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Lecture 11 — Complex Numbers
Holden Lee
1 Introduction
1.1 Definitions
The set of real numbers R—from the familiar integers—0, 1 , − 1 , 2 , 2 ,... to rational num- bers 12 , 23 ,... to irrational numbers
2 , π, e are familiar to us and have a clear place in the world around us. But math is not quite complete without the complex numbers. In the reals there is no solution to the polynomial equation x^2 = − 1 ,
as the square of any real number is positive. However, we can “create” a solution: let’s adjoin a number to R whose square is −1. Call it i, and let the resulting number system be C. Since we want addition and multiplication to work out, we have to include in C all numbers of the form z = a + bi, where a and b are real.
Definition 1.1: The set of numbers in the form a + bi, where a and b are real, is called the complex numbers C. We call a the real part of z and b the imaginary part, denoted by <(z) = Re(z) and =(z) = Im(z), respectively.
A good way to imagine complex numbers is to think of them as points on a plane (the complex plane), by graphing a + bi as the point (a, b). The x-axis consists of the real numbers a ∈ R, while the y-axis consists of the imaginary numbers bi, b ∈ R. However, the complex numbers have much more structure than just ordered pairs of real numbers—for instance, we can multiply complex numbers as we show below. Let’s see how basic operations work in C. To add or subtract complex numbers, we simply add the real and imaginary components: (1+2i)−(3− 5 i) = (1−3)i+(2−(−5))i = −2 + 7i. We multiply complex numbers by the distributive law, and recall that i^2 = −1: for example (1 + 2i)(3 − 5 i) = 1 · 3 + 1 · − 5 i + 2i · 3 + 2i · − 5 i = 3 + 6i − 10 i^2 = 13 + 6i. To define division, we need the following:
Definition 1.2: The conjugate of a complex number z = a + bi, where a, b are real, is z¯ = a − bi.
Note that the product of a complex number and its conjugate is always real: (a + bi)(a − bi) = a^2 − (bi)^2 = a^2 + b^2.
This allows us to divide complex numbers: to evaluate ac++dibi we multiply both the numer- ator and the denominator by the complex conjugate of c + di, c − di. For example,
1 + 2i 3 − 5 i
(1 + 2i)(3 + 5i) (3 − 5 i)(3 + 5i)
3 + 11i + 10i^2 32 + 5^2
i.
Note that complex conjugation preserves addition and multiplication, i.e. ¯z 1 +¯z 2 = z 1 + z 2 and ¯z 1 z¯ 2 = ¯z 1 z¯ 2. Indeed, let z 1 = a + bi and z 2 = c + di. Then
¯z 1 + ¯z 2 = (a − bi) + (c − di) = (a + c) − (b + d)i = z 1 + z 2. z ¯ 1 z¯ 2 = (a − bi)(c − di) = (ac − bd) − (ad + bc)i = (ac − bd) + (ad + bc)i = z 1 z 2.
The absolute value of z = a + bi is defined by
|z| =
a^2 + b^2.
It is the distance from z to 0 when we plot z in the complex plane. Now negative numbers have square roots too, since (
ai)^2 = −a. With this, we can solve any quadratic equation by completing the square or using the quadratic formula. But what about general polynomials equations? Do we have to adjoin more elements to C so that every cubic polynomial ax^3 + bx^2 + cx + d = 0 where a, b, c, d ∈ C, has a solution? Due to the following miracle, we do not:
Theorem 1.3 (Fundamental Theorem of Algebra): Every nonconstant polynomial with coefficients in C has a zero in C.
(See lecture 8 for a proof.) This shows that C is a very “natural” number system to do algebra over.
1.2 Motivations
So what are complex numbers good for? Who cares about polynomial equations that can’t be solved over the reals—after all, complex solutions don’t make sense in real life, right? Complex numbers were first developed to solve cubic polynomial equations—the for- mula hinged on the use of complex numbers even when the solutions of the equation were real. At first mathematicians were hesitant to adopt complex numbers, but complex numbers soon proved their use in a variety of other ways. A math problem (for exam- ple, a differential equation) might require as an intermediate step to solve a polynomial equation that may not have real solutions. But if we know how to work with complex numbers, we can proceed just as we would in the real case, and the simplified answer may in fact not have complex numbers at all, but would be hard to obtain otherwise. Calculus over the complex numbers turns be much nicer—so that it is often advanta- geous to extend the domain of real functions to the complex numbers, and then look at their properties. This is the case with the infamous zeta function, which has applications to number theory. Many problems that don’t look like they involve complex numbers turn out to be much easier if we use them. For instance, to find Pythagorean triples we want to solve
a^2 + b^2 = c^2
over the integers. We may rewrite this as c^2 − b^2 = a^2 and factor as (c − b)(c + b) = a^2 , and then use number theory. Suppose we want to do the same with
a^3 + b^3 = c^3.
