In formal terms, a complex sequence is a function whose domain is the positive integers and whose range is a subset of the complex numbers. The following are examples of sequences:
\begin{align}
f(n) \amp = \left(2-\frac{1}{n}\right) + \left(5+\frac{1}{n}\right)i \text{ for } n=1,\,2,\,3,\,\ldots;\tag{4.1.1}\\
g(n) \amp = e^{i\frac{\pi n}{4}} \text{ for } n=1,\,2,\,3,\,\ldots;\tag{4.1.2}\\
h(k) \amp = 5+3i + \left(\frac{1}{1+i}\right)^{\!k} \text{ for } k=1,\,2,\,3,\,\ldots \text{ and }\tag{4.1.3}\\
r(n) \amp = \left(\frac{1}{4}+\frac{i}{2}\right)^{\!n} \text{ for } n=1,\,2,\,3,\,\ldots\text{.}\tag{4.1.4}
\end{align}
For convenience, at times we use the term sequence rather than complex sequence. If we want a function \(s\) to represent an arbitrary sequence, we can specify it by writing \(s(1)=z_1, \, s(3)=z_3\text{,}\) and so on. The values \(z_1, \, z_2, \, z_3, \ldots\text{,}\) are called the terms of a sequence, and mathematicians, being generally lazy when it comes to such things, often refer to \(z_1, \, z_2, \, z_3 \ldots\) as the sequence itself, even though they are really speaking of the range of the sequence when they do so. You will usually see a sequence written as \(\{z_n\}_{n=1}^{\infty}, \; \{z_n\}_1^{\infty}\text{,}\) or, when the indices are understood, as \(\{z_n\}\text{.}\) Mathematicians are also not so fussy about starting a sequence at \(z_1\) so that \(\{z_n\} _{n=-1}^{\infty}\text{,}\)\(\{z_k\} _{k=0}^{\infty}\text{,}\)\(\ldots\) would also be acceptable notation provided all terms were defined. For example, the sequence \(r\) given by Equation (4.1.4) could be written in a variety of ways:
The sequences \(f\) and \(g\) given by Equations (4.1.1) and (4.1.2) behave differently as \(n\) gets larger. The terms in Equation (4.1.1) approach \(2+5i=(2,5)\text{,}\) but those in Equation (4.1.2) do not approach any particular number, as they oscillate around the eight eighth roots of unity on the unit circle. Informally, the sequence \(\{z_n\}_1^\infty\) has \(\zeta\) as its limit as \(n\) approaches infinity, provided the terms \(z_n\) can be made as close as we want to \(\zeta\) by making \(n\) large enough. When this happens, we write
\begin{equation}
\lim_{n \to \infty} z_n = \zeta \text{ or } z_n \to \zeta \text{ as } n \to \infty\text{.}\tag{4.1.5}
\end{equation}
If \(\lim_{n \to \infty}z_n = \zeta\text{,}\) we say that the sequence \(\{z_n\}_1^\infty\) converges to \(\zeta\text{.}\)
We need a rigorous definition for Statement (4.1.5), however, if we are to do honest mathematics.
Definition4.1.1.Limit of a Sequence.
\(\lim_{n \to \infty}z_n = \zeta\) means that for any real number \(\varepsilon>0\) there corresponds a positive integer \(N_\varepsilon\) (which depends on \(\varepsilon\)) such that \(z_n \in D_\varepsilon(\zeta )\) whenever \(n>N_\varepsilon\text{.}\) That is, \(|z_n - \zeta|\lt \varepsilon\) whenever \(n>N_\varepsilon\text{.}\)
Remark4.1.2.
The reason that we use the notation \(N_\varepsilon\) is to emphasize the fact that this number depends on our choice of \(\varepsilon\text{.}\) Sometimes, for convenience, we drop the subscript.
Figure4.1.3.A sequence that converges to \(\zeta\)
In form, Definition Definition 4.1.1 is exactly the same as the corresponding definition for limits of real sequences. In fact, a simple criterion casts the convergence of complex sequences in terms of the convergence of real sequences.
Theorem4.1.4.
