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It turns out that if afunction f (z) is differentiable in a neighborhood of a point z0 ∈ C, then it is analyticat that point. This is certainly an amazing fact, since it then follows from the theoremjust proved that if a function f (z) has one derivative f (z) in a neighborhood of apoint, it also has derivatives of all orders in that neighborhood.At first sight this result is just as surprising as the fact that by adjoining to R aroot i of the one particular equation z2 = −1 we obtain a field C in which everyalgebraic polynomial P (z) has a root.
We intend to make use of the fact that analgebraic equation P (z) = 0 has a solution in C, and for that reason we shall prove itas a good illustration of the elementary concepts of complex numbers and functionsof a complex variable introduced in this section.2805Differential CalculusFig. 5.325.5.5 Algebraic Closedness of the Field C of Complex NumbersIf we prove that every polynomial P (z) = c0 +c1 z +· · ·+cn zn , n ≥ 1, with complexcoefficients has a root in C, then there will be no need to enlarge the field C becausesome algebraic equation is not solvable in C.
In this sense the assertion that everypolynomial P (z) has a root establishes that the field C is algebraically closed.To obtain a clear idea of the reason why every polynomial has a root in C whilethere can fail to be a root in R, we use the geometric interpretation of complexnumbers and functions of a complex variable.We remark thatc0c1cn−1P (z) = zn n + n−1 + · · · ++ cn ,zzzso that P (z) = cn zn + o(zn ) as |z| → ∞. Since we are interested in finding a rootof the equation P (z) = 0, dividing both sides of the equation by cn , we may assumethat the leading coefficient cn of P (z) equals 1, and hence P (z) = zn + o zn as |z| → ∞.(5.131)If we recall (Example 15) that the circle of radius r maps to the circle of radius r nwith center at 0 under the mapping z → zn , we see that for sufficiently large valuesof r the image of the circle |z| = r under the mapping w = P (z) will be, with smallrelative error, the circle |w| = r n in the w-plane (Fig.
5.32). What is important isthat, in any case, it will be a curve that encloses the point w = 0.If the disk |z| ≤ r is regarded as a film stretched over the circle |z| = r, thisfilm is mapped into a film stretched over the image of that disk under the mappingw = P (z). But, since the latter encloses the point w = 0, some point of that filmmust coincide with w = 0, and hence there is a point z0 in the disk |z| < r that mapsto w = 0 under the mapping w = P (z), that is, P (z0 ) = 0.This intuitive reasoning leads to a number of important and useful concepts oftopology (the index of a path with respect to a point, and the degree of a mapping),by means of which it can be made into a complete proof that is valid not onlyfor polynomials, as one can see.
However, these considerations would unfortunatelydistract us from the main subject we are now studying. For that reason, we shall giveanother proof that is more in the mainstream of the ideas we have already mastered.5.5 Complex Numbers and Elementary Functions281Theorem 2 Every polynomialP (z) = c0 + c1 z + · · · + cn znof degree n ≥ 1 with complex coefficients has a root in C.Proof Without loss of generality, we may obviously assume that cn = 1.c0Let μ = infz∈C |P (z)|. Since P (z) = zn (1 + cn−1z + · · · + zn ), we haveP (z) ≥ |z|n 1 − |cn−1 | − · · · − |c0 | ,|z||z|nand obviously |P (z)| > max{1, 2μ} for |z| > R if R is sufficiently large. Consequently, the points of a sequence {zk } at which 0 < |P (zk )| − μ < k1 lie inside thedisk |z| ≤ R.We shall verify that there is a point z0 in C (in fact, in this disk) at which|P (z0 )| = μ.
To do this, we remark that if zk = xk + iyk , then max{|xk |, |yk |} ≤|zk | ≤ R and hence the sequences of real numbers {xk } and {yk } are bounded.Choosing first a convergent subsequence {xkl } from {xk } and then a convergentsubsequence {yklm } from {ykl }, we obtain a subsequence zklm = xklm + iyklm ofthe sequence {zk } that has a limit limm→∞ zklm = limm→∞ xklm + i limm→∞ yklm =x0 + iy0 = z0 , and since |zklm | → |z0 | as m → ∞, it follows that |z0 | ≤ R. So as toavoid cumbersome notation, and not have to pass to subsequences, we shall assumethat the sequence {zk } itself converges.
It follows from the continuity of P (z) atz0 ∈ C that limk→∞ P (zk ) = P (z0 ). But then26 |P (z0 )| = limk→∞ |P (zk )| = μ.We shall now assume that μ > 0, and use this assumption to derive a contra0)diction. If P (z0 ) = 0, consider the polynomial Q(z) = PP(z+z(z0 ) . By construction0 )|Q(0) = 1 and |Q(z)| = |P|P(z+z(z0 )| ≥ 1.Since Q(0) = 1, the polynomial Q(z) has the formQ(z) = 1 + qk zk + qk+1 zk+1 + · · · + qn zn ,where |qk | = 0 and 1 ≤ k ≤ n. If qk = ρeiψ , then for ϕ = π−ψk we shall have qk ·(eiϕ )k = ρeiψ ei(π−ψ) = ρeiπ = −ρ = −|qk |. Then, for z = reiϕ we obtain iϕ Q re ≤ 1 + qk zk + qk+1 zk+1 + · · · + qn zn == 1 − r k |qk | + r k+1 |qk+1 | + · · · + |qn |r n−k−1 == 1 − r k |qk | − r|qk+1 | − · · · − r n−k |qn | < 1,26 Observe that on the one hand we have shown that from every sequence of complex numberswhose moduli are bounded one can extract a convergent subsequence, while on the other hand wehave given another possible proof of the theorem that a continuous function on a closed intervalhas a minimum, as was done here for the disk |z| ≤ R.2825Differential Calculusif r is sufficiently close to 0.
