9.3. THE FUNDAMENTAL THEOREM OF ALGEBRA 209

where pk (z1, · · · ,zn) is the kth elementary symmetric polynomial. Note this shows

an−k = pk (z1, · · · ,zn)(−1)k , a real number. (9.21)

There is another polynomial which has coefficients which are sums of real numberstimes the pk raised to various powers and it is

qt (x)≡ ∏1≤i< j≤n

(x− (zi + z j + tziz j)) , t ∈ R

I need to verify this is really the case for qt (x). When you switch any two of the zi inqt (x) the polynomial does not change. Thus the coefficients of qt (x) must be symmetricpolynomials in the zi with real coefficients. Hence by Proposition 9.1.4 these coefficientsare real polynomials in terms of the elementary symmetric polynomials pk. Thus by 9.21the coefficients of qt (x) are real polynomials in terms of the ak of the original polynomial.Recall these were all real. It follows, and this is what was wanted, that qt (x) has all realcoefficients.

Note that the degree of qt (x) is

(n2

)because there are this number of ways to pick

i < j out of {1, · · · ,n}. Now for some m,(n2

)=

n(n−1)2

= 2k−1m(

2km−1)= 2k−1 (odd)

and so by induction, for each t ∈ R,qt (x) has a complex root.There must exist s ̸= t such that for a single pair of indices i, j, with i < j,

(zi + z j + tziz j) ,(zi + z j + sziz j)

are both complex. Here is an explanation why. Let A(i, j) denote those t ∈ R such that(zi + z j + tziz j) is complex. It was just shown that every t ∈ R must be in some A(i, j).There are infinitely many t ∈ R and so some A(i, j) contains two of them.

Now for that t,s,

zi + z j + tziz j = a ∈ Czi + z j + sziz j = b ∈ C

where t ̸= s and so by Cramer’s rule,

zi + z j =

∣∣∣∣∣ a tb s

∣∣∣∣∣∣∣∣∣∣ 1 t1 s

∣∣∣∣∣∈ C

and also

ziz j =

∣∣∣∣∣ 1 a1 b

∣∣∣∣∣∣∣∣∣∣ 1 t1 s

∣∣∣∣∣∈ C