Solution Set 2: Complex Numbers
In our meeting we identified several problem-solving techniques and themes that were very helpful guides:
• The Principle of Mathematical Laziness. This showed up, for example, in computing Taylor Series. We balked at directly multiplying the series for exp(ax) by the series for cos(bx) because it would be a lot of work. This caused us to search for a simpler solution.
• The Principle of Mathematical Laziness also led us to eschew characterizing equilateral triangles in terms of equal side lengths. That’s because calculating and comparing lengths involves lots of sums of squares of differences, which is generally an unappetizing calculation. It also often forces us to look separately at all the components of a complex number, automatically at least doubling the complexity of the calculations. Instead, we were led by this principle to favor comparing angles, because that can be done by suitable multiplications that are easily handled algebraically (see the last problem). Thus the PML is emerging as a kind of aesthetic guide for selecting problem-solving strategies.
• Symmetry. We looked for and exploited symmetry in our exploration of many
of the problems. For instance, at one point we wanted to invert a 4 x 4 matrix and found that using some symmetries led us to rewrite in a way that made the inverse almost immediate to compute. (This exploration is not detailed further below, although it is lurking behind the last demonstration below that roots of unity sum to zero.)
• Problem reduction. In the geometric situations, for instance, we can arbitrarily
choose an origin, a scale, and a direction for the x-axis when setting up a coordinate system. This lets us reduce complex problems to simpler ones with fewer variables.
• Exploitation of transformations. We characterized equilateral triangles in
terms of the transformations (rotations by 120 degrees) that preserve them. We characterized circles as being translations and dilations of a standard circle, the unit circle centered at the origin. Even in the simple warm-up problem #1, we used a transformation (rotation by 180 degrees) to characterize the midpoint of two vertices, leading to a very quick, elegant demonstration.
• Good choice of notation. We consistently resisted using complicated symbols, choosing plain z instead of x + iy for a complex number or ω or ζ to stand in for exp(2πi/n). This seems like a little thing—it is—but it can make a big difference by hiding unnecessary details and revealing the structure of a problem. In
particular, it highlights how much doing algebra with complex numbers is just like doing algebra with real numbers. We might dub this the “Sweep it Under the
• We avoided doing detailed calculations at first, attempting only to see how to get a solution. Once we saw our way, then we paid attention to details.
• We tried to follow up algebraic calculations with geometric interpretations. In general, a good way to check one’s work and one’s understanding is to find another completely different way to arrive at a solution.
Here are the problems and their solutions. Don’t peek until you’ve tried them all yourself! Warm-up #1: Show that the midpoint of z and w is (z + w)/2.
The rotation by 180 degrees around (z + w)/2 is accomplished by translating by -(z + w)/2, rotating (that’s just multiplication by -1), then translating back, giving the transformation
a à (a - (z + w)/2)*(-1) + (z + w)/2 = z + w – a.
Applying this transformation to z gives w and applying it to w gives z, demonstrating their midpoint is (z + w)/2.
Warm-up #2: Show that the circle centered at z (a complex number) of radius r (a positive
real number) consists of the points z + r exp(iθ) where θ ranges from 0 to 2π.
Such a circle is obtained by dilating the unit circle by r and translating it to z. Points on the unit circle are all of the form exp(iθ); dilation is accomplished via multiplication by r, and translation via addition of z, whence the desired expression.
Warm-up #3: Show that the function w à z + exp(iα)(w – z) rotates the complex plane by
α radians around the point z (α is a real number).
The function can be rewritten as w à exp(iα)w + (1 - exp(iα))z. Multiplication by exp(iα) certainly rotates w by α radians. The point z goes to the point z + exp(iα)(z – z) = z + 0 = z, showing that z is the center of the rotation.
Warm-up #4: Let ω = exp(2πi/3). Note that ω 3 = 1. Show that 1 + ω + ω 2 = 0. When p
is a polynomial (with real or complex coefficients), use this identity to show that
[p(x) + p(ωx) + p(ω 2x)]/3 picks out every third term of p (starting with the constant term).
What do you get when you let x = 1?
In general, if ω = exp(2πi/n) for some whole number n > 1, then 1 + ω + ω 2 + … + ω n-1 =
0. There are many demonstrations. Here are some, ordered approximately from the most prosaic to the more far-fetched and surprising.