Let \(z_n=x_n+iy_n\) and \(\zeta =u+iv\text{.}\) Then,
First we assume that Statement (4.1.6) is true and then deduce the truth of Statement (4.1.7). Let \(\varepsilon\) be an arbitrary positive real number. To establish Statement (4.1.7), we must show (1) that there is a positive integer \(N_\varepsilon\) such that the inequality \(|x_n-u|\lt \varepsilon\) holds whenever \(n>N_\varepsilon\) and (2) that there is a positive integer \(M_\varepsilon\) such that the inequality \(|y_n -v|\lt \varepsilon\) holds whenever \(n>M_\varepsilon\text{.}\) Because we are assuming Statement (4.1.6) to be true, we know (according to Definition 4.1.1) that there is a positive integer \(N_\varepsilon\) such that \(z_n \in D_\varepsilon(\zeta)\) if \(n>N_\varepsilon\text{.}\) Recall that \(z_n\in D_\varepsilon(\zeta )\) is equivalent to the inequality \(|z_n - \zeta|\lt \varepsilon\text{.}\) Thus, whenever \(n > N_\varepsilon\text{,}\) we have
Similarly, we can show that there is a number \(M_\varepsilon\) such that \(|y_n-v|\lt \varepsilon\) whenever \(n>M_\varepsilon\text{,}\) which proves Statement (4.1.7).
To complete the proof of this theorem, we must show that the conditions of Statement (4.1.7) imply Statement (4.1.6). Let \(\varepsilon>0\) be an arbitrary real number. By Statement (4.1.7), there exist positive integers \(N_\varepsilon\) and \(M_{\varepsilon}\) such that
We needed to show the strict inequality \(|z_n - \zeta|\lt \varepsilon\text{,}\) and the next-to-last line in the proof gives us precisely that. Note also that we have been speaking of the limit of a sequence. Strictly speaking, we are not entitled to use this terminology because we haven’t proved that a complex sequence can have only one limit. The proof, however, is almost identical to the corresponding result for real sequences, and we leave it as an exercise.
Example4.1.5.
Find \(\lim\limits_{n \to \infty}z_n\) if \(z_n=\frac{\sqrt{n} + i(n+1)}{n}\text{.}\)
Solution.
We write \(z_n = x_n+iy_n = \frac{1}{\sqrt{n}} +i\frac{n+1}{n}\text{.}\) Using results concerning sequences of real numbers, we find that \(\lim\limits_{n \to \infty}x_n = \lim\limits_{n \to \infty} \frac{1}{\sqrt{n}} = 0\) and \(\lim\limits_{n \to \infty}y_n = \lim\limits_{n \to \infty} \frac{n+1}{n} = 1\text{.}\) Therefore, \(\lim\limits_{n \to \infty}z_n = \lim\limits_{n \to \infty} \frac{\sqrt{n} +i(n+1)}{n}=i\text{.}\)
The real sequences \(\{(\sqrt{2})^n \cos \frac{n\pi}{4}\}\) and \(\{(\sqrt{2})^n \sin \frac{n\pi}{4}\}\) both diverge, so we conclude that the sequence \(\{(1+i)^n\}\) diverges.
Definition4.1.7.Bounded Sequence.
A complex sequence \(\{z_n\}\) is bounded provided that there exists a positive real number \(R\) and an integer \(N\) such that \(|z_n|\lt R\) for all \(n>N\text{.}\) In other words, for \(n>N\text{,}\) the sequence \(\{z_n\}\) is contained in the disk \(D_R(0)\text{.}\)
Bounded sequences play an important role in some newer developments in complex analysis that are discussed in Section 4.2. A theorem from real analysis stipulates that convergent sequences are bounded. The same result holds for complex sequences.
Theorem4.1.8.
If \(\{z_n\}\) is a convergent sequence, then \(\{z_n\}\) is bounded.
Proof.
The proof is left as an exercise.
As with the real numbers, we also have the following definition.
Definition4.1.9.Cauchy Sequence.
The sequence \(\{z_n\}\) is a Cauchy sequence if for every \(\varepsilon>0\) there is a positive integer \(N_\varepsilon\)\ such that if \(n,\,m>N_\varepsilon\text{,}\) then \(|z_n-z_m|\lt \varepsilon\text{,}\) or, equivalently, \(z_n - z_m \in D_\varepsilon(0)\text{.}\)
The following theorem should now come as no surprise.