But |Q(z)| ≥ 1 for z ∈ C. This contradiction showsthat P (z0 ) = 0.Remark 1 The first proof of the theorem that every algebraic equation with complexcoefficients has a solution in C (which is traditionally known as the fundamentaltheorem of algebra) was given by Gauss, who in general breathed real life into theso-called “imaginary” numbers by finding a variety of profound applications forthem.Remark 2 A polynomial with real coefficients P (z) = a0 + · · · + an zn , as we know,does not always have real roots. However, compared with an arbitrary polynomialhaving complex coefficients, it does have the unusual property that if P (z0 ) = 0,then P (z0 ) = 0 also.
Indeed, it follows from the definition of the complex conjugateand the rules for adding complex numbers that (z1 + z0 ) = z1 + z2 . It follows fromthe trigonometric form of writing a complex number and the rules for multiplyingcomplex numbers that(z1 · z2 ) = r1 eiϕ1 · r2 eiϕ2 = r1 r2 ei(ϕ1 +ϕ2 ) == r1 r2 e−i(ϕ1 +ϕ2 ) = r1 e−iϕ1 · r2 e−iϕ2 = z1 · z2 .Thus,P (z0 ) = a0 + · · · + an z0n = a 0 + · · · + a n zn0 = a0 + · · · + an zn0 = P (z0 ),and if P (z0 ) = 0, then P (z0 ) = P (z0 ) = 0.Corollary 1 Every polynomial P (z) = c0 + · · · + cn zn of degree n ≥ 1 with complexcoefficients admits a representation in the formP (z) = cn (z − z1 ) · · · (z − zn ),(5.132)where z1 , . .
. , zn ∈ C (and the numbers z1 , . . . , zn are not necessarily all distinct).This representation is unique up to the order of the factors.Proof From the long division algorithm for dividing one polynomial P (z) by another polynomial Q(z) of lower degree, we find that P (z) = q(z)Q(z) + r(z), whereq(z) and r(z) are polynomials, the degree of r(z) being less than the degree m ofQ(z). Thus if m = 1, then r(z) = r is simply a constant.Let z1 be a root of the polynomial P (z). Then P (z) = q(z)(z − z1 ) + r, andsince P (z1 ) = r, it follows that r = 0.
Hence if z1 is a root of P (z), we have therepresentation P (z) = (z − z1 )q(z). The degree of the polynomial q(z) is n − 1, andwe can repeat the reasoning with q(z) if n − 1 ≥ 1. By induction we find that P (z) =c(z − z1 ) · · · (z − zn ). Since we must have czn = cn zn , it follows that c = cn .Corollary 2 Every polynomial P (z) = a0 + · · · + an zn with real coefficients can beexpanded as a product of linear and quadratic polynomials with real coefficients.5.5 Complex Numbers and Elementary Functions283Proof This follows from Corollary 1 and Remark 2, by virtue of which for anyroot zk of P (z) the number z̄k is also a root. Then, carrying out the multiplication(z − zk )(z − z̄k ) in the product (5.132), we obtain the quadratic polynomial z2 −(zk + z̄k )z + |zk |2 with real coefficients.
The number cn , which equals an , is a realnumber in this case and can be moved inside one of the sets of parentheses withoutchanging the degree of that factor.By multiplying out all the identical factors in (5.132), we can rewrite that product:P (z) = cn (z − z1 )k1 · · · (z − zp )kρ .(5.133)The number kj is called the multiplicity of the root zj .Since P (z) = (z − zj )kj Q(z), where Q(zj ) = 0, it follows thatP (z) = kj (z − zj )kj −1 Q(z) + (z − zj )kj Q (z) = (z − zj )kj −1 R(z),where R(zj ) = kj Q(zj ) = 0. We thus arrive at the following conclusion.Corollary 3 Every root zj of multiplicity kj > 1 of a polynomial P (z) is a root ofmultiplicity kj − 1 of the derivative P (z).Not yet being in a position to find the roots of the polynomial P (z), we can usethis last proposition and the representation (5.133) to find a polynomial p(z) = (z −z1 ) · · · (z − zp ) whose roots are the same as those of P (z) but are of multiplicity 1.Indeed, by the Euclidean algorithm, we first find the greatest common divisorq(z) of P (z) and P (z).
By Corollary 3, the expansion (5.133), and Theorem 2,the polynomial q(z) is equal, apart from a constant factor, to (z − z1 )k1 −1 · · · (z −zp )kp −1 . Hence by dividing P (z) by q(z) we obtain, apart from a constant factorthat can be removed by dividing out the coefficient of zp , a polynomial p(z) =(z − z1 ) · · · (z − zp ).P (x)of two polynomials, where Q(x) ≡ const.
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