Solution. We can write z = a + 164i where a is real. Then
a + 164i (a + n) + 164i
= 4i.
Clearing the denominator gives
a + 164i = −656 + 4(a + n)i.
Hence matching the real and imaginary parts, a = −656 and 164 = 4(a+n) = 4(−656+n). We get 41 = −656 + n or n = 697.
1.4 Problem Set 1
- [1] Simplify ac++dibi , where a, b, c, d are real and c + di 6 = 0.
- [1] Factor x^2 + y^2 and x^3 + y^3 over the complex numbers.
- [1] Find i^4017.
- [1.5] (AMC 12B, 2004/16) A function f is defined by f (z) = iz¯. How many values of z satisfy |z| = 5 and f (z) = z?
- [1.5] a and b are complex numbers such that x a + b
x a
x b is an identity, true for all values of x. Find all possible values of ab.
- [1.5] (AMC 12A, 2007/18) The polynomial f (x) = x^4 + ax^3 + bx^2 + cx + d has real coefficients, and f (2i) = f (2 + i) = 0. What is a + b + c + d?
- [1.5] (AMC 12B, 2008/19) A function f is defined by f (z) = (4 + i)z^2 + αz + γ for all complex numbers z, where α and γ are complex numbers and i^2 = −1. Suppose that f (1) and f (i) are both real. What is the smallest possible value of |α| + |γ|?
- [1.5] (AIME1 2007/3) The complex number z is equal to 9 + bi, where b is a positive real number. Given that the imaginary parts of z^2 and z^3 are equal, find b.
- [2] (AIME1 2002/12) Let F (z) = zz+−ii for all complex numbers z 6 = i, and let zn = F (zn− 1 ) for all positive integers n. Given that z 0 = 1371 + i and z 2002 = a + bi, where a and b are real, find a + b.
- [2.5] Let Λ∗^ = {a + bi|a, b ∈ Z, (a, b) 6 = 0}. Evaluate ∑
λ∈Λ∗
λ^6
(Note: You may assume the sum is absolutely convergent, i.e. well-defined.)
- [4] (USAMO 1989/3) Let P (z) = zn^ + c 1 zn−^1 + · · · + cn be a polynomial in z, with real coefficients ck. Suppose that |P (i)| < 1. Prove that there exist real numbers a and b such that P (a + bi) = 0 and (a^2 + b^2 + a)^2 < 4 b^2 + 1.
2 Polar form
2.1 Basic Facts and Examples
Consider the complex plane. When we write a complex number as a + bi, we graph the point by going horizontally a units and vertically b units. This is called the rectangular form of the complex number. Addition of complex numbers corresponds nicely to addition of vectors. Conjugation corresponds to reflection over the x-axis (real axis). But what does multiplication and division? To answer this question, we need to write our complex numbers in a different form. We could instead specify a location z on the complex plane by the distance r from 0, and the angle θ made with the positive real axis. Drawing a triangle, we see that
z = r(cos θ + i sin θ)
which we abbreviate as z = r cis θ. This is called the polar form, z is called the modulus, and θ is called the argument. Let’s try to multiply complex numbers in this form. Let z 1 = r 1 cis α and z 2 = r 2 cis β. Then using the addition identities for sine and cosine,
z 1 z 2 = r 1 r 2 (cos α + i sin α)(cos β + i sin β) = r 1 r 2 [(cos α cos β − sin α sin β) + i(cos α sin β + cos β sin α)] = r 1 r 2 [cos(α + β) + i sin(α + β)].
We state this as a theorem.
Theorem 2.1 (De Moivre):
(r 1 cis α)(r 2 cis β) = r 1 r 2 cis(α + β).