• 1 + ω + ω 2 + … + ω n-1 is a geometric series. By the usual formula, it sums to
(1 - ω n) / (1 - ω) = (1 – 1)/( 1 - ω) = 0 (because ω ≠ 1 follows from n > 1).
• The points 1, ω, ω 2, …, ω n-1 form the vertices of a regular n-gon inscribed in
the unit circle. Its barycenter (1 + ω + ω 2 + … + ω n-1)/n coincides with the
center of the circle, which is 0. QED. (Not rigorous, but memorable.)
• Because ω n = 1, it follows that (ω 2)n = ω 2n = (ωn)2 = 12 = 1 and similarly (ω k)n =
1 for any integer k. Thus each of 1, ω, ω 2, …, ω n-1 is a root of the polynomial
xn – 1. Clearly (look at a diagram in the Complex plane) all these roots are
distinct; therefore, they are all of the n roots of the polynomial. The sum of the roots of any polynomial (with leading term 1) is the negative of its n-1st term,
which in this case is 0. QED.
• The points 0, 1, 1 + ω, 1 + ω + ω 2, …, 1 + ω + ω 2 + … + ω n-1 describe a
polygonal path in the complex plane. The distance between successive vertices is the length of ωk (k = 0, 1, …, n-1), which is a constant 1. The exterior angle
between successive vertices is the argument of ωk / ωk-1 (k = 0, …, n-1), which is
a constant 2π/n. Therefore this path describes a regular n-gon. After the nth
step, it must return where it began: at 0. Whence 1 + ω + ω 2 + … + ω n-1 = 0.
QED.
• Let z = 1 + ω + ω 2 + … + ω n-1. Note that ωz = ω(1 + ω + ω 2 + … + ω n-1) =
ω + ω 2 + … + ω n = ω + ω 2 + … + ω n-1 + 1 = 1 + ω + ω 2 + … + ω n-1 = z.
Therefore (ω - 1)z = 0. Since ω - 1 is nonzero, z must be zero. QED.
• Write a = 1 + ω + ω 2 + … + ω n-1. Consider the complex polynomial q(z) = az.
Its value at 1 is a while its value at ω is aω = ω + ω 2 + … + ω n = a because
ω n = 1. Since n > 1, 1 and ω are distinct. Therefore q(z) must be a constant (it
has the same value at more distinct points than its degree). What constant is it? We have already seen that it’s a. But since q(0) = a*0 = 0, we must have a = 0, QED.
• Because n > 1, we can form the permutation matrix P = (pij) with pn1 = p12 = p23 =
… = pn-1,n = 1 and all other entries zero. When applied to any n-vector wk of the
form (1, ω k, ω 2k, …, ω (n-1)k)tr it yields Pw
k = (ω k, ω 2k, …, ω (n-1)k, 1)tr = ω kwk
because ω nk = 1. Thus w
0, w1, …, wn-1 are all eigenvectors of P with eigenvalues
1, ω, …, and ω n-1, respectively. Because there are n of them and their
eigenvalues are all distinct, they form a basis. The trace of a matrix is the sum of its diagonal elements. It is, however, a property of the linear transformation: it does not change when the basis is changed. In the w basis the trace of P is the sum of the eigenvalues. In the original basis all diagonal elements are zero, so the trace there is zero. Therefore 1 + ω + ω 2 + … + ω n-1 = 0, QED.
Turn now to the polynomial p of the warm-up problem. Let n be any natural number. Consider the monomial xn. We compute xn + (ωx)n + (ω 2x)n = xn(1 +ω n + ω 2n). When n is
have so copiously shown. Consequently, [xn + (ωx)n + (ω 2x)n]/3 = xn when n is a multiple
of 3 and equals zero otherwise. The assertion about general polynomials (and power series, too) follows immediately. When x = 1, this gives a proposition about the sum of the polynomial’s coefficients, because all powers of x are 1.
