Complex Analysis: #7 Standard Theorems of Complex Analysis

• Complex Analysis: #7 Some Standard Theorems of Complex Analysis

Combining the last two theorems, we have:

Corollary (Goursat’s Theorem)
The derivative of every analytic function is again analytic. Thus every analytic function has arbitrarily many continuous derivatives.

We can also complete the statement of theorem 3


Theorem 10 (Morera’s Theorem)
Let G ⊂ ℂ be a region and let f : G → ℂ be continuous such that ∫_γ f(z)dz = 0, for all closed paths which are the boundaries of triangles completely contained within G. Then f is analytic.

Proof
According to theorem 3, there exists an antiderivative F : G → ℂ, with F' = f. Thus, by Goursat’s Theorem, f is also analytic.



Theorem 11 (Cauchy’s estimate for the Taylor coefficients)
Again, let f : G → ℂ be analytic, z₀ ∈ G, r > 0 is such that D(z₀, r) = {z : |z − z₀| ≤ r} ⊂ G, and

for all z ∈ D(z₀, r). Since f is continuous and D(z₀, r) is compact, we must have |f| being bounded in D(z₀, r). Let M > 0 be such that |f(z)| ≤ M for all z ∈ D(z₀, r). Then we have

Definition 5
Let the function f : ℂ → ℂ be defined throughout the whole complex plane, and let it be analytic everywhere. Then we say that f is an entire function.


Theorem 12
A bounded entire function is constant.

Proof
Assume that the entire function f : ℂ → ℂ is bounded with |f(z)| ≤ M say, for all z ∈ ℂ, where M > 0 is fixed. Thus |c_n| ≤ M/rⁿ = 0 for all r > 0. This can only be true if c_n = 0 for all n > 0.

Definition 6
A field is called algebraically closed if every polynomial within the field of degree greater than or equal to one has a root.


Theorem 13 (The Fundamental Theorem of Algebra)
ℂ is algebraically closed.

Proof
Let f(z) = ∑a_kz^k, (with k = 0, . . .,n), with n ≥ 1 and a_n ≠ 0 be a polynomial of degree n. Looking for a contradiction, we assume that there is no root, that is, f(z) ≠ 0 for all z ∈ ℂ.

For z ≠ 0, we have

Note that for a and b arbitrary numbers, we have |a| = |a + b − b| ≤ |a + b| + |b| or |a + b| ≥ |a| − |b|, and more generally, |a + b₁ + · · · + b_n| ≥ |a| − |b₁| − · · · − |b_n|.

Since |a_n|/2 remains constant, |zⁿ| · |a_n|/2 becomes arbitrarily large, as |z| → ∞. Therefore |f(z)| → ∞ when |z| → ∞. That is to say, if M > 0 is given, then there exists an r > 0 such that |f(z)| > M for all z with |z| > r. That is, |1/f(z)| < 1/M for |z| > r. Now, since f(z) ≠ 0 always, and f (being a polynomial) is an entire function, we have that 1/f is also an entire function. It is bounded outside the closed disc D(0, r), but since the function is continuous, and D(0, r) is compact, it is also bounded on D(0, r). Thus it is bounded throughout ℂ, and is therefore constant, by theorem 12. Therefore, the polynomial f itself is a constant function. This contradicts the assumption that f is of degree greater than zero.