In particular, (r cis θ)n^ = rn^ cis(nθ). Using this theorem we can find nth roots of a complex number z, i.e. find the solutions to xn^ = z. Writing z = r cis θ and x = s cis φ, this is equivalent to (s cis φ)n^ = r cis θ. Using De Moivre’s Theorem, this is equivalent to
sn^ cis nφ = r cis θ.
Hence we need sn^ = r and nφ equal to θ as angles. This means they are allowed to differ by a multiple of 2πi. Hence r = n
s and nφ = θ + 2πik
φ =
θ n
2 πik n
for some integer k. Taking k = 0, 1 ,... , n − 1—the n possibilities for k modulo n—give n distinct possible values of φ (other values differ from one of these by a multiple of 2πi). Thus each nonzero complex number has n nth roots. For example, the third roots of i = 8 cis π 2 are 2 cis
(π 6 +^
2 πk 3
for k = 0, 1 , 2, i.e.
2 cis
π 6
i
2 cis
5 π 6
i
2 cis
3 π 2
= −i.
Proof. Since a polynomial of degree at most n − 1 has at most n − 1 zeros, it suffices to show that there are n − 1 pure imaginary roots. Let x = yi and θk(y) = tan−^1 yk. Making this substitution and rewriting in polar form, Pn(x) = 0 is equivalent to
(iy + n)(iy + n − 1) · · · (iy + 1) = (iy − 1)(iy − 2) · · · (iy − n) ⇐⇒ cis θn(y) · · · cis θ 1 (y) = cis(π − θ 1 (y)) · · · cis(π − θn(y)) ⇐⇒ cis(θ 1 (y) + · · · θn(y)) = cis(nπ − (θ 1 (y) + · · · + θn(y))) ⇐⇒ 2(θ 1 (y) + · · · + θn(y)) ≡ nπ (mod 2π). (1)
Note that each θk is a continuous function of y, and that
2(θ 1 (y) + · · · + θn(y)) → − 2 n ·
π 2
= −nπ as y → −∞
2(θ 1 (y) + · · · + θn(y)) → 2 n ·
π 2
= nπ as y → ∞
Thus 2(θ 1 (y) + · · · + θn(y)) attains the value nπ − 2 kπ for k = 1, 2 ,... , n − 1. Hence there are n − 1 values of y such that (1) holds and hence there are n − 1 purely imaginary roots of Pn, as needed.
2.2 Trigonometry
The polar form of complex numbers can be used to simplify trig expressions and prove trig identities.
Example 2.5: Show that cos 3θ = 4 cos^3 θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin^3 θ.
Solution. By De Moivre’s Theorem,
cos 3θ + i sin 3θ = (cos θ + i sin θ)^3 = cos^3 θ + 3 cos^2 θ(i sin θ) + 3 cos θ(i sin θ)^2 + (i sin θ)^3 = (cos^3 θ − 3 cos θ sin^2 (θ)) + (3 cos^2 θ sin θ − sin^3 θ)i
Matching the real and imaginary parts,
cos 3θ = cos^3 θ − 3 cos θ sin^2 (θ) = cos^3 θ − 3 cos θ(1 − cos^2 (θ)) = 4 cos^3 θ − 3 cos θ
and
sin 3θ = 3 cos^2 θ sin θ − sin^3 θ = 3(1 − sin^2 θ) sin θ − sin^3 θ = 3 sin θ − 4 sin^3 θ.
Example 2.6: Show that cos 0◦^ + cos 1◦^ + cos 2◦^ + · · · + cos 89◦^ = 1+cot^.^5
◦
Solution. Let ω = cos 1◦^ + i sin 1◦. Then ωn^ = cos n◦^ + i sin n◦. Hence the desired sum equals the real part of 1 + ω + ω^2 + · · · + ω^89. Using the geometric series formula,
1 + ω + ω^2 + · · · + ω^89 =
ω^90 − 1 ω − 1
i − 1 (cos 1◦^ − 1) + i sin 1◦
=
(i − 1)((cos 1◦^ − 1) − i sin 1◦) (cos 1◦^ − 1)^2 + sin^2 1 ◦
=
(i − 1)((cos 1◦^ − 1) − i sin 1◦) 2 − 2 cos 1◦
=
1 − cos 1◦^ + sin 1◦^ + (cos 1◦^ − 1 + sin 1◦)i 2 − 2 cos 1◦^
This has real part
1 − cos 1◦^ + sin 1◦ 2 − 2 cos 1◦^
sin 1◦ 1 − cos 1◦
1 + cot. 5 ◦ 2
where we used the trig identity cot θ 2 = (^1) −sincos^ θ θ.
