• Complex Analysis: #6 Power Series

Definition 4
A power series is a sum of the form

where (a_n)_n∈ℕ0 is some arbitrary sequence of complex numbers and z₀ is a given complex number. (ℕ₀ is the set of non-negative integers)

So the question is, for which z does the power series converge? Well it obviously converges for z = z₀. But more generally, we can say the following.


Theorem 8
Let the power series ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) be given. Then there exists 0 ≤ R ≤ ∞, the radius of convergence, such that

1. The series is absolutely convergent for |z − z₀| < R, and uniformly convergent for |z − z₀| ≤ ρ, for 0 ≤ ρ < R fixed. 
2. It diverges for |z − z₀| > R.
3. The radius of convergence is given by 1/R = lim_n→∞ sup ⁿ√|a_n|. 
4. The function given by f(z) = ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) is analytic in the region |z − z₀| < R. For each such z, the derivative is given by the series f '(z) = ∑na_n(z − z₀)^n-1, (with n = 1, . . .,∞), and the radius of convergence of this derivative series is also R. 

Proof
Parts 1 and 2 are proved in the analysis lecture. For 3, let |z − z₀| < ρ < R with 1/R = lim_n→∞ sup ⁿ√|a_n|. Thus there exists some N₀ ∈ ℕ with √|a_n| < 1/R for all n ≥ N₀. That is, |a_n| < 1/Rⁿ. Therefore

with ρ/R < 1. This is a geometric series which, as is well known, converges. On the other hand, if |z − z₀| ≥ ρ > R then there exist arbitrarily large n with ⁿ√|a_n|  > 1/R. That is,  |a_n| > 1/Rⁿ or

So the series cannot possibly converge, since the terms of the series do not converge to zero.

As far as part 4 is concerned, it is clear that lim_n→∞ sup ⁿ√|na_n| = lim_n→∞ sup ⁿ√|a_n| since lim_n→∞ sup ⁿ√n = 1. So let f₁(z) = ∑na_n(z − z₀)^n-1, (with n = 1, . . .,∞), be the function which is defined in the region |z − z₀| < R. We must show that f is analytic here, with f ' = f₁. To simplify the notation, let us assume from now on that z₀ = 0. Choose some complex number w with |w| < R. We must show that the derivative of f exists at w, and it equals f₁(w).

To begin with, we write

So take some ρ with |w| < ρ < R and we restrict ourselves to examining complex numbers z with |z| < ρ. Furthermore, choose ∈ > 0. We must show that there exists a δ > 0 such that if 0 < |z −w| < δ then

(Remember that the series is absolutely and uniformly convergent in the closed disc with radius ρ.) Thus for some N₁ ∈ ℕ, the “tail” of the series beyond N₁ sums to something less than ∈ /3. Similarly, the series defining f₁ is absolutely and uniformly convergent in this disc. Therefore take N₂ to be sufficiently large that

for all n ≥ N₂. Let N be the larger of N₁ and N₂. Finally we must determine the number δ. For this, we note that since S_n is just a polynomial, and thus analytic, we have a δ > 0 such that

for all z with |z − w| < δ. In particular, if necessary, we can choose a smaller δ to ensure that such z are in our disc of radius ρ. The fact that

Theorem 9
Let f : G → ℂ be an analytic function defined in a region G, and let z₀ ∈ G be given. Then there exists a unique power series ∑a_n(z − z₀)ⁿ, (with n = 0, . . .,∞) whose radius of convergence is greater than zero, and which converges to f(z) in a neighborhood of z₀.

Proof
Let r > 0 be sufficiently small that B(z₀, r) = {z ∈ ℂ : |z − z₀| < r} ⊂ G. In fact, we will also assume the r is sufficiently small that z ∈ G for all z with |z − z₀| = r. Once again, in order to simplify the notation, we will assume that z₀ = 0. That is to say, we will imagine that we are dealing with the function f(z − z₀) rather than the function f(z). But obviously if the theorem is true for this simplified function, then it is also true for the original function. According to theorem 7, for |z| < r we then have

Here are a few points to think about in this proof.

• The third equation is true since |z/ζ| < 1, and thus the sum is absolutely convergent. 
• The fifth equation is true since the partial sums are uniformly convergent, thus the sum and integral operations can be exchanged. 
• Although the function f(ζ)/ζⁿ⁺¹ is not differentiable at zero, it is defined and continuous on the (compact) circle |ζ| = r. Thus, although c_n is not always zero, still it is always a well defined complex number, for all n. 
• It looks like c_n might vary with r. But this is not the case. Theorem 8 implies that f⁽ⁿ⁾(0) = n!c_n, for all n, and this is certainly independent of r. 
• The power series converges to f(z) at all points of B(z₀, r).