7  Sequences in \(\mathbb{C}\)

The theory for sequences in \(\mathbb{C}\) is very similar to that of sequences in \(\mathbb{R}\). In \(\mathbb{R}\), we said that a sequence \((a_n)\) converges to some number \(a \in \mathbb{R}\) if for all \(\varepsilon>0\), it holds \[ |a_n - a| < \varepsilon \] for all \(n\) suffieciently large. The definition of convergence in \(\mathbb{C}\) is essentially the same, with the absolute value replaced by the complex modulus.

7.1 Definition and convergence

First of all, let us give the formal definition of sequence in \(\mathbb{C}\).

Definition 1: Sequence of Complex numbers
A sequence \(a\) in \(\mathbb{C}\) is a function \[ a \colon \mathbb{N} \to \mathbb{C} \,. \] For \(n \in \mathbb{N}\), we denote the \(n\)-th element of the sequence \(a\) by \[ a_{n}=a(n) \] and write the sequence as \[ \left(a_{n}\right)_{n \in \mathbb{N}} \quad \mbox{ or } \quad (a_n) \,. \]

In the following we define convergent sequences in \(\mathbb{C}\).

Definition 2: Convergent sequence in \(\mathbb{C}\)
We say that a sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) in \(\mathbb{C}\) converges to \(a \in \mathbb{C}\), or equivalently has limit \(a\), denoted by \[ \lim_{n \rightarrow \infty} a_{n}=a \quad \mbox{ or } \quad a_n \to a \,, \] if for all \(\varepsilon \in \mathbb{R}\) with \(\varepsilon>0\), there exists \(N \in \mathbb{N}\) such that for all \(n \in \mathbb{N}, n \geq N\) it holds that \[ \left|a_{n}-a\right| < \varepsilon \,. \] Using quantifiers, we can write this as \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\left|a_{n}-a\right| < \varepsilon \,. \]

If there exists \(a \in \mathbb{C}\) such that \(\lim _{n \rightarrow \infty} a_{n}=a\), then we say that the sequence \(\left(a_{n}\right)_{n \in \mathbb{N}}\) is convergent.

Important
In Definition 2 we still take \(\varepsilon\) to be real. This makes sense, since \[ |z| = \sqrt{ x^2 + y^2 } \in \mathbb{R} \] for all \(z = x + i y \in \mathbb{C}\).

Example 3
Using Definition 2, prove that \[ \lim_{n \rightarrow \infty} \frac{(3+i) n-7 i}{n}=3+i \,. \]

Part 1. Rough Work. Let \(\varepsilon >0\). We would like to understand for which values of \(n\) the following holds: \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| < \varepsilon \,. \] We have \[\begin{align*} \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| & = \left|\frac{(3+i) n-7 i-(3+i) n}{n}\right| \\ & = \frac{|-7 i|}{n} \\ & = \frac{7}{n} \,.\\ \end{align*}\] Therefore \[ \frac{7}{n} < \varepsilon \quad \iff \quad n > \frac{7}{\varepsilon} \,. \]

Part 2. Formal Proof. We want to prove that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\, \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| < \varepsilon \,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{7}{\varepsilon} \,. \] The above is equivalent to \[ \frac{7}{N} < \varepsilon \,. \] For \(n \geq N\) we have \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| = \frac{7}{n} \leq \frac{7}{N} < \varepsilon \,. \]

7.2 Boundedness

Boundedness plays an important role for complex sequences.

Definition 4: Bounded sequence in \(\mathbb{C}\)
A sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) is called bounded if there exists a constant \(M \in \mathbb{R}\), with \(M>0\), such that \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N} \,. \]

As it happens in \(\mathbb{R}\), we have that complex sequences which converge are also bounded.

Theorem 5
If a sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) converges, then the sequence is bounded.

The proof is identical to the one in \(\mathbb{R}\), and is hence omitted. Similarly to real sequences, we can define divergent complex sequences.

Definition 6: Divergent sequences in \(\mathbb{C}\)
We say that a sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) is divergent if it is not convergent.

As a corollary of Theorem 5 we have the following.

Corollary 7
Let \(\left(a_{n}\right)\) be a complex sequence. If \(\left(a_{n}\right)\) is not bounded, then it is divergent.

