208 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

which cannot happen if ln(a) is algebraic according to the above theorem. If a is algebraicand sin(a) ̸= 0, then sin(a) is transcendental because

12i

eia− 12i

e−ia +(−1)sin(a)e0 = 0

which cannot occur if sin(a) is algebraic. There are doubtless other examples of numberswhich are transcendental by this amazing theorem. For example, π is also transcendental.This is because 1+eiπ = 0. This couldn’t happen if π were algebraic because then so wouldbe iπ .

9.3 The Fundamental Theorem of AlgebraThis is devoted to a mostly algebraic proof of the fundamental theorem of algebra. Itdepends on the interesting results about symmetric polynomials which are presented above.I found it on the Wikipedia article about the fundamental theorem of algebra. You google“fundamental theorem of algebra” and go to the Wikipedia article. It gives several otherproofs in addition to this one. According to this article, the first completely correct proofof this major theorem is due to Argand in 1806. Gauss and others did it earlier but theirarguments had gaps in them.

You can’t completely escape analysis when you prove this theorem. The necessaryanalysis due to Bolzano in about 1817 is in the following lemma.

Lemma 9.3.1 Suppose p(x) = xn +an−1xn−1 + · · ·+a1x+a0 where n is odd and the coef-ficients are real. Then p(x) has a real root.

Proof: This follows from the intermediate value theorem from calculus.Next is an algebraic consideration. First recall some notation.

m

∏i=1

ai ≡ a1a2 · · ·am

Recall a polynomial in {z1, · · · ,zn} is symmetric only if it can be written as a sum ofelementary symmetric polynomials raised to various powers multiplied by constants.

The following is the main part of the theorem. In fact this is one version of the funda-mental theorem of algebra which people studied earlier in the 1700’s.

Lemma 9.3.2 Let p(x) = xn +an−1xn−1 + · · ·+a1x+a0 be a polynomial with real coeffi-cients. Then it has a complex root.

Proof: It is possible to write n = 2km. where m is odd. If n is odd, k = 0. If n is even,keep dividing by 2 until you are left with an odd number. If k = 0 so that n is odd, it followsfrom Lemma 9.3.1 that p(x) has a real, hence complex root. The proof will be by inductionon k, the case k = 0 being done. Suppose then that it works for n = 2lm where m is odd andl ≤ k−1 and let n = 2km where m is odd. Let {z1, · · · ,zn} be the roots of the polynomialp(x) in a splitting field, the existence of this field being given by the above proposition. Ineed to show that at least one of these z j is complex. Then

p(x) =n

∏j=1

(x− z j) =n

∑k=0

(−1)k pk (z1, · · · ,zn)xk (9.20)