# 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 z

_{0}is a given complex number. (ℕ

_{0}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 = z

_{0}. But more generally, we can say the following.

**Theorem 8**Let the power series ∑a

_{n}(z − z

_{0})

^{n}, (with n = 0, . . .,∞) be given. Then there exists 0 ≤ R ≤ ∞, the radius of convergence, such that

- The series is absolutely convergent for |z − z
_{0}| < R, and uniformly convergent for |z − z_{0}| ≤ ρ, for 0 ≤ ρ < R ﬁxed. - It diverges for |z − z
_{0}| > R. - The radius of convergence is given by 1/R = lim
_{n→∞}sup^{n}√|a_{n}|. - The function given by f(z) = ∑a
_{n}(z − z_{0})^{n}, (with n = 0, . . .,∞) is analytic in the region |z − z_{0}| < R. For each such z, the derivative is given by the series f '(z) = ∑na_{n}(z − z_{0})^{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 − z

_{0}| < ρ < R with 1/R = lim

_{n→∞}sup

^{n}√|a

_{n}|. Thus there exists some N

_{0}∈ ℕ with √|a

_{n}| < 1/R for all n ≥ N

_{0}. That is, |a

_{n}| < 1/R

^{n}. Therefore

with ρ/R < 1. This is a geometric series which, as is well known, converges. On the other hand, if |z − z

_{0}| ≥ ρ > R then there exist arbitrarily large n with

^{n}√|a

_{n}| > 1/R. That is, |a

_{n}| > 1/R

^{n}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 f

_{1}(z) = ∑na

_{n}(z − z

_{0})

^{n-1}, (with n = 1, . . .,∞), be the function which is defined in the region |z − z

_{0}| < R. We must show that f is analytic here, with f ' = f

_{1}. To simplify the notation, let us assume from now on that z

_{0}= 0. Choose some complex number w with |w| < R. We must show that the derivative of f exists at w, and it equals f

_{1}(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 N

_{1}∈ ℕ, the “tail” of the series beyond N

_{1}sums to something less than ∈ /3. Similarly, the series defining f

_{1}is absolutely and uniformly convergent in this disc. Therefore take N

_{2}to be sufficiently large that

for all n ≥ N

_{2}. Let N be the larger of N

_{1}and N

_{2}. 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 z

_{0}∈ G be given. Then there exists a unique power series ∑a

_{n}(z − z

_{0})

^{n}, (with n = 0, . . .,∞) whose radius of convergence is greater than zero, and which converges to f(z) in a neighborhood of z

_{0}.

*Proof*

Let r > 0 be sufficiently small that B(z

_{0}, r) = {z ∈ ℂ : |z − z

_{0}| < r} ⊂ G. In fact, we will also assume the r is sufficiently small that z ∈ G for all z with |z − z

_{0}| = r. Once again, in order to simplify the notation, we will assume that z

_{0}= 0. That is to say, we will imagine that we are dealing with the function f(z − z

_{0}) 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)/ζ
^{n+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(z
_{0}, r).

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