To see this, first note that, for any natural numberk,
cos2k
n Cisin2k n
n
Dcos2kCisin2k
by De Moivre’s Theorem (9.2.6). Since cos2kCisin2k D 1, this shows that each of cos2kn Cisin2kn is annthroot of unity. To show that these are the onlynth roots of unity we proceed as follows. Suppose thatzDcosCisinandznD1.
Then cosnCisinn D 1, so cosn D 1and sinn D 0. Thus,n D 2kfor some integerk. It follows that D 2kn . Takingk D 0; 1; : : : ; n1gives thenth roots that we have listed. Taking other values ofkgives different values for 2kn , but each of them differs from one of the listed values by a multiple of2and therefore gives a value for cosCisinthat we already have. Thus, thenroots that we listed are all of thenthroots of unity.
Roots of other complex numbers can also be computed.
Example 9.2.12. All of the solutions of the equationz3 D 1Ci can be found as follows. First note that j1Cij D p
2 and the argument of 1Ci is 4. That is, 1CiDp
2
cos4 Cisin4
. Suppose thatzDr.cosCisin /andz3D1Ci. Thenz3 D r3.cos3Cisin3 /. Thereforer3 D p
2, sor D 216, and3 is 4 or
4 C2or 4 C4. Therefore, itself can be12,34 , or 1712 . This gives the three solutions of the equationz3D1Ci:216
cos12 Cisin12 ,216
cos34 Cisin34 , and216
cos1712 Cisin1712 .
9.3 The Fundamental Theorem of Algebra
One reason for introducing complex numbers was to provide a root for the polynomialx2 C1. There are many other polynomials that do not have any real roots. For example, ifp.x/ is any polynomial, then the polynomial obtained by writing out
p.x/2C1has no real roots, since its value is at least 1 for every value ofx.
Does every such polynomial have a complex root? More generally, does every polynomial have a complex root? There is a trivial sense in which the answer to this question is “no,” since constant polynomials other than 0 clearly do not have any roots of any kind. For other polynomials, the answer is not so simple. It is a remarkable and very useful fact that every non-constant polynomial with real coefficients, or even with complex coefficients, has a complex root.
The Fundamental Theorem of Algebra 9.3.1. Every non-constant polynomial with complex coefficients has a complex root.
80 9 The Complex Numbers
There are a number of different proofs of the Fundamental Theorem of Algebra.
They all rely on mathematical concepts that we do not develop in this book. We will therefore simply discuss implications of this theorem without proving it.
How many roots does a polynomial have?
Example 9.3.2. The only root of the polynomialp.z/Dz26zC9iszD3. This follows from the fact thatp.z/D.z3/.z3/. Since the product of two complex numbers is 0 only if at least one of the numbers is 0, the only solution top.z/D0 isz D 3. In some sense, however, this polynomial has 3 as a “double root”; we’ll discuss this a little more below.
To explore the question of the number of roots that a polynomial can have, we need to use the concept of the division of one polynomial by another. This concept of division is very similar to “long division” of one natural number into another.
Actually, we only need a special case of this concept, the case where the polynomial divisor is linear (i.e., has degree 1). We begin with an example.
Example 9.3.3. To dividez3intoz4C5z32zC1we proceed as follows:
z3 C8z2C24z C70 z3
z4C5z3 2z C1
z4 3z3
8z3 2z C1
8z324z2
24z2 2z C1 24z2 72z
70z C1 70z210 211
What this tabulation shows, like with long division of numbers, is that z4C5z32zC1D.z3/.z3C8z2C24zC70/C211
The only consequence of the division of one polynomial by another that we need for present purposes is the following.
Theorem 9.3.4. Ifr is a complex number andp.z/is a non-constant polynomial with complex coefficients, then there exists a polynomialq.z/and a constantcsuch that
p.z/D.zr/q.z/Cc
9.3 The Fundamental Theorem of Algebra 81
Proof. We will proceed by using the Principle of Complete Mathematical Induction (2.2.1) on the degree of the polynomialp.z/. Sincep.z/is non-constant, the base case of our induction proof is when the degree ofp.z/is 1. In other words,p.z/D azCb, whereaandbare complex numbers anda¤0. Letrbe a complex number.