2.3 Calculus viewpoint
Using calculus we can write cos θ+i sin θ in a more suggestive form. The Taylor expansions of cos and sin are
cos x = 1 −
x^2 2!
x^4 4!
x^6 6!
sin x = x −
x^3 3!
x^5 5!
x^7 7!
Thus
cos x + i sin x = 1 + ix −
x^2 2!
− i
x^3 3!
x^4 4!
x^5 5!
= 1 + (ix) +
(ix)^2 2!
(ix)^3 3!
(ix)^4 4!
(ix)^5 5!
= eix.
Therefore we can write cis θ as eiθ. Thus De Moivre’s Theorem simply corresponds to the fact that eiθ^1 eiθ^2 = ei(θ^1 +θ^2 )—the familiar rule for exponents.
2.4 Problem set 2
- [1] Show that
cos θ =
eiθ^ + e−iθ 2
sin θ =
eiθ^ − e−iθ 2 i
- [2] Prove that
cos nθ = cosn^ θ −
n 2
cosn−^2 θ sin^2 θ + · · ·
b ∑n/ 2 c
k=
(−1)k
n 2 k
cosn−^2 k^ θ sin^2 k^ θ.
sin nθ =
n 1
cosn−^1 θ −
n 3
cosn−^3 θ sin^3 θ + · · ·
b(n ∑−1)/ 2 c
k=
(−1)k
n 2 k + 1
cosn−^2 k−^1 θ sin^2 k+1^ θ.
- [4] (MOSP 2007) Let a, b 1 ,... , bn, c 1 ,... , cn be real numbers such that x^2 n^ + ax^2 n−^1 + · · · + ax + 1 = (x^2 + b 1 x + c 1 )(x^2 + b 2 x + c 2 ) · · · (x^2 + bnx + cn). Prove that c 1 = c 2 = · · · = cn = 1.
- [5] (APMO 2009/5) Larry and Rob are two robots traveling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorith: Larry makes a 90◦^ left turn after every l kilometers driving from the start; Rob makes a 90◦^ right turn after every r kilometers driving from the start, where l and r are relatively prime positive integers. In the event of both turns occuring simultaneously, the car will keep going straight without changing direction. Assume that the ground is flat and the car can move in any direction. Let the car start from Argovia facing towards Zillis. For which choices of the pair (l, r) is the car guaranteed to reach Zillis, regardless of how far it is from Argovia?
3 Roots of Unity
The nth roots of unity, i.e. roots of 1, satisfy the equation
xn^ − 1 = 0.
The nth roots of unity, not counting 1, satisfy
xn^ − 1 x − 1
= xn−^1 + · · · + x + 1 = 0.
If we plug in x = cis (^2) nπ , the terms of this sum are all the nth roots of unity, and we get Example 2.2. This simple idea—the sum of the nth roots of unity satisfy the equations above, and that their sum is zero—can be very useful.
Example 3.1: n points Q 1 ,... , Qn are equally spaced on a circle of radius 1 centered at O. Point P is on ray OQ 1 so that OP = 2. Find the product
∏^ n
k=
P Qk
in closed form, in terms of n.
Solution. Letting ray OQ 1 be the positive real axis, Qi represent the nth roots of unity ωi in the complex plane. Hence P Qi equals | 2 − ωi|. The roots of xn^ − 1 = 0 are just the nth roots of unity, so xn^ − 1 =
∏n− 1 i=0 (x^ −^ ω i). Plugging in x = 2 gives ∏n k=1 |P Qi|^ = 2 n (^) − 1. Plugging in roots of unity can help in factoring polynomials and establishing divisi- bility, as the following example shows:
Example 3.2: Find all integers n ≥ 1 such that xn+1^ + xn^ + 1 is divisible by x^2 + x + 1.
Solution. In order for xn+1^ + xn^ + 1 to be divisible by x^2 ± x + 1, we must have that all zeros of x^2 + x + 1 are zeros of xn+1^ + xn^ + 1. The zeros are exactly the two third roots of unity not equal to 1. Let ω be such a root of unity. Then we need ωn+1^ + ωn^ + 1 = 0. However, we know ω^3 = 1, so the above equation is equivalent to
ω(n+1) mod 3^ + ωn^ mod 3^ + 1 = 0.
