27  Complex power series

A power series is an infinite series of the form

\[ \sum_{n=0}^\infty a_n z^n = a_0+a_1z+a_2z^2+\cdots, \]

where the variable \(z\) may be a complex number. We can think of a power series as a polynomial with infinitely many terms. As with any infinite series, an important question is whether the partial sums approach a finite limit. Typically, a power series converges when \(z\) lies inside some circle \(|z|<R\) in the complex plane, where \(R\) is called the radius of convergence.

The simplest example is the geometric series

\[ 1+z+z^2+z^3+\cdots=\frac{1}{1-z}, \qquad |z|<1. \]

Thus, inside the unit circle, the function \(1/(1-z)\) can be represented exactly by an infinite sum of powers of \(z\). Truncating the series after finitely many terms gives a polynomial approximation to the function (the MacLaurin series).

Many important functions can be represented by power series. In particular, the complex exponential is defined by

\[ e^z=\sum_{n=0}^\infty\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots. \]

Unlike the geometric series, this series converges for every complex number \(z\)! Substituting imaginary arguments into the exponential series leads directly to the familiar power series for sine and cosine and provides another way of understanding Euler’s formula \(e^{ix}=\cos x+i\sin x\).

Power series are especially useful because they can often be manipulated much like ordinary polynomials. We can add and multiply them, substitute expressions for their variables, and use them to investigate functions that may initially appear problematic. For example, the expressions \((e^z-1)/z\) and \(\sin z/z\) appear to be undefined at \(z=0\), but their power-series representations reveal what happens there and show how the apparent singularity can be removed.

Exercises

Exercise 27.1 (Partial sums of the geometric series) Consider the geometric series

\[ 1+z+z^2+z^3+\cdots. \]

For a finite integer \(N\geq 0\), define the partial sum

\[ S_N(z)=\sum_{n=0}^N z^n=1+z+z^2+\cdots+z^N. \]

1. Write down \(S_0(z)\), \(S_1(z)\), \(S_2(z)\), and \(S_3(z)\).

2. Multiply \(S_N(z)\) by \(z\) and write down the resulting expression.

3. Subtract \(zS_N(z)\) from \(S_N(z)\) and show that

   \[ (1-z)S_N(z)=1-z^{N+1}. \]

4. Hence show, for \(z\neq 1\), that

   \[ S_N(z)=\frac{1-z^{N+1}}{1-z}. \]

Exercise 27.2 (Convergence of the geometric series) From Exercise 27.1,

\[ S_N(z)=\frac{1-z^{N+1}}{1-z}. \]

1. Let \(z=\frac12\). What happens to \(z^{N+1}\) as \(N\to\infty\)? Find the limit of \(S_N(z)\).

2. Let \(z=\frac{i}{2}\). What happens to \(|z^{N+1}|\) as \(N\to\infty\)?

3. Show that if \(|z|<1\), then

   \[ |z^{N+1}|=|z|^{N+1}\longrightarrow 0. \]

4. Conclude that

   \[ \sum_{n=0}^{\infty}z^n=\frac{1}{1-z},\qquad |z|<1. \]

5. What happens when \(z=1\)? What happens when \(z=-1\)? Do these series converge?

6. The condition \(|z|<1\) describes a region in the complex plane. What does this region look like?

Exercise 27.3 (The exponential, sine, and cosine series) The complex exponential is represented by the power series

\[ e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots. \]

Euler’s formula states that, for real \(x\),

\[ e^{ix}=\cos x+i\sin x. \]

1. Substitute \(z=ix\) into the power series for \(e^z\) and write out the first seven terms.

2. Use

   \[ i^0=1,\qquad i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1 \]

   to separate the real and imaginary parts of the series.

3. By comparing with Euler’s formula, show that

   \[ \cos x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} \]

   and

   \[ \sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. \]

4. Write out the first four nonzero terms of each series.

5. Use the series to obtain polynomial approximations to \(\cos x\) and \(\sin x\) containing terms up to and including order \(x^5\).

