Let p(z)p(z) be a nonconstant polynomial of a complex variable. Then there exists a
complex number z0z0 such that p(z0)=0.p(z0)=0.
That is, every nonconstant polynomial of a complex variable has a root in the complex numbers.
We prove this theorem by contradiction.
Thus, suppose that pp is a nonconstant polynomial of degree n≥1,n≥1, and that p(z)p(z) is never 0.
Set f(z)=1/p(z),f(z)=1/p(z),
and observe that ff is defined and differentiable at every point z∈C.z∈C.
We will show that ff is a constant function, implying that p=1/fp=1/f is a constant,
and that will give the contradiction.
We prove that ff is constant by showing that its derivative is identically 0,
and we compute its derivative by using the Cauchy Integral Formula for the derivative.
From part (4) of (Reference), we recall that there exists a B>0B>0 such that
cn2zn≤p(z),cn2zn≤p(z),
for all zz for which z≥B,z≥B, and where cncn is the (nonzero) leading coefficient of the polynomial p.p.
Hence, f(z)≤Mznf(z)≤Mzn
for all z≥B,z≥B, where we write MM for 2/cn.2/cn.
Now, fix a point c∈C.c∈C.
Because ff is differentiable on the open set U=C,U=C,
we can use the corollary to (Reference) to
compute the derivative of ff at cc by using any of the curves CrCr that bound
the disks Br(c),Br(c),
and we choose an rr large enough so that c+reit≥Bc+reit≥B
for all 0≤t≤2π.0≤t≤2π. Then,

f
'
(
c
)

=

1
2
π
i
∫
C
r
f
(
ζ
)
(
ζ

c
)
2
d
ζ

=
1
2
π

∫
0
2
π
f
(
c
+
r
e
i
t
)
(
c
+
r
e
i
t

c
)
2
i
r
e
i
t
d
t

≤
1
2
π
r
∫
0
2
π

f
(
c
+
r
e
i
t
)

d
t
≤
1
2
π
r
∫
0
2
π
M

c
+
r
e
i
t

n
d
t
≤
M
r
B
n
.

f
'
(
c
)

=

1
2
π
i
∫
C
r
f
(
ζ
)
(
ζ

c
)
2
d
ζ

=
1
2
π

∫
0
2
π
f
(
c
+
r
e
i
t
)
(
c
+
r
e
i
t

c
)
2
i
r
e
i
t
d
t

≤
1
2
π
r
∫
0
2
π

f
(
c
+
r
e
i
t
)

d
t
≤
1
2
π
r
∫
0
2
π
M

c
+
r
e
i
t

n
d
t
≤
M
r
B
n
.
(1)Hence, by letting rr tend to infinity, we get that

f
'
(
c
)

≤
lim
r
→
∞
M
r
B
n
=
0
,

f
'
(
c
)

≤
lim
r
→
∞
M
r
B
n
=
0
,
(2)and the proof is complete.