For n ≡ 0 (mod 3) this equals ω + 2, for n ≡ 1 (mod 3) this equals ω^2 + ω + 1 = 0, and for n ≡ 2 (mod 3) this equals ω + 2. Hence the answer is all n with n ≡ 2 (mod 3).
Example 3.3: [MOSP 2007] Let n be a positive integer which is not a prime power. Prove that there exists an equiangular polygon whose side lengths are 1,... , n in some order.
Proof. Since n is not a prime power, we can write n = pq, where p and q are relatively prime. Let ω = cis (^2) nπ. The existence of such an equiangular polygon is equivalent to the existence of a permutation a 1 ,... , an of the numbers 1,... , n such that
∑^ n−^1
k=
akωk^ = 0 (2)
Indeed, given such an equiangular polygon, place it on the complex plane so that the real axis be parallel to one of its sides. Think of the sides of the polygon as vectors forming a loop, and translate them to the origin. Then we get n vectors equally spacced apart; they are in the direction of the nth roots of unity and have lengths 1,... , n in some order. The sum of the vectors must be 0 because they formed a loop. Hence we get the equation above. Conversely, we can easily reverse the above construction. We use the two factors of n to cause “double” cancelation. Note that
(pr + 1)ωrp^ + (pr + 2)ωrp+q^ + · · · + (pr + p)ωrp+(p−1)q^ (3) = prωrp(1 + ωq^ + · · · + ω(p−1)q) + ωrp(1 + 2ω + · · · + pωp−^1 ) = ωrp(1 + 2ωq^ + · · · + pω(p−1)q).
We used the fact 1 + ωq^ + · · · + ω(p−1)q^ = 0 as the terms are the pth roots of unity. Now adding the above expression for r = 1,... , q gives
∑^ q
r=
ωrp(1 + 2ωq^ + · · · + pω(p−1)q) = (1 + ωp^ + · · · + ω(q−1)p)(1 + 2ωq^ + · · · + pω(p−1)q) = 0
becuase 1 + ωp^ + · · · + ω(q−1)p^ = 0 as the terms are the qth roots of unity. However, adding up (3) without simplifying for r = 1,... , q, we get the pq exponents of ω range over the numbers rp + kq, where 1 ≤ r ≤ q and 0 ≤ k ≤ q. Since p, q are relatively prime, these numbers are all distinct modulo n = pq. (If r 1 p + k 1 q = r 2 p + k 2 q then (r 1 − r 2 )p = (k 2 − k 1 )q with |r 1 − r 2 | < q, |k 2 − k 1 | < p so they must equal 0.) Thus taking the exponents modulo pq, we get a sum of the form (2), as needed.
3.1 Problems
- [1] Factor x^5 + x + 1.
- [2] (AIME 1996/11) Let P be the product of the roots of z^6 + z^4 + z^3 + z^2 + 1 = 0 that have positive imaginary part, and suppose that P = r(cos θ◦^ + i sin θ◦) where 0 < r and 0 ≤ θ < 360. Find θ.
- [2.5] (AIME 1997/14) Let v and w be distinct, randomly chosen roots of the equation z^1997 − 1 = 0. Let mn be the probability that
3 ≤ |v + w|, where m and n are relatively prime positive integers. Find m + n.
Solution. If n|a or n|b then the rectangle can obviously be tiled. Now suppose that an a × b rectangle can be tiled by 1 × n blocks. Label the squares of the rectangle (x, y) with 0 ≤ x ≤ a − 1, 0 ≤ y ≤ b − 1. Let ω be a primitive nth root of unity. Label (x, y) with ωx+y. Each 1 × n tile covers numbers of the form ωt, ωt+1,... , ωt+n−^1
for some t; these numbers sum to 0. However the sum of all numbers in the board is
(1 + ω + · · · + ωa−^1 )(1 + ω + · · · + ωb−^1 ) =
ωa^ − 1 ω − 1
ωb^ − 1 ω − 1
This is 0 only if ωa^ = 1 or ωb^ = 1, i.e. n|a or n|b.
Example 4.2 (TST 2004/2): Assume n is a positive integer. Consider sequences a 0 , a 1 ,... , an for which ai ∈ { 1 , 2 ,... , n} for all i and an = a 0.
