Complex Power Series

From Complex Series to Power Series

In the previous tutorial, convergence of a complex series was defined through the sequence of its partial sums. A power series is a particular kind of complex series: its terms depend on a complex variable through successive powers. The central question is how the coefficients control the set of points where those partial sums converge. The answer is organized by a radius: convergence is guaranteed inside it, while outside it the terms fail to approach zero.

Definition (Complex Power Series): Let \(a_0,a_1,a_2,\ldots\) be complex numbers and let \(z_0\in\mathbb{C}\). The power series with coefficients \(a_n\) and center \(z_0\) is \(\sum_{n=0}^{\infty}a_n(z-z_0)^n\). At a fixed \(z\in\mathbb{C}\), this is a complex series whose \(n\)th term is \(a_n(z-z_0)^n\). The series converges at \(z\) if its partial sums converge in \(\mathbb{C}\).

The center matters because the size of the \(n\)th term depends on \(|z-z_0|\). If \(z=z_0\), all terms with \(n\geq1\) are zero, so the series converges to \(a_0\). Away from the center, both the coefficient sizes and the distance from the center determine convergence.

To express how quickly the coefficients grow, use the limit superior of their \(n\)th roots. Set \(\alpha=\limsup_{n\to\infty}|a_n|^{1/n}\), allowing \(\alpha=+\infty\). The coefficient \(a_0\) does not affect this quantity or convergence away from the center.

Definition (Radius of Convergence): For a power series \(\sum_{n=0}^{\infty}a_n(z-z_0)^n\), its radius of convergence is \(R=1/\alpha\), where \(\alpha=\limsup_{n\to\infty}|a_n|^{1/n}\). We use the conventions \(1/0=+\infty\) and \(1/(+\infty)=0\). Thus \(R\) may be zero, a positive finite number, or \(+\infty\).

The Radius-of-Convergence Theorem

The root growth of the coefficients gives a geometric comparison for the terms at each point. This leads to the Cauchy–Hadamard formula and the basic convergence classification.

Theorem (Cauchy–Hadamard Formula and Convergence Regions): Let \(\alpha=\limsup_{n\to\infty}|a_n|^{1/n}\) and \(R=1/\alpha\) with the conventions above. If \(|z-z_0|<R\), then \(\sum_{n=0}^{\infty}a_n(z-z_0)^n\) converges absolutely. If \(|z-z_0|>R\), then its terms do not tend to zero, so the series diverges. This statement imposes no general conclusion at points where \(|z-z_0|=R\).

Proof. Write \(w=z-z_0\). First suppose \(|w|<R\). If \(w=0\), the series reduces to \(a_0\), so it converges absolutely. Otherwise, the inequality \(|w|<1/\alpha\) implies \(\alpha<1/|w|\). Choose a real number \(q\) with \(\alpha<q<1/|w|\). By the definition of limit superior, \(|a_n|^{1/n}<q\) for all sufficiently large \(n\). For those \(n\),

|a_n w^n| =|a_n|\,|w|^n < (q|w|)^n.

Since \(0\leq q|w|<1\), the geometric series \(\sum_{n=0}^{\infty}(q|w|)^n\) converges. Comparison shows that the tail of \(\sum |a_nw^n|\) converges, and adding its finitely many initial terms preserves convergence. Thus the power series converges absolutely.

Now suppose \(|w|>R\). This case can occur only when \(R\) is finite. The inequality implies \(\alpha>1/|w|\). Choose \(c\) such that \(1/|w|<c<\alpha\). Since \(c\) is below the limit superior, infinitely many \(n\) satisfy \(|a_n|^{1/n}>c\). For each such \(n\),

|a_n w^n| =|a_n|\,|w|^n >(c|w|)^n.

Here \(c|w|>1\), so the right-hand side does not tend to zero along those indices. Therefore the terms \(a_nw^n\) do not tend to zero. By the Term Test for Convergence, proved in the previous tutorial, the series diverges. This also covers \(R=0\): every \(w\neq0\) lies outside the radius. At \(|w|=R\), neither comparison gives a conclusion, so boundary convergence must be checked separately. \(\square\)

The theorem identifies a disk centered at \(z_0\), not necessarily including its boundary, on which absolute convergence is guaranteed. Outside that disk there is divergence. On the boundary circle, the coefficient growth estimate alone is insufficient: some power series converge at boundary points, and others do not.

Worked Example: A Geometric Power Series with a Shifted Center

Consider \(\sum_{n=0}^{\infty}(2-i)^n(z-1)^n\), centered at \(z_0=1\). Its coefficients are \(a_n=(2-i)^n\), so

|a_n|^{1/n}=|2-i|=\sqrt{2^2+(-1)^2}=\sqrt{5} \qquad(n\geq1).

Thus \(\alpha=\sqrt{5}\) and \(R=1/\sqrt{5}\). The series converges absolutely when \(|z-1|<1/\sqrt{5}\) and diverges when \(|z-1|>1/\sqrt{5}\). This is also a geometric series with ratio \(r=(2-i)(z-1)\), since its \(n\)th term is \(r^n\). For example, at \(z=1\), the ratio is zero and the series has sum \(1\). At \(z=1+i\), the ratio is \((2-i)i=1+2i\), whose modulus is \(\sqrt{5}>1\). The terms \(r^n\) then have modulus \((\sqrt{5})^n\), so they do not tend to zero and the series diverges.