Warm-up #5: Show that e axcos(bx) is the real part of exp([a+ib]x). Use this to compute the
first few terms of the Taylor series for e axcos(bx) around x = 0. Rewrite a+ib in polar form
as re iθ (you don’t need to do this explicitly; just remember that such an r and θ always exist)
and recompute the Taylor series in terms of r, θ, and x. What simplifications occur when θ = 0? π? π/2? Are there any other such “nice” values of θ?
exp([a+ib]x) = exp(ax)exp(ibx) = e ax [cos(bx) + isin(bx)] by DeMoivre’s Theorem. Its real
part is e axcos(bx) because e axcos(bx) is real and e axsin(bx) is also real. The Taylor series is
1 + [a+ib]x + ([a+ib]x)2/2! + … + ([a+ib]x)n/n! + …
= 1 + re iθ x + (re iθ x)2/2! + … + (re iθ x)n/n! + …
= 1 + re iθ x + r 2e 2iθ x2/2! + … + r nen iθ xn/n! + … .
The “nice” values of θ are those for which the complex terms en iθ eventually start repeating:
that is, for some n, en iθ = 1. These are the roots of unity.
2004 B4: “Let n be a positive integer, n ≥ 2, and put θ = 2π/n. Define points Pk = (k, 0) in
the xy-plane, for k = 1, 2, . . . , n. Let Rk be the map that rotates the plane counterclockwise
by the angle θ about the point Pk. Let R denote the map obtained by applying, in order, R1,
then R2, . . . , then Rn. For an arbitrary point (x, y), find, and simplify, the coordinates of R(x, y).”
Let ω = exp(2πi/n). In Complex coordinates, the rotation Rk is achieved by translating by
-k, multiplying by ω, then translating back by k. In other words, Rk(z) = ω(z – k) + k.
Define S1(z) = R1(z) and, recursively, Sk(z) = Rk(Sk-1(z)). Thus Sn = Rn. We will prove by
induction on k that Sk(z) = ω k(z) – (ω + ω 2 + … + ω k) + k. This evidently is true for S1,
because ω(z – 1) + 1 = ωz – (ω 1) + 1. Assuming it true for k ≥ 1, we will show it’s true for
k+1 by applying Rk+1:
Sk+1(z) = Rk+1(Sk(z)) = ω (Sk(z) – (k+1)) + k+1
= ω (ω k(z) – (ω + ω 2 + … + ω k) + k – (k+1)) + k+1
= ω k+1(z) – (ω + ω 2 + … + ω k + ω k+1) + k+1,
as required. Letting k = n and using the identities ω n = 1 and ω + ω 2 + … + ω n = 0 gives
1996 A2 : “Let C1 and C2 be circles whose centers are 10 units apart and whose radii are 1
and 3. Find, with proof, the locus of all points M for which there exist points X on C1 and
Y on C2 such that M is the midpoint of line segment XY.”
Let C3 be the circle centered at the midpoint of the centers of C1 and C2 having radius 1 and
let C4 be concentric with C3 with radius 2. The closed annulus bounded by C3 and C4 is the
desired locus.
On the one hand, if x = z + exp(iθ ) is a point on C1 and y = w + 3exp(iφ ) is a point on C2,
then (a) the midpoint of x and y is m = (z + exp(iθ) + w + 3exp(iφ))/2 and (b) the midpoint of the centers is (z + w)/2. The distance between m and the midpoint of the centers is therefore |(exp(iθ) + 3exp(iφ))/2|, which cannot exceed (|exp(iθ)| +|3exp(iφ)|)/2 = (1+3)/2 = 2, and cannot be less than (-|exp(iθ)| +|3exp(iφ)|)/2 = (-1 + 3)/2 = 1. Thus the desired locus lies within the annular region.
There are many ways to show that the annular region lies within the locus. Here’s an elegant and relatively painless way. We can show explicitly that the boundaries C3 and C4 of this
annular region lie in the locus, using points in the form θ =φ (entailing
|(exp(iθ) + 3exp(iφ))/2| = 2| exp(iθ) | = 2) and θ =-φ (giving |(exp(iθ) + 3exp(iφ))/2| = | -2exp(iθ)/2 | = 1). But because the map from [0, 2π] × [0, 2π] given by (θ, φ) à
|(exp(iθ) + 3exp(iφ))/2| is continuous (being a composition of continuous functions), it must attain all possible values between 1 and 2 (for otherwise its image would be
disconnected while its domain is connected). This shows that the annular region is a subset of the locus.
The conclusions of the preceding two paragraphs imply the annular region and the locus are identical, QED.
1974 B6 : Find a closed formula (with proof) for
∑
≡ 3 0 mod k k n(the sum of every third binomial coefficient) where n is any positive integer.