7.3 Algebra of limits in \(\mathbb{C}\)

Most of the results about limits that we have shown in \(\mathbb{R}\) also hold in \(\mathbb{C}\). The first result is the Algebra of Limits.

Theorem 8: Algebra of limits in \(\mathbb{C}\)

Let \(\left(a_{n}\right)\) and \(\left(b_{n}\right)\) be sequences in \(\mathbb{C}\). Suppose that \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \,, \] for some \(a,b \in \mathbb{C}\). Then,

  1. Limit of sum is the sum of limits: \[ \lim_{n \rightarrow \infty}\left(a_{n} \pm b_{n}\right)=a \pm b \]

  2. Limit of product is the product of limits: \[ \lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right) = a b \]

  3. If \(b_{n} \neq 0\) for all \(n \in \mathbb{N}\) and \(b \neq 0\), then \[ \lim_{n \rightarrow \infty} \left(\frac{a_{n}}{b_{n}}\right)=\frac{a}{b} \]

The proof of Theorem 8 follows word by word the proof of the Algebra of Limits for sequences in \(\mathbb{R}\): one just needs to replace the absolute value by the complex modulus.

We can use the Algebra of Limits to compute limits of complex sequences.

Example 9
Let \(\left(a_{n}\right)\) be the sequence defined by \[ a_{n}:=\frac{(2-i) n^{2}+6 i n-5-3 i}{(6+3 i) n^{2}+11 i} \,. \] The largest power of \(n\) in the denominator is \(n^{2}\). We hence divide by \(n^{2}\) to obtain \[ a_{n}=\frac{(2-i) n^{2}+6 i n-5-3 i}{(6+3 i) n^{2}+11 i}= \frac{(2-i)+\dfrac{6 i}{n}- \dfrac{5}{n^2}- \dfrac{3 i}{ n^{2}} }{(6+3 i)+\dfrac{11 i}{n^{2}}} \] By the Algebra of Limits the right hand side converges to \[ \frac{(2-i)+0-0-0}{(6+3 i)+0}=\frac{2-i}{6+3 i} \] By performing the complex division, we can write the limit in the form \(x + i y\): \[\begin{align*} \lim_{n \rightarrow \infty} a_{n} & =\frac{2-i}{6+3 i} \\ & =\frac{(2-i)(6-3 i)}{(6+3 i)(6-3 i)} \\ & =\frac{12-6 i-6 i+3 i^{2}}{36-9 i^{2}} \\ & =\frac{9-12 i}{45} \\ & =\frac{1}{5}-\frac{4}{15} i \end{align*}\]

7.4 Convergence to zero

One of the results that cannot hold for complex sequences is the Squeeze Theorem. Indeed the chain of inequalities \[ b_{n} \leq a_{n} \leq c_{n} \] would not make sense in \(\mathbb{C}\), since there is no order relation.

We can however prove the following (weaker) result.

Theorem 10
Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\) and suppose that \[ \lim_{n \rightarrow \infty}\left|a_{n}\right|=0 \,. \] Then \[ \lim_{n \rightarrow \infty} a_{n}=0 \,. \]

Proof
Assume that \(|a_n| \to 0\). We need to show that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\, \left|a_{n}-0\right| < \varepsilon \,. \] Let \(\varepsilon>0\). Since \(\left|a_{n}\right| \rightarrow 0\), there exists \(N \in \mathbb{N}\) such that \[ \left||a_{n}|-0 \right| < \varepsilon \,, \quad \forall \, n \geq N \,. \] Let \(n \geq N\). Then, \[\begin{align*} \left|a_{n}-0\right| & = \left|a_{n}\right| \\ & = \left|a_{n}\right|-0 \\ & = || a_{n}|-0| \\ & < \varepsilon \,. \end{align*}\]

Note that the sequence \(|a_n|\) is real. Therefore the convergence of \(|a_n|\) can be studied using convergence results in \(\mathbb{R}\).

Example 11

Consider the complex sequence \[ a_{n}=\left(\frac{1}{2}+\frac{1}{3} i\right)^{n} \,. \] Prove that \(a_n \to 0\).