Theorem4.1.10.
If \(\{z_n\}\) is a Cauchy sequence, \(\{z_n\}\) converges.
Proof.
Let \(z_n = x_n+iy_n\text{.}\) Using the techniques of Theorem 4.1.4, we can easily show that both \(\{x_n\}\) and \(\{y_n\}\) are Cauchy sequences of real numbers. Since Cauchy sequences of real numbers are convergent, we know that
for some real numbers \(x_0\) and \(y_0\text{.}\) By Theorem 4.1.4, then, \(\lim_{n \to \infty} z_n = z_0\text{,}\) where \(z_0 = x_0+iy_0\text{.}\) In other words, the sequence \(\{z_n\}\) converges to \(z_0\text{.}\)
One of the most important notions in analysis (real or complex) is a theory that allows us to add up infinitely many terms. To make sense of such an idea we begin with a sequence \(\{z_n\}\text{,}\) and form a new sequence \(\{S_n\}\text{,}\) called the sequence of partial sums, as follows.
The formal expression \(\sum\limits_{k=1}^\infty z_k = z_1 + z_2 + \cdots +z_n + \cdots\) is called an infinite series, and \(z_1, \, z_2, \ldots\) are called the terms of the series.
Definition4.1.12.Convergent Series.
If there is a complex number \(S\) for which \(S=\lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} \sum\limits_{k=1}^n z_k\text{,}\) we say that the infinite series \(\sum\limits_{k=1}^{\infty}z_k\) converges to \(S\text{,}\) and that \(S\) is the sum of the infinite series. When convergence occurs, we write \(S=\sum\limits_{k=1}^{\infty}z_k\text{.}\)
Definition4.1.13.Absolutely Convergent Series.
The series \(\sum\limits_{k=1}^{\infty}z_k\) is said to be absolutely convergent provided that the (real) series of magnitudes \(\sum\limits_{k=1}^{\infty}|z_k|\) converges.
Definition4.1.14.Divergent Series.
If a series does not converge, we say that it diverges.
Remark4.1.15.
The first finitely many terms of a series do not affect its convergence or divergence and, in this respect, the beginning index of a series is irrelevant. Thus, we will without comment conclude that if a series \(\sum\limits_{k=N+1}^\infty z_k\) converges, then so does \(\sum\limits_{k=0}^\infty z_k\text{,}\) where \(z_0, \, z_1, \ldots,
z_N\) is any finite collection of terms. A similar remark applies to determining divergence of a series.
As you might expect, many of the results concerning real series carry over to complex series. We now give several of the more standard theorems for complex series, along with examples of how they are used.
Theorem4.1.16.
Let \(z_n=x_n+iy_n\) and \(S=U+iV\text{.}\) Then
\begin{align*}
S \amp = \sum\limits_{n=1}^\infty z_n=\sum\limits_{n=1}^{\infty}(x_n+iy_n) \text{ iff }\\
U \amp = \sum\limits_{n=1}^\infty x_n \text{ and } V=\sum\limits_{n=1}^{\infty}y_n\text{.}
\end{align*}
Proof.
Let \(U_n=\sum\limits_{k=1}^n x_k\text{,}\)\(V_n=\sum\limits_{k=1}^ny_k\text{,}\) and \(S_n=U_n+iV_n\text{.}\) We use Theorem 4.1.4 to conclude that \(\lim\limits_{n \to \infty}S_n=\lim\limits_{n \to \infty}(U_n+iV_n) = U+iV=S\) iff both \(\lim\limits_{n \to \infty} U_n=U\) and \(\lim\limits_{n \to \infty} V_n=V\text{.}\) The completion of the proof now follows from Definition 4.1.11.
Theorem4.1.17.
If \(\sum\limits_{n=1}^\infty z_n\) is a convergent complex series, then \(\lim\limits_{n \to \infty} z_n=0\text{.}\)
Proof.
The proof is left as an exercise.
Example4.1.18.