(a) Call a sequence good if for all i = 1, 2 ,... , n, ai − ai− 1 6 ≡ i (mod n). Suppose that n is odd. Find the number of good sequences.
(b) Call a sequence great if for all i = 1, 2 ,... , n, ai − ai− 1 6 ≡ i, 2 i (mod n). Suppose that n is an odd prime. Find the number of great sequences.
Solution. Let f be a function from { 1 , 2 ,... , n} to the set of subsets of { 0 , 1 ,... , n}. Define a non-f sequence to be a sequence a 0 , a 1 ,... , an such that
- ai ∈ { 1 , 2 ,... , n} for i = 0,... , n,
- an = a 0 , and
- ai − ai− 1 is not congruent to a number in f (i) modulo n.
Claim 4.3: The number of non-f sequences is n times the number of sequences d 1 , d 2 ,... , dn such that
- di ∈ { 0 , 1 , 2 ,... , n − 1 } for each i,
- d 1 + · · · dn ≡ 0 (mod n), and
- di 6 ∈ f (i).
Proof. Given a non-f sequence ai, associate with it a sequence as above, with di = (ai − ai− 1 ) mod n. Each sequence di satisfying the above is associated with n non-f sequences: a 0 can be chosen arbitrarily (there are n choices), and once ai− 1 has been defined, ai must be the unique integer in { 0 , 1 ,... , n − 1 } so that ai ≡ ai− 1 + di (mod n).
Claim 4.4: Let P (x) = 1 + x + · · · + xn−^1 , and define bk such that
∏^ n
i=
P (x) −
j∈f (i)
xj
k≥ 0
bkxk.
Then the number of valid sequences {di}ni=1 is
j≥ 0 bjn.
Proof. Note that P (x) −
j∈f (i) x
j (^) contains only powers of x whose exponents are allow-
able values for di. Take a term in the expansion of the left-hand-side, suppose it takes the term xdi^ from P (x) −
j∈f (i) x
j (^). Then the term is xd 1 +d 2 +···+dn (^). Now {di}n i=1 is a valid sequence iff d 1 + d 2 + · · · + dn ≡ 0 (mod n). Hence summing the coefficients of xk for n|k gives the number of valid sequences.
Combining this with the previous claim, the number of non-f sequences in n
j≥ 0 bjn. (a) A good sequence is exactly a non-f sequence for f (i) = {i mod n}. Let
Q(x) =
∏^ n
i=
P (x) −
j∈f (i)
xj
n∏− 1
i=
[
P (x) − xi
]
Now
Q(1) =
n∏− 1
i=
(P (1) − 1) =
n∏− 1
i=
(n − 1) = (n − 1)n
and for any ω 6 = 1 a nth root of unity,
Q(ω) =
n∏− 1
i=
(P (ω) − ωi)
n∏− 1
i=
(−ωi)
= −ω
n(n−1) (^2) = − 1
since the fact that n is odd gives n|n(n 2 − 1). Now Q(x) + 1 is zero for all nth roots of unity ω 6 = 1, so is divisible by P (x). Writing Q(x) + 1 =
k≥ 0 ckx
k, the fact that P (x)|Q(x) gives that the sums Sr =
k≡r (mod n) ck^ are all equal (fill in the details—basically any term times P (x) is spread among terms whose exponents make up a complete residue class modulo n). Hence, since the sum of coefficients of Q(x) + 1 is Q(1) + 1, ∑
k≥ 0
ckn =
Q(1) + 1
n
(n − 1)n^ + 1 n
Then the desired sum for Q(x) is (n−1)
n+ n −^ 1 and multiplying by^ n^ we get there are (n − 1)n^ − n + 1 good sequences.
(b) A great sequence is a non-f sequence for f (i) = {i mod n, 2 i mod n}. The calcula- tions are left to the reader.
The following lemma will be useful below.
Lemma 4.5: If ω 6 = 1 is pth root of unity, and a 0 ,... , ap− 1 are rational numbers such that a 0 + a 1 ω + · · · + ap− 1 ωp−^1 = 0
then a 0 = · · · = ap− 1.
This lemma follows from the fact that 1 + x +... + xp−^1 is the irreducible (minimal) polynomial of ω, which we will prove in a later lecture.