Exercise 27.4 (Approximating sine and cosine by polynomials) Define the partial sums

\[ C_N(x)=\sum_{n=0}^{N}(-1)^n\frac{x^{2n}}{(2n)!}, \qquad S_N(x)=\sum_{n=0}^{N}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. \]

1. Write down \(C_0(x)\), \(C_1(x)\), and \(C_2(x)\) explicitly.

2. Write down \(S_0(x)\), \(S_1(x)\), and \(S_2(x)\) explicitly.

3. Plot \(\cos x\) together with \(C_0(x)\), \(C_1(x)\), \(C_2(x)\), and \(C_3(x)\) over the interval \(-2\pi\leq x\leq2\pi\).

4. Make a corresponding plot for \(\sin x\) and \(S_0(x)\), \(S_1(x)\), \(S_2(x)\), and \(S_3(x)\).

5. Describe what happens to the quality of the approximation as more terms are included. Is the approximation equally good close to \(x=0\) and far away from \(x=0\)?

Exercise 27.5 (A removable singularity) Consider

\[ f(z)=\frac{e^z-1}{z},\qquad z\neq0. \]

At first sight, this expression is not defined at \(z=0\).

1. Write down the power series for \(e^z-1\).

2. Divide the series by \(z\) and show that, for \(z\neq0\),

   \[ f(z)=1+\frac{z}{2!}+\frac{z^2}{3!}+\frac{z^3}{4!}+\cdots. \]

3. Write the result using summation notation.

4. What value does the power series suggest that we should assign to \(f(0)\)?

5. Define \(f(0)\) using your answer to part d. Explain why the apparent singularity at \(z=0\) is then removed.

Exercise 27.6 (The function \(\sin z/z\)) Consider the function

\[ f(z)= \begin{cases} \dfrac{\sin z}{z}, & z\neq0,\\[4pt] 1, & z=0. \end{cases} \]

1. Start from the power series

   \[ \sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots \]

   and divide by \(z\).

2. Show that

   \[ f(z)=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots. \]

3. Write this series in summation notation.

4. Use the series to explain why the choice \(f(0)=1\) is natural.

5. What is the coefficient of \(z^2\) in the power series? What is the coefficient of \(z^4\)?

6. Show directly from the power series that \(f(-z)=f(z)\). Thus \(f\) is an even function.

Exercise 27.7 (The Cauchy product of power series) Consider two power series

\[ A(z)=\sum_{n=0}^{\infty}a_nz^n,\qquad B(z)=\sum_{n=0}^{\infty}b_nz^n. \]

Their product can be found in the same way as the product of two polynomials, by multiplying terms and collecting equal powers of \(z\).

1. Write out the terms of \(A(z)\) and \(B(z)\) up to order \(z^3\). Multiply the two expressions and show that the first four coefficients of \(A(z)B(z)\) are

   \[ a_0b_0,\qquad a_0b_1+a_1b_0,\qquad a_0b_2+a_1b_1+a_2b_0,\qquad a_0b_3+a_1b_2+a_2b_1+a_3b_0. \]

2. In general, show that the coefficient \(c_n\) of \(z^n\) is

   \[ c_n=\sum_{k=0}^n a_kb_{n-k}. \]

   This multiplication rule is called the Cauchy product.

3. Apply the Cauchy product to two copies of the exponential series

   \[ e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}. \]

   Show that the coefficient of \(z^n\) in \(e^ze^z\) is

   \[ c_n=\sum_{k=0}^n\frac{1}{k!(n-k)!}. \]

4. Use

   \[ \frac{1}{k!(n-k)!}=\frac{1}{n!}\binom{n}{k} \]

   and the identity

   \[ \sum_{k=0}^n\binom{n}{k}=2^n \]

   to simplify \(c_n\).

5. Use your result to show directly from the power series that

   \[ e^ze^z=e^{2z}. \]