Proof. We have \[\begin{align*} \left|a_{n}\right| & = \left|\left(\frac{1}{2}+\frac{1}{3} i\right)^{n}\right| \\ & = \left|\frac{1}{2}+\frac{1}{3} i\right|^{n} \\ & = \left(\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}}\right)^{n} \\ & = \left(\sqrt{\frac{13}{36}}\right)^{n} \,. \end{align*}\] Since \[ \left|\sqrt{\frac{13}{36}} \right| < 1 \,, \] by the Geometric Sequence Test for real sequences, we conclude that \[ \left|a_{n}\right| \rightarrow 0 \,. \] Hence \(a_n \to 0\) by Theorem 10.

Although the Squeeze Theorem cannot be used for complex sequences, sometimes it can be used to deal with real terms in a complex sequence.

Example 12

Consider the sequence \[ a_{n}:=\frac{2 i \cos (3 n) n+(7-i) n^{2}}{3 n^{2}+2 i n+\sin (2 n)} \,. \] Prove that \[ \lim_{n \to \infty} a_n = \frac{7}{3}-\frac{1}{3} i \,. \]

Proof. We divide by the largest power in the denominator, to get \[ a_{n} = \frac{\dfrac{2 i \cos (3 n)}{ n} + (7-i) }{3+ \dfrac{2 i}{n} + \dfrac{\sin (2 n)}{n^{2}}} \,. \] Notice that \[ - 1 \leq \cos (3 n) \leq 1 \,, \quad \forall \, n \in \mathbb{N} \,, \] and thus \[ - \frac{2}{n} \leq \frac{2 \cos (3 n)}{n} \leq \frac{2}{n} \,, \quad \forall \, n \in \mathbb{N} \,. \] Since \[ -\frac{2}{n} \longrightarrow 0 \,, \quad \frac{2}{n} \longrightarrow 0 \,, \] by the Squeeze Theorem we conclude that also \[ \frac{2 \cos (3 n)}{n} \to 0 \,. \] In particular we have shown that \[ \left| \dfrac{2 i \cos (3 n)}{ n} \right| = \left| \dfrac{2 \cos (3 n)}{ n} \right| \to 0 \,. \] Using Theorem 10 we infer \[ \dfrac{2 i \cos (3 n)}{ n} \to 0 \,. \] Similarly, \[ - \frac{1}{n^{2}} \leq \frac{\sin (2 n)}{n^{2}} \leq - \frac{1}{n^{2}} \,, \quad \forall \, n \in \mathbb{N} \,. \] Since \[ - \frac{1}{n^{2}} \longrightarrow 0 \,, \quad \frac{1}{n^{2}} \longrightarrow 0 \,, \] by the Squeeze Theorem we conclude \[ \frac{\sin (2 n)}{n^2} \longrightarrow 0 \,. \] Finally, we have \[ \left| \frac{2i}{n} \right| = \frac{2}{n} \longrightarrow 0 \,, \] and therefore \[ \frac{2i}{n} \longrightarrow 0 \] by Theorem 10. Using the Algebra of Limits in \(\mathbb{C}\) we conclude \[ a_{n} = \frac{\dfrac{2 i \cos (3 n)}{ n} + (7-i) }{3+ \dfrac{2 i}{n} + \dfrac{\sin (2 n)}{n^{2}}} \longrightarrow \frac{ 0+ (7-i) }{3+0+0}=\frac{7}{3}-\frac{1}{3} i \,. \]

7.5 Geometric sequence Test and Ratio Test in \(\mathbb{C}\)

The Geometric Sequence Test and Ratio Test can be generalized to complex sequences.

Theorem 13: Geometric sequence Test in \(\mathbb{C}\)

Let \(x \in \mathbb{C}\) and let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be the geometric sequence in \(\mathbb{C}\) defined by \[ a_{n}:=x^{n} \,. \] We have:

  1. If \(|x|<1\), then \[ \lim_{n \to \infty} a_{n} = 0 \,. \]

  2. If \(|x|>1\), then sequence \(\left(a_{n}\right)\) is unbounded, and hence divergent.

The proof can be obtained as in the real case, replacing the absolute value by the modulus.

