Number Theory: An Introduction to Mathematics

44 I The Expanding Universe of Numbers

discB={z∈C:|z−ζ|≤δ}is contained inGand contains no point ofEexceptζ.If
S={z∈C:|z−ζ|=δ}is the boundary ofB, there exists an open discUwith centre
ωthat contains no point off(S). It follows that ifA={z∈C:0<|z−ζ|<δ}
is the annulusB(S∪{ζ}),thenU{ω}is the union of the disjoint nonempty open
setsU∩{C\f(B)}andU∩f(A).SinceU{ω}is a connected set (because it is
path-connected), this is a contradiction.

From Proposition 29 we readily obtain

Theorem 30If

f(z)=zn+a 1 zn−^1 +···+an

is a polynomial of degree n≥ 1 with complex coefficients a 1 ,...,an,then f(ζ)= 0
for someζ∈C.

Proof Since

f(z)/zn= 1 +a 1 /z+···+an/zn→1as|z|→∞,

we can chooseR>0 so large that

|f(z)|>|f( 0 )| for allz∈Csuch that|z|=R.

Since the closed discD={z∈C:|z|≤R}is compact, the continuous function|f(z)|
assumes its minimum value inDat a pointζin the interiorG={z∈C:|z|<R}.
The functionf(z)is differentiable inGand the setEof all points ofGat which
the derivativef′(z)vanishes is finite. (In factEcontains at mostn−1 points, by
Proposition II.15.) Hence, by Proposition 29, f(G)is an open subset ofC.Since
|f(z)|≥|f(ζ)|for allz∈G, this impliesf(ζ)=0.

The first ‘proof’ of the fundamental theorem of algebra was given by d’Alembert
(1746). Assuming the convergence of what are now called Puiseux expansions, he
showed that if a polynomial assumes a valuew=0, then it also assumes a valuew′
such that|w′|<|w|. A much simpler way of reaching this conclusion, which required
only the existence ofk-th roots of complex numbers, was given by Argand (1814).
Cauchy (1820) gave a similar proof and, with latter-day rigour, it is still reproduced
in textbooks. The proof we have given rests on the same general principle, but uses
neither the existence ofk-th roots nor the continuity of the derivative. These may be
calleddifferential calculus proofs.
The basis for analgebraic proofwas given by Euler (1749). His proof was com-
pleted by Lagrange (1772) and then simplified by Laplace (1795). The algebraic proof
starts from the facts thatRis an ordered field, that any positive element ofRhas a
square root inRand that any polynomial of odd degree with coefficients fromRhas
a root inR. It then shows that any polynomial of degreen≥1 with coefficients from
C=R(i),wherei^2 =−1, has a root inCby using induction on the highest power of
2 which dividesn.
Gauss (1799) objected to this proof, because itassumed that there were ‘roots’ and
then proved that these roots were complex numbers. The difficulty disappears if one