As in Example9.3.3, we will use long division to dividezrintop.z/:
a zr
az Cb
az ar
arCb
This shows thatp.z/DazCbD.zr/aC.arCb/. So settingq.z/Daand cDarCbgives us the desired result when the degree ofp.z/is 1. Thus, the base case of the induction is complete.
Now assume that the theorem is true for all polynomials of degree less than or equal to n 1. Using this assumption we will show that the theorem holds for any polynomial of degree equal tonC1. Letp.z/be a polynomial with complex coefficients with degreenC1. That is,
p.z/DanC1znC1CanznCan1zn1C Ca1zCa0
where eachaiis a complex number andanC1is nonzero. Letrbe a complex number.
Once again we use long division to dividezrintop.z/:
anC1zn zr
anC1znC1 CanznCan1zn1C Ca1zCa0
anC1znC1 ranC1zn
.anCranC1/znCan1zn1C Ca1zCa0
To simplify the notation, letpn.z/D.anCranC1/znCan1zn1C Ca1zCa0. Then the above long division tells us thatp.z/D.zr/.anC1zn/Cpn.z/. Sincepn.z/is a polynomial of degree less than or equal ton, the induction hypothesis tells us that there exists a polynomialqn.z/and a constantcsuch thatpn.z/D.zr/qn.z/Cc.
Thus,
p.z/ D .zr/.anC1zn/Cpn.z/ D .zr/.anC1zn/C.zr/qn.z/Cc D .zr/
anC1znCqn.z/
Cc Therefore, settingq.z/DanC1znCqn.z/gives us the desired result when the degree ofp.z/isnC1. This completes the proof by induction. ut
82 9 The Complex Numbers
Definition 9.3.5. The polynomialf .z/is afactorof the polynomialp.z/if there exists a polynomialq.z/such thatp.z/Df .z/q.z/.
The Factor Theorem 9.3.6. The complex numberris a root of a polynomialp.z/
if and only if zris a factor ofp.z/.
Proof. If.zr/is a factor ofp.z/, thenp.z/D .zr/q.z/implies thatp.r/ D .r r/q.r/ D 0q.r/ D 0. Conversely, suppose thatr is a root ofp.z/. By Theorem9.3.4,p.z/D.zr/q.z/Ccfor some constantc. Substitutingrforzand using the fact thatris a root gives0D.rr/q.r/Cc, so0D0Cc, from which it follows thatcD0. Hence,p.z/D.zr/q.z/andzris a factor ofp.z/. ut Example 9.3.7. The complex number2i is a root of the polynomializ3Cz24 (as can be seen by simply substituting2iforzin the expression for the polynomial and noting that the result is 0). It follows from the Factor Theorem that z2i is a factor of the given polynomial. Doing “long division” gives iz3 Cz2 4 D .z2i /.iz2z2i /.
We can use the Factor Theorem to determine the maximum number of roots that a polynomial may have.
Theorem 9.3.8. A polynomial of degreenhas at mostncomplex roots; if “multi- plicities” are counted, it has exactlynroots.
Proof. Let p.z/ be a polynomial of degree n. If n is at least 1, then p.z/ has a root, say r1, by the Fundamental Theorem of Algebra (9.3.1). By the Factor Theorem (9.3.6), there exists a polynomialq1.z/such thatp.z/D.zr1/q1.z/. The degree ofq1is clearlyn1. Ifn1 > 0, thenq1.z/has a root, sayr2. It follows from the Factor Theorem that there is a polynomialq2.z/such thatq1.z/D.zr2/q2.z/.
The degree ofq2.z/isn2, and
p.z/D.zr1/.zr2/q2.z/
This process can continue (a formal proof can be given using mathematical induction) until a quotient is simply a constant, sayk. Then,
p.z/Dk.zr1/.zr2/ .zrn/
If theri are all different, the polynomial will haven roots. If some of the ri
coincide, collecting all the terms whereri is equal to a givenrproduces a factor of the form.zr/m, wheremis the number of times thatroccurs in the factorization.
In this situation, we say thatr is aroot of multiplicitymof the polynomial. Thus, a polynomial of degree n has at most n distinct roots. If the roots are counted according to their multiplicities, then a polynomial of degree n has exactly n
roots. ut