Show that the series \(\sum\limits_{n=1}^{\infty}\frac{1+in(-1)^n}{n^2}=\sum\limits_{n=1}^{\infty}\left[\frac{1}{n^2}+i\frac{(-1)^n}{n}\right]\) is convergent.
Solution.
Recall that the real series \(\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\) and \(\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}\) are convergent. Hence, Theorem 4.1.16 implies that the given complex series is convergent.
Example4.1.19.
Show that the series \(\sum\limits_{n=1}^{\infty}\frac{(-1)^n+i}{n} =\sum\limits_{n=1}^{\infty}\left[\frac{(-1)^n}{n} + i\frac{1}{n}\right]\) is divergent.
Solution.
We know that the real series \(\sum\limits_{n=1}^{\infty}\frac{1}{n}\) is divergent. Hence, Theorem 4.1.16 implies that the given complex series is divergent.
Example4.1.20.
Show that the series \(\sum\limits_{n=1}^{\infty}(1+i)^n\) is divergent.
Solution.
Here we set \(z_n=(1+i)^n\) and observe that \(\lim\limits_{n \to \infty}|z_n|= \lim\limits_{n \to \infty}(\sqrt{2})^n =\infty\text{.}\) Thus \(\lim\limits_{n \to \infty}z_n \ne 0\text{,}\) and Theorem 4.1.17 implies that the series is not convergent; hence it is divergent.
Theorem4.1.21.
Let \(\sum\limits_{n=1}^{\infty}z_n\) and \(\sum\limits_{n=1}^{\infty}w_n\) be convergent series and let \(c\) be a complex number. Then
Let \(\sum\limits_{n=0}^{\infty} a_n\) and \(\sum\limits_{n=0}^{\infty}b_n\) be convergent series, where \(a_n\) and \(b_n\) are complex numbers. The Cauchy product of the two series is defined to be the series \(\sum\limits_{n=0}^{\infty}c_n\text{,}\) where \(c_n=\sum \limits_{k=0}^na_kb_{n-k}\text{.}\)
The proof can be found in a number of texts—for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956).
Theorem4.1.24.Comparison test.
Let \(\sum\limits_{n=1}^{\infty}M_n\) be a convergent series of real nonnegative terms. If \(\{z_n\}\) is a sequence of complex numbers and \(|z_n|\le M_n\) holds for all \(n\text{,}\) then the infinite series \(\sum\limits_{n=1}^{\infty}z_n=\sum\limits_{n=1}^{\infty}(x_n+iy_n)\) converges.
Proof.
Using Equations (1.3.2), we determine that \(|x_n|\le |z_n|\le M_n\) and \(|y_n|\le |z_n|\le M_n\) holds for all \(n\text{.}\) By the comparison test for real series, we conclude that \(\sum\limits_{n=1}^{\infty}|x_n|\) and \(\sum\limits_{n=1}^{\infty}|y_n|\) are convergent. An absolutely convergent real series is convergent, so \(\sum\limits_{n=1}^{\infty}x_n\) and \(\sum\limits_{n=1}^{\infty}y_n\) are convergent. With these results, together with Theorem 4.1.16, we conclude that the infinite series \(\sum\limits_{n=1}^{\infty}z_n = \sum\limits_{n=1}^{\infty}x_n +i\sum\limits_{n=1}^{\infty}y_n\) is convergent.
Corollary4.1.25.
If \(\sum\limits_{n=1}^{\infty}|z_n|\) converges, then \(\sum\limits_{n=0}^{\infty}z_n\) converges. In other words, absolute convergence implies convergence for complex series as well as for real series.
Proof.
The proof is left as an exercise.
Example4.1.26.
Show that \(\sum\limits_{n=1}^{\infty} \frac{(3+4i)^n}{5^n n^2}\) converges.
Solution.
We calculate \(|z_n| = \left|\frac{(3+4i)^n}{5^n n^2}\right|=\frac{1}{n^2} = M_n\text{.}\) Using the comparison test and the fact that \(\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\) converges, we determine that \(\sum\limits_{n=1}^{\infty}\left|\frac{(3+4i)^n}{5^n n^2}\right|\) converges, and hence so does \(\sum\limits_{n=1}^{\infty} \frac{(3+4i)^n}{5^n n^2}\text{.}\)
Show that \(\lim\limits_{n \to \infty}(i)^{\frac{1}{n}}=1\text{,}\) where \((i)^{\frac{1}{n}}\) is the principal value of the \(n\)th root of \(i\text{.}\)
3.
