Number Theory: An Introduction to Mathematics

5 Complex Numbers 43

f(z)=a 0 zn+a 1 zn−^1 +···+an,

wherea 0 ,a 1 ,...,an∈C,n≥1anda 0 =0, has a complex root. Thus by adjoining
to the real fieldRa root of the polynomialz^2 +1 we ensure that every non-constant
polynomial has a root. Today the fundamental theorem of algebra is considered to be-
long to analysis, rather than to algebra. It is useful to retain the name, however, as a
reminder that our own pronouncements may seem equally quaint in the future.
Our proof of the theorem will use the fact that any polynomial is differentiable,
since sums and products of differentiable functions are again differentiable, and hence
also continuous. We first prove

Proposition 29Let G⊆Cbe an open set and E a proper subset (possibly empty) of
G such that each point of G has a neighbourhood containing at most one point of E.
If f:G→Cis a continuous map which at every point of G\E is differentiable and
has a nonzero derivative, then f(G)is an open subset ofC.

Proof EvidentlyG\Eis an open set. We show first thatf(G\E)is also an open set.
Letζ∈G\E.Thenfis differentiable atζandρ=|f′(ζ)|>0. We can chooseδ> 0
so that the closed discB={z∈C:|z−ζ|≤δ}contains no point ofE, is contained
inGand

|f(z)−f(ζ)|≥ρ|z−ζ|/2foreveryz∈B.

In particular, ifS={z∈C:|z−ζ|=δ}is the boundary ofB,then

|f(z)−f(ζ)|≥ρδ/2foreveryz∈S.

Choosew∈Cso that|w−f(ζ)|<ρδ/4 and consider the minimum in the com-
pact setBof the continuous real-valued functionφ(z)=|f(z)−w|. On the boundary
Swe have

φ(z)≥|f(z)−f(ζ)|−|f(ζ)−w|≥ρδ/ 2 −ρδ/ 4 =ρδ/ 4.

Sinceφ(ζ) < ρδ/4, it follows thatφattains its minimum value inBat an interior
pointz 0 .Sincez 0 ∈/E, we can take

z=z 0 −h[f′(z 0 )]−^1 {f(z 0 )−w},

whereh>0 is so small that|z−ζ|<δ.Then

f(z)−w=f(z 0 )−w+f′(z 0 )(z−z 0 )+o(h)=( 1 −h){f(z 0 )−w}+o(h).

Iff(z 0 )=wthen, for sufficiently smallh>0,

|f(z)−w|≤( 1 −h/ 2 )|f(z 0 )−w|<|f(z 0 )−w|,

which contradicts the definition ofz 0. We conclude that f(z 0 )=w. Thusf(G\E)
contains not onlyf(ζ), but also an open disc{w∈C:|w−f(ζ)|<ρδ/ 4 }surround-
ing it. Since this holds for everyζ∈G\E, it follows thatf(G\E)is an open set.
Now letζ∈Eand assume thatf(G)does not contain any open neighbourhood of
ω:=f(ζ).Thenf(z)=ωfor everyz∈G\E. Chooseδ>0 so small that the closed