show that z 7→ αz + β z̄ has determinant |α|2 − |β|2 as an R-linear map.
Consequently, it is injective (and hence bijective) if and only if α and β have different
moduli.
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UNIVERSITY OF TRENTO – DISCRETE FOURIER ANALYSIS (2012/13) – 14 MARCH 2013
Hint: For the first part, expand each of z, α, β into real and imaginary parts. Alternatively,
prove that the automorphisms z 7→ z and z 7→ z̄ of the complex field C are C-linearly independent, and then argue by dimension. (With this second proof it is a special case of Artin’s
lemma on independence of characters, a fact which is used in Galois theory.)
Exercise 20. Fix an integer k > 2. Prove that if each polygon of k sides is similar to its
derived polygon, then k = 3.
Hint: If k > 3 one can always produce a polygon z of k sides which is not similar to Dz,
explain how.
Exercise 21. (This problem might be slightly harder than the rest.) In the lectures
we have studied the polygons z which are similar to their derived polygon under the
assumption that the similarity is proper, that is, it preserves orientation. Now change
that assumption to an improper similarity. Analytically, such a similarity is represented
by a map of the form z 7→ αz̄. Moreover, apart from degenerate cases, the similarity
brings vertices of z in ascending order to vertices of Dz in descending order. Reversing
the order of the vertices of a polygon z is achieved by passing to z ◦ (x) := z(−x) (which
is a dilation of a factor −1). Therefore, a polygon z is improperly similar to its derived
polygon if and only if there are y ∈ Z/kZ and α ∈ C such that Dz = αδy ∗ z̄ ◦ . Note that
if k is odd one may assume y = 0 after a renumbering of the vertices, but not if k is even.
As in the case of a proper similarity seen in the lectures, prove that this can only occur
if ẑ(a) = 0 for a 6= ±b, for some b.
The regular polygons, that is, those with ẑ(a) = 0 for a 6= b, are certainly one possibility, and y can be chosen arbitrarily in this case. To see if there are more polygons
assume that ẑ(±b) 6= 0. Assume also that (b, k) = 1. Under these assumptions, prove
that z is improperly similar to Dz if and only if the difference between the (complex)
arguments of ẑ(b) and ẑ(−b) is an odd multiple of π/k (which really means an arbitrary
multiple if k is odd).
Interpret the answer geometrically in the cases k = 3 and k = 4.
Exercise 22. Prove that the polygons z which are (properly) similar to their second
derived polygon D2 z are exactly the images of all regular polygons (with k even or odd)
under arbitrary linear maps.
Remark: Note that for odd k the solutions of this problem coincide with the polygons similar
to their (first) derived polygon, but this is not the case for even k.
Exercise 23. Make rigorous and then prove the following statement:
For a ‘random’ or ‘generic’ polygon, the ‘shape’ of the successive derived
polygons tends to that of a regular polygon.
First of all you should find a sufficient and necessary condition on z (or ẑ) such that
this is really true.
Can you also describe geometrically those polygons for which the statement fails?
Exercise 24. Determine all polygons with an even number k of vertices, such that
all even-numbered sides (for a given cyclic numbering of the sides) can be individually
translated in order to form a polygon of k/2 sides. (The same then clearly holds for the
odd-numbered sides.)
Hint: Your polygons are exactly those such that the alternating sum of all sides (i.e., the
sum of the even-numbered sides minus the sum of the odd-numbered sides) equals zero. You
should find the condition ẑ(k/2) = 0.