Suppose that \(\lim\limits_{n \to \infty}z_n=z_0\text{.}\) Show that \(\lim\limits_{n \to \infty}\overline{z_n}=\overline{z}_0\text{.}\)
Solution.
Let \(\varepsilon >0\) be given. Since \(\lim\limits_{n \to \infty}z_n=z_0\text{,}\) there exists \(N_{\varepsilon}\) such that if \(n>N_{\varepsilon}\) then \(z_n\in D_{e}(z_0)\text{,}\)i.e., \(|z_n-z_0|\lt \varepsilon\text{.}\) But since \(|\bar{z}_n-\bar{z}_0|=|z_n-z_0|\text{,}\) we see that if \(n>N_{\varepsilon}\text{,}\) then \(\bar{z_n}\in D_{\varepsilon}(\bar{z}_0)\text{.}\)
4.
Suppose that the complex series \(\{z_n\}\) converges to \(\zeta\text{.}\) Show that \(\{z_n\}\) is bounded in two ways. \label {4.1.4}
(a)
Write \(z_n = x_n+iy_n\) and use the fact that convergent series of real numbers are bounded.
(b)
For \(\varepsilon=1\text{,}\) use Definitions 4.1.1 and Definition 4.1.7 to show that there is some integer \(N\) such that, for \(n>N\text{,}\)\(|z_n| = |\zeta +(z_n-\zeta)| \le |\zeta| + 1\text{.}\) Then set \(R=\max\{|z_1|, \, |z_2|, \ldots, |z_N|, \, \zeta +1\}\text{.}\)
5.
Show that \(\sum\limits_{n=0}^{\infty}(\frac{1}{n+1+i}-\frac{1}{n+i}) = i\text{.}\)
Solution.
This is a “telescoping sum” and we have for the \(n^{th}\) partial sum \(S_n=-\frac{1}{i}+\frac{1}{n+i}\) (show the details for this). Then \(\lim\limits_{n \to \infty}S_n = \lim\limits_{n \to \infty}(-\frac{1}{i}+\frac{1}{n+i}) = \lim\limits_{n \to \infty}(i+\frac{n-i}{n^2+1}) = i+0 = i\text{.}\)
6.
Suppose that \(\sum\limits_{n=1}^{\infty}z_n=S\text{.}\) Show that \(\sum\limits_{n=1}^{\infty}\overline{z_n}=\overline{S}\text{.}\)
7.
Does \(\lim\limits_{n \to \infty}(\frac{1+i}{\sqrt{2}})^n\) exist? Why or why not?
Solution.
No. In polar form we have \(\lim\limits_{n \to \infty}(e^{i\frac{\pi}{4}})^n = \lim\limits_{n \to \infty}e^{i\frac{n\pi}{4}}\text{.}\) These points oscillate around the eight roots of unity as Example (4.1.2) indicated.
8.
Let \(z_n=r_ne^{i\theta _n}\ne 0\text{,}\) where \(\theta _n = \mathrm{Arg}(z_n)\text{.}\)
(a)
Suppose \(\lim\limits_{n \to \infty}r_n=r_0\) and \(\lim\limits_{n \to \infty}\theta _n=\theta _0\text{.}\) Show that \(\lim\limits_{n \to \infty}r_ne^{i\theta _n}=r_0e^{i\theta_0}\text{.}\)
(b)
Find an example where \(\lim\limits_{n \to \infty}z_n = z_0 = r_0e^{i\theta _0}, \; \lim\limits_{n \to \infty}r_n = r_0\text{,}\) but \(\lim\limits_{n \to \infty}\theta _n\) does not exist.
(c)
If \(\{z_n\} = \{r_ne^{i\theta_n}\}\text{,}\) is it possible to have \(\lim\limits_{n \to \infty}z_n = z_0 = r_0e^{i\theta_0}\text{,}\) but \(\lim\limits_{n \to \infty}r_n\) does not exist?
9.
