T heorem: If P (z) is a polynomial of degree n with complex valued coefficients, then P (z)
has exactly n roots.
Proof. Let P (z) be an arbitrary polynomial of degree n < ∞ and choose R > 0 large enough
such that every root of P (z) lies in B(0, R). Consider the following integral, which counts
the total number of roots of P (z) which lie in B(0, r):
Z
1
P 0 (z)
dz = N
2πi |z|=r P (z)
for any r > R (see lemma). We’ll show that N = n and hence the number of roots of P (z)
is equal to the highest power of P (z). Notice that:
Z
Z
Z
1
1
n
1
1
n
dz = 1 =⇒
dz =
dz = n
n(|z| = r; 0) =
2πi |z|=r z
2πi |z|=r z
2πi |z|=r z
Then, consider the difference between this integral and the one derived above (via lemma):
Z
Z
1
1
P 0 (z)
n
N −n=
dz −
dz =
2πi |z|=r P (z)
2πi |z|=r z
1
=
2πi
Z
|z|=r
P 0 (z) n
1
− dz =
P (z)
z
2πi
Z
|z|=r
zP 0 (z) − nP (z)
dz
zP (z)
Notice that the degree of the numerator is less than n and the degree of the denominator is
n + 1, that is deg(zP (z)) ≥ deg(zP 0 (z) − nP (z)) + 2, then by Jordan’s Lemma (with k = 0
and therefore eikz = 1) it follows that both the upper and lower semi-circles described by
|z| = r goes to 0 as r → ∞, from which we see that:
Z
1
zP 0 (z) − nP (z)
lim (N − n) = lim
dz = 0 =⇒ N = n
r→∞
r→∞ 2πi |z|=r
zP (z)
Then the total number of roots of P (z) is exactly n.
lemma: Let p(z) be a polynomial of degree n and let R > 0 be sufficiently large so that p
never vanishes in {z : |z| ≥ R}. If γ(t) = Reit , 0 ≤ t ≤ 2π, then:
Z 0
p (z)
dz = 2πin
γ p(z)
Q
Proof. To begin, clearly p splits over C, so we can write p(z) = c ni=1 (z − ai ) where c is
some constant and ai ∈ C are the zeroes of the polynomial p. Next, notice that γ(t) = Reit ,
≤ t ≤ 2π is a closed, rectifiable curve and that when R > 0 is large enough that p does not
1
2
vanish in {z : |z| ≥ R} the zeroes, a1 , a2 , . . . an are not points on the path γ and further,
n(γ, ai ) = 1 for each i ∈ [1, n]. Then, differentiating p, we get:
n
d Y
d
0
p (z) = [c (z − ai )] = c [(z − a1 )(z − a2 ) . . . (z − an )] =
dz i=1
dz
= c[(z−a1 )0 (z−a2 ) . . . (z−an )+(z−a1 )(z−a2 )0 . . . (z−an )+. . .+(z−a1 )(z−a2 ) . . . (z−an )0 ] =
= c[1(z −a2 ) . . . (z −an )+(z −a1 )1(z −a3 ) . . . (z −an )+. . .+(z −a1 )(z −a2 ) . . . (z −an−1 )1] =
n
n
n Y
X
Y
X
Y
X
0
= c[ (z − ai )
(z − aj )] = c[
1 (z − aj )] = c[
(z − aj )]
i=1
j6=i
i=1
j6=i
i=1 j6=i
Then, using the above expressions, we have:
Z Pn Q
Z Pn Q
(z − aj )
c[ i=1 j6=i (z − aj )]
i=1
Qn
Qn j6=i
=
=
c i=1 (z − ai )
γ
γ
i=1 (z − ai )
Z
Z
(z − a2 )(z − a3 ) . . . (z − an )
(z − a1 )(z − a2 ) . . . (z − an−1 )
Qn
Qn
=
+ ... +
=
γ
γ
i=1 (z − ai )
i=1 (z − ai )
Z
Z
Z
n
X
1
1
1
=
+
+ ... +
=
2πi n(γ, ai ) = 2πin
γ (z − a1 )
γ (z − a2 )
γ (z − an )
i=1