Example 14

  • Let \[ a_{n}=\frac{(-1+4 i)^{n}}{(7+3 i)^{n}} \,. \] We first rewrite \[ a_{n}=\frac{(-1+4 i)^{n}}{(7+3 i)^{n}}=\left(\frac{-1+4 i}{7+3 i}\right)^{n} \] Then, we compute \[\begin{align*} \left|\frac{-1+4 i}{7+3 i}\right| & =\frac{|-1+4 i|}{|7+3 i|} \\ & = \frac{\sqrt{(-1)^{2}+4^{2}}}{\sqrt{7^{2}+3^{2}}} \\ & = \frac{\sqrt{17}}{\sqrt{58}} \\ & = \sqrt{\frac{17}{58}} \\ & < 1 \, \end{align*}\] By the Geometric Sequence Test \(a_{n} \rightarrow 0\).

  • Let \[ b_{n}=\frac{(-5+12 i)^{n}}{(3-4 i)^{n}} \,. \] We first rewrite \[ b_{n}=\frac{(-5+12 i)^{n}}{(3-4 i)^{n}}=\left(\frac{-5+12 i}{3-4 i}\right)^{n} \,. \] We compute \[\begin{align*} \left|\frac{-5+12 i}{3-4 i}\right| & = \frac{|-5+12 i|}{|3-4 i|} \\ & = \frac{\sqrt{5^{2}+(-12)^{2}}}{\sqrt{3^{2}+(-4)^{2}}} \\ & = \frac{13}{5} \\ & > 1 \,. \end{align*}\] By the Geometric Sequence Test, the sequence \((b_{n})\) does not converge.

  • Let \[ c_{n}=e^{\frac{i \pi}{2} n} \,. \] We have \[ \left|c_{n}\right|=\left|e^{\frac{i \pi}{2} n}\right|=1 \,, \] and hence the Geometric Sequence Test cannot be applied. However, we can see that \[ c_n = (i,-1,-i, 1, i,-1,-i, 1, \ldots ) \,, \] that is, \(c_n\) assumes only the values \(\{i,-1,-i,1\}\), and each of them is assumed infinitely many times. Thus \(c_n\) is oscillating and it is divergent.

We now provide the statement of the Ratio Test in \(\mathbb{C}\).

Theorem 15: Ratio Test in \(\mathbb{C}\)

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\) such that \[ a_{n} \neq 0 \,, \quad \forall \, n \in \mathbb{N} \,. \]

  1. Suppose that the following limit exists: \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Then,

    • If \(L<1\) we have \[ \lim_{n \to\infty} a_{n}=0 \,. \]

    • If \(L>1\), the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

  2. Suppose that there exists \(N \in N\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Then the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

The proof of Theorem 15 follows word by word the proof of the Ratio Test Theorem in \(\mathbb{R}\), and only two minor modifications are needed:

  • Replace the absolute value with the complex modulus,
  • Instead of the Squeeze Theorem, use Theorem 10.
Example 16

Let \[ a_{n}=\frac{(4-3 i)^{n}}{(2 n) !} \,. \] Prove that \(a_n \to 0\).

Proof. We compute \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & =\left|\frac{(4-3 i)^{n+1}}{(2(n+1)) !} \frac{(2 n) !}{(4-3 i)^{n}}\right| \\ & = \frac{|4-3 i|^{n+1}}{|4-3 i|^{n}} \cdot \frac{(2 n) !}{(2 n+2) !} \\ & = \frac{|4 - 3i|}{(2 n+2)(2 n+1)} \\ & =\frac{\sqrt{4^{2}+(-3)^{2}}}{(2 n+2)(2 n+1)} \\ & = \frac{5}{(2 n+2)(2 n+1)} \\ & = \frac{ \dfrac{5}{n^{2}}}{ \left(2+ \dfrac{2}{n} \right) \left(2+ \dfrac{1}{n} \right) } \longrightarrow L = 0 \,. \end{align*}\] Since \(L=0 < 1\), by the Ratio Test in \(\mathbb{C}\) we infer \(a_{n} \to 0\).

7.6 Convergence of real and imaginary part

A complex number \(z \in \mathbb{C}\) can be written as \[ z=a+b i \] for some \(a, b \in \mathbb{R}\), where

  • \(a\) is the real part of \(z\),
  • \(b\) the imaginary part of \(z\).

We can prove that a complex sequence converges if and only if both the real parts and the imaginary parts converge.