Show that, if \(\sum\limits_{n=1}^{\infty}z_n\) converges, then \(\lim\limits_{n \to \infty}z_n=0\text{.}\) \hint{\(z_n=S_n-S_{n-1}\text{.}\)}
Solution.
Since \(\sum\limits_{n=1}^{\infty}z_n\) converges, \(\lim\limits_{n \to \infty}S_n=S\text{,}\) where \(S\) is a complex number. But then \(\lim\limits_{n \to \infty}S_{n-1}=S\text{,}\) so \(\lim\limits_{n \to \infty}z_n=\lim\limits_{n \to \infty }(S_n-S_{n-1}) = \lim\limits_{n \to \infty }S_n-\lim\limits_{n \to \infty}S_{n-1}=0\text{.}\)
10.
State (with justification) whether the following series converge.
\(\sum\limits_{n=1}^{\infty}(a+ib) (x_n+iy_n) = \sum\limits_{n=1}^{\infty}[(ax_n-by_n)+i(bx_n+ay_n)]\text{.}\) By Theorem 4.1.16, this expression equals \((au-bv)+i(bu+av) = (a+ib)(u+iv)\text{.}\) Explain why in detail.
12.
If \(\sum\limits_{n=0}^{\infty}z_n\) converges, show that \(\left|\sum\limits_{n=0}^{\infty}z_n\right| \le \sum\limits_{n=0}^{\infty}|z_n|\text{.}\)
13.
Complete the proof of Theorem 4.1.4. In other words, suppose that \(\lim\limits_{n \to \infty}z_n=\zeta\text{,}\) where \(z_n=x_n+iy_n\) and \(\zeta = u+iv\text{.}\) Prove that \(\lim\limits_{n \to \infty}y_n=v\text{.}\)
Solution.
Duplicate the part of the Theorem that shows \(\lim\limits_{n \to \infty}x_n=u\text{,}\) but replace \(x_n\) with \(y_n\) and \(u\) with \(v\text{.}\)
14.
A side comment asked you to justify the first inequality in the proof of Theorem 4.1.4. Give a justification.
15.
Prove that a sequence can have only one limit. \hint{Suppose that there is a sequence \(\{z_n\}\) such that \(z_n \to \zeta _1\) and \(z_n \to \zeta _2\text{.}\) Show this assumption implies \(\zeta _1=\zeta _2\) by proving that for all \(\varepsilon>0\text{,}\)\(|\zeta _1-\zeta _2| \lt \varepsilon\text{.}\)}
Solution.
Following the hint, for \(\varepsilon >0\) there exists numbers \(N_{\varepsilon}\) and \(M_{\varepsilon}\) such that \(n>N_{\varepsilon}\) implies \(|z_n-\zeta _1|\lt \frac{ \varepsilon}{2}\text{,}\) and \(n>M_{\varepsilon}\) implies \(|z_n-\zeta_2|\lt \frac{\varepsilon}{2}\text{.}\) Let \(L_{\varepsilon}=\max\{N_{\varepsilon},\,M_{\varepsilon}\}\text{.}\) It follows that if \(n>L_{\varepsilon}\text{,}\) then \(|\zeta _1-\zeta _2|=|\zeta _1-z_n+z_n-\zeta _2| \le |\zeta _1-z_n|+|z_n-\zeta _2|\lt \varepsilon\text{.}\)
Prove that \(\lim\limits_{n \to \infty}z_n=0\) iff \(\lim\limits_{n \to \infty}|z_n|=0\text{.}\)
Solution.
Let \(\varepsilon>0\) and suppose \(\lim\limits_{n \to \infty}z_n=0\text{.}\) This means there exists \(N_{\varepsilon}\) such that \(n>N_{\varepsilon}\) implies \(z_n \in D_{\varepsilon}(0)\text{,}\) that is, \(|z_n-0|\lt \varepsilon\text{.}\) But then \(||z_n|-0|=|z_n-0|\lt \varepsilon\text{,}\) so also we have \(|z_n|\in D_{\varepsilon}(0)\text{.}\) Therefore, \(\lim\limits_{n \to \infty}|z_n|=0\text{.}\) The other direction is similar. Show the details.