Worked Example: A Series That Converges at Every Point on Its Boundary

Consider \(\sum_{n=1}^{\infty}z^n/n^2\), centered at zero. Its coefficients are \(a_n=1/n^2\). Since \((1/n^2)^{1/n}\to1\), we have \(\alpha=1\) and \(R=1\). The radius theorem gives absolute convergence for \(|z|<1\) and divergence for \(|z|>1\).

On the boundary, \(|z|=1\), each term has modulus \[ \left|\frac{z^n}{n^2}\right|=\frac{|z|^n}{n^2}=\frac1{n^2}. \] The real series \(\sum_{n=1}^{\infty}1/n^2\) converges, so the Absolute Convergence Theorem for a Complex Series implies convergence at every boundary point. In fact, the convergence there is absolute. This example shows why the theorem’s boundary clause is necessary: the conclusion on the circle can be more favorable than the outside-divergence result might suggest.

Worked Example: Polynomial Coefficients and Boundary Divergence

Consider \(\sum_{n=0}^{\infty}(n+1)z^n\). For \(n\geq1\), its coefficient root is \((n+1)^{1/n}\to1\), so its radius of convergence is \(R=1\). The series converges absolutely for \(|z|<1\).

If \(|z|=1\), then the modulus of its \(n\)th term is \[ |(n+1)z^n|=(n+1)|z|^n=n+1. \] These term moduli do not tend to zero, so the Term Test shows divergence at every point of the boundary circle. If \(|z|>1\), then \((n+1)|z|^n\geq n+1\), so the terms again fail to tend to zero. Here the boundary diverges, unlike the preceding example, even though both series have the same radius.

Uniform Convergence on Smaller Disks

Pointwise convergence says that for each fixed \(z\), the partial sums eventually approach their limit. A stronger statement is uniform convergence: one choice of how far out in the partial sums works simultaneously for every \(z\) in a specified set. Power series have this stronger property on every closed disk strictly inside their radius.

Theorem (Uniform Convergence Inside the Radius): Let a power series have radius of convergence \(R\). For every \(r\) with \(0\leq r<R\), the series converges uniformly and absolutely on the closed disk \(\{z\in\mathbb{C}:|z-z_0|\leq r\}\).

Proof. If \(r=0\), the disk consists only of \(z_0\), where all terms except \(a_0\) vanish, and the assertion follows. Suppose \(0<r<R\). Choose \(s\) such that \(r<s<R\). By the radius theorem, the power series converges absolutely at the point \(z=z_0+s\), because its distance from \(z_0\) is \(s\). Hence

\sum_{n=0}^{\infty}|a_n|s^n<\infty.

For every \(z\) in the closed disk, \(|z-z_0|\leq r\), and therefore \[ |a_n(z-z_0)^n|\leq |a_n|r^n\leq |a_n|s^n. \] The tails of the convergent nonnegative series \(\sum |a_n|s^n\) tend to zero. Given \(\varepsilon>0\), choose \(N\) such that \[ \sum_{n=N+1}^{\infty}|a_n|s^n<\varepsilon. \] For any \(z\) in the disk and any \(p>q\geq N\), the triangle inequality gives

\left|\sum_{n=q+1}^{p}a_n(z-z_0)^n\right| \leq\sum_{n=q+1}^{p}|a_n|r^n \leq\sum_{n=N+1}^{\infty}|a_n|s^n <\varepsilon.

The bound is independent of \(z\). Thus the tails of the partial sums are uniformly small throughout the disk, which is the uniform Cauchy condition. Since \(\mathbb{C}\) is complete, these partial sums converge at each \(z\), and the same tail estimate shows that the convergence is uniform. The displayed comparison also gives absolute convergence at each point of the disk. \(\square\)

The strict inequality \(r<R\) matters. The proof chooses an intermediate \(s\) between \(r\) and \(R\), and that step is not available when \(r=R\). Uniform convergence up to the boundary is therefore not guaranteed by the radius alone. Nor should a radius be mistaken for a claim about its boundary: as the examples show, the behavior there depends on the particular coefficients.

Using the Radius Correctly

The radius of convergence is a compact summary of coefficient growth, but it does not tell the whole story. Inside the radius, absolute convergence makes the complex tails controllable by a convergent real series. Outside it, the terms fail to approach zero. On the boundary, each point requires a separate argument, such as comparison of moduli or the Term Test.

When calculating a radius, the limit superior handles coefficients that do not have a convenient ratio limit. If the ordinary limit of \(|a_n|^{1/n}\) exists, it may be used in place of the limit superior. But a ratio calculation that suggests a radius should not be used to decide boundary convergence automatically. The examples with coefficients \(1/n^2\) and \(n+1\) both have radius \(1\), yet their boundary behavior is opposite.

Finally, uniform convergence on smaller closed disks is useful because it supplies one tail estimate for all points in the disk, not just a separate estimate at each point. This distinction will matter when studying how power series behave as functions of a complex variable.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What is the radius of convergence in terms of \(\limsup |a_n|^{1/n}\), including the conventions for zero and infinity?
Why does a power series converge absolutely when \(|z-z_0|<R\)? Identify the comparison series used in the proof.
What does the Term Test establish when \(|z-z_0|>R\), and why does it not determine boundary convergence?
For \(\sum_{n=1}^{\infty}z^n/n^2\), what happens when \(|z|=1\), and what estimate proves it?
Why does the uniform convergence proof require a disk of radius \(r<R\), rather than simply taking \(r=R\)?

What You'll Learn