Theorem 17
Let \(\left(z_{n}\right)_{n \in \mathbb{N}}\) be a sequence in \(\mathbb{C}\). For \(n \in \mathbb{N}\), let \(a_{n}, b_{n} \in \mathbb{R}\) such that \[ z_{n}=a_{n}+b_{n} i \,. \] Let \[ z=a+b i \] with \(a, b \in \mathbb{R}\). Then \[ \lim_{n \rightarrow \infty} z_{n}=z \quad \iff \quad \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \text {. } \]

Proof
Part 1. Suppose that \[ \lim_{n \rightarrow \infty} z_{n}=z \,. \] To prove that \(a_{n} \to a\) we need to show that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\, \left|a_{n}-a\right| < \varepsilon \,. \]

Let \(\varepsilon>0\). Since \(z_{n} \rightarrow z\), there exists \(N \in \mathbb{N}\) such that \[ \left|z_{n}-z\right| < \varepsilon \,, \quad \forall \, n \geq N \,. \] Let \(n \geq N\). Then \[\begin{align*} \left|a_{n}-a\right| & =\sqrt{\left(a_{n}-a\right)^{2}} \\ & \leq \sqrt{\left(a_{n}-a\right)^{2}+\left(b_{n}-b\right)^{2}} \\ & =\left|\left(a_{n}-a\right)+\left(b_{n}-b\right) i\right| \\ & = \left|\left(a_{n}+b_{n} i\right)-(a+b i)\right| \\ & = \left|z_{n}-z\right| \\ & < \varepsilon \,. \end{align*}\] The proof for \(b_{n} \to b\) is similar.

Part 2. Suppose that \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \,. \] To prove that \(z_{n} \to z\) we need to show that \[ \forall \, \varepsilon>0 \,, \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\, \left|z_{n}-z\right| < \varepsilon \,. \] Let \(\varepsilon>0\). Since \(a_{n} \rightarrow a\), there exists \(N_1 \in \mathbb{N}\) such that \[ \left|a_{n}-a\right| < \frac{\varepsilon}{2} \,, \quad \forall \, n \geq N_1 \,. \] Since \(b_{n} \rightarrow b\), there exists \(N_2 \in \mathbb{N}\) such that \[ \left|b_{n}-b\right| < \frac{\varepsilon}{2} \,, \quad \forall \, n \geq N_2 \,. \] Let \[ N := \max \left\{ N_1, N_2 \right\} \,. \] Let \(n \geq N\). By the triangle inequality, \[\begin{align*} \left|z_{n}-z\right| & =\left|\left(a_{n}+b_{n} i\right)-(a+b i)\right|\\ & =\left|\left(a_{n}-a\right)+\left(b_{n}-b\right) i\right| \\ & \leq\left|a_{n}-a\right|+\left|b_{n}-b\right| \cdot|i| \\ & =\left|a_{n}-a\right|+\left|b_{n}-b\right| \\ & < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ & = \varepsilon \,. \end{align*}\]

Example 18

Consider the complex sequence \[ z_{n}:=\frac{\left(4 n+3 n^{2} i\right)\left(2 n^{2}+i\right)}{5 n^{4}} \] Show that \[ \lim_{n \to \infty} z_n = \frac65 i \,. \]

Proof. Let us find the real and imaginary parts of \(z_n\) \[\begin{align*} z_n & = \frac{\left(4 n+3 n^{2} i\right)\left(2 n^{2}+i\right)}{5 n^{4}} \\ & = \frac{8 n^{3}+4 n i + 6 n^{4} i + 3 n^{2}i^2}{5 n^{4}} \\ & = \frac{8 n^{3}-3 n^{2}}{5 n^{4}} + \frac{6 n^{4}+4 n}{5 n^{4}} i \\ & = a_n + b_n i \,. \end{align*}\] Using the Algebra of Limits for real sequences we have that \[ a_n = \frac{8 n^{3}-3 n^{2}}{5 n^{4}}=\frac{ \dfrac{8}{n} - \dfrac{3}{n^2} }{5} \longrightarrow \frac{0-0}{5}=0 \,, \] and \[ b_n = \frac{6 n^{4}+4 n}{5 n^{4}}=\frac{6+ \dfrac{4}{n^3} }{5} \longrightarrow \frac{6+0}{5}=\frac{6}{5} \,. \] By Theorem 17 we conclude \[ \lim _{n \rightarrow \infty} z_{n} = \lim_{n \to \infty} a_n + i \, \lim_{n \to \infty} b_n = 0+\frac{6}{5} i=\frac{6}{5} i \,. \]