Let p(x) = n n k k x x k n ) ( + =
∑
= 1 0(by the Binomial Theorem). Let ω be a primitive third root
of unity. Then [p(x) + p(ωx) + p(ω 2x)]/3 =
∑
≡ 3 0 mod k k x k n . Consequently∑
≡ 3 0 mod k k n = [p(1) + p(ω1) + p(ω 21)]/3 = [(1+1)n + (1+ω)n + (1+ω 2)n]/3 = [2n + (-ω)n + (-ω 2)n]/3{because 1 + ω + ω 2 = 0}. This is a “closed formula.” However, both -ω and -ω 2 are sixth
roots of unity so we compute the six possible cases to show explicitly it gives a real-valued result (indeed, an integral result) for all n.
n mod 6 (-ω)n + (-ω 2)n 0 2 1 1 2 -1 3 -2 4 -1 5 1
1970 A1 : Show that the Taylor series for e ax cos(bx) around 0 has either none or infinitely many zero coefficients.
Writing a + bi = re iθ yields e ax cos(bx) = the real part of (1 + re iθ x + r 2e 2iθ x2/2! + … +
r nen iθ xn/n! + … ) = 1 + rcos(θ)x + r 2 cos(2θ)x2/2! + … + r n cos(nθ)xn/n! + … .
If for some n, cos(nθ) = 0, then nθ must be an odd multiple of π/2. Consequently, mθ will be an odd multiple of π/2 whenever m is an odd multiple of n, and cos(mθ) will also be zero. Because 1/n! is never zero, there is only one other way a coefficient r ncos(nθ)/n! can be
zero: r must be zero. In either case, if at least one coefficient is zero, infinitely many must be zero. QED.
1967 B1 : A convex hexagon is inscribed in a circle. The length of every other edge equals
the radius of the circle. Prove that the triangle formed by the midpoints of the remaining three edges is equilateral.
Introduce Complex coordinates where 0 is the origin of the circle. the radius has length 2, a vertex of the hexagon is located at the point 2, and as one proceeds counterclockwise around the circle, the other vertices are found at angles of α, α + π/3, β, β + π/3 and γ , in that order. (The angle π/3 arises because the central angle subtended by a chord equal to the radius is always π/3. This follows by observing that the chord along with the two radii terminating at its endpoints forms an equilateral triangle.) These vertices therefore correspond to the complex numbers 2*exp(iα), 2*exp(i[α + π/3]), etc. Introduce the complex number ζ = exp(iπ/3). This convenience lets us rewrite exp(i[α + π/3]) as ζ*exp(iα) and exp(i[β + π/3]) as ζ*exp(iβ). Completing the way around the circle (which we can do because of the convexity of the hexagon), note that 1 = ζ*exp(iγ).
The midpoints of the “irregular” sides are therefore (2ζ*exp(iγ) + 2*exp(iα))/2 =
ζ*exp(iγ) + exp(iα), ζ*exp(iα) + exp(iβ), and ζ*exp(iβ) + exp(iγ). (The symmetry should be apparent! You now see why we chose the radius of the circle to be 2: it cancels the divisions by 2 in the midpoint formula.) The vector from the first to the second vertex is their difference, ζ*exp(iα) + exp(iβ) – [ζ*exp(iγ) + exp(iα)] = (ζ-1)exp(iα) + exp(iβ) - ζ*exp(iγ). To make this look more symmetric, use the identities -ζ = ζ 4 and ζ-1 = ζ 2, giving
Side A = ζ 4exp(iγ) + ζ 2exp(iα) + exp(iβ).
Side B =ζ*exp(iβ) + exp(iγ) – [ζ*exp(iα) + exp(iβ)] = ζ 4exp(iα) + ζ 2exp(iβ) + exp(iγ).
Finally, exploiting the fact that ζ 6 = 1, we compute that multiplying Side A by ζ 2 gives
Side B:
ζ 2[ζ 4exp(iγ) + ζ 2exp(iα) + exp(iβ)] = ζ 6exp(iγ) + ζ 4exp(iα) + ζ 2exp(iβ)
= ζ 4exp(iα) + ζ 2exp(iβ) + exp(iγ).
Geometrically this means that a rotation of 120 degrees (multiplication by ζ 2) makes one
side of the triangle coincide with the other. These sides therefore have equal length and form an exterior angle of 120 degrees. Only equilateral triangles have this property, QED.
