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 $\epsilon >0$, it holds $$|a_n-a|<\epsilon$$ 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.

Definition 1: Sequence of Complex numbers

A sequence $a$ in $\mathbb{C}$ is a function $$a:\mathbb{N}\to \mathbb{C}.$$ 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 $${\left(a_n\right)}_{n\in \mathbb{N}}\text{ or }(a_n).$$

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

Definition 2: Convergent sequence in $\mathbb{C}$

We say that a sequence $\left(a_n\right)$ in $\mathbb{C}$ converges to $a\in \mathbb{C}$, or equivalently has limit $a$, denoted by $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a\text{ or }{a}_n\to a,$$ if it holds: $$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N}\text{ s.t. }\mathrm{\forall }n\ge N,|a_n-a|<\epsilon .$$

If there exists $a\in \mathbb{C}$ such that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$, we say that the sequence $\left(a_n\right)$ is convergent.

Important

In Definition 2 we still take $\epsilon$ to be real. This makes sense, since $$|z|=\sqrt{x^2+y^2}\in \mathbb{R}$$ for all $z=x+iy\in \mathbb{C}$.

Example 3

Question. Using Definition 2, prove that $$\underset{n\to \mathrm{\infty }}{lim}\frac{(3+i)n-7i}{n}=3+i.$$

Solution.

Part 1. Rough Work. Let $\epsilon >0$. We need to clarify for which values of $n$ the following holds: $$|\frac{(3+i)n-7i}{n}-(3+i)|<\epsilon .$$ We have $$|\frac{(3+i)n-7i}{n}-(3+i)|=\frac{|-7i|}{n}=\frac{7}{n}.$$ Therefore $$\frac{7}{n}<\epsilon ⟺n>\frac{7}{\epsilon }.$$

Part 2. Formal Proof. We want to prove that for all $\epsilon >0$ there exists $N\in \mathbb{N}$ such that $$|\frac{(3+i)n-7i}{n}-(3+i)|<\epsilon ,\mathrm{\forall }n\ge N.$$ Let $\epsilon >0$. Choose $N\in \mathbb{N}$ such that $$N>\frac{7}{\epsilon }.$$ The above is equivalent to $$\frac{7}{N}<\epsilon .$$ For $n\ge N$ we have $$|\frac{(3+i)n-7i}{n}-(3+i)|=\frac{7}{n}\le \frac{7}{N}<\epsilon .$$

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 $$\left|a_n\right|\le M,\mathrm{\forall }n\in \mathbb{N}.$$

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.

7.1 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 $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a,\underset{n\to \mathrm{\infty }}{lim}{b}_n=b,$$ for some $a,b\in \mathbb{C}$. Then,

1. Limit of sum is the sum of limits: $$\underset{n\to \mathrm{\infty }}{lim}(a_n\pm b_n)=a\pm b$$

2. Limit of product is the product of limits: $$\underset{n\to \mathrm{\infty }}{lim}\left(a_n{b}_n\right)=ab$$

3. If $b_n\ne 0$ for all $n\in \mathbb{N}$ and $b\ne 0$, then $$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{a_n}{b_n}\right)=\frac{a}{b}$$

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

Question. Compute the limit of $$a_n=\frac{(2-i)n^2+6in-5-3i}{(6+3i)n^2+11i}.$$

Solution. Factor $n^2$, the largest power of $n$ in the denominator, $$a_n=\frac{(2-i)+{\displaystyle \frac{6i}{n}}-{\displaystyle \frac{5}{n^2}}-{\displaystyle \frac{3i}{n^2}}}{(6+3i)+{\displaystyle \frac{11i}{n^2}}}⟶\frac{2-i}{6+3i},$$ where we used the Algebra of Limits. Finally, $$\frac{2-i}{6+3i}=\frac{(2-i)(6-3i)}{(6+3i)(6-3i)}=\frac{1}{5}-\frac{4}{15}i.$$

7.2 Convergence to zero

One of the results that cannot hold for complex sequences is the Squeeze Theorem. Indeed the chain of inequalities $$b_n\le a_n\le c_n$$ 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 $$\underset{n\to \mathrm{\infty }}{lim}\left|a_n\right|=0.$$ Then $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$$

Proof

Assume that $|a_n|\to 0$. We need to show that $$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N}\text{ s.t. }\mathrm{\forall }n\ge N,|a_n-0|<\epsilon .$$ Let $\epsilon >0$. Since $\left|a_n\right|\to 0$, there exists $N\in \mathbb{N}$ such that $$||a_n|-0|<\epsilon ,\mathrm{\forall }n\ge N.$$ Let $n\ge N$. Then, $$\begin{array}{rl}|a_n-0|& =\left|a_n\right|\\ & =\left|a_n\right|-0\\ & =||a_n|-0|\\ & <\epsilon .\end{array}$$

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

Question. Prove that $a_n\to 0$, where $$a_n={(\frac{1}{2}+\frac{1}{3}i)}^n.$$

Solution. We have $$\begin{array}{rl}\left|a_n\right|& =\left|{(\frac{1}{2}+\frac{1}{3}i)}^n\right|\\ & ={|\frac{1}{2}+\frac{1}{3}i|}^n\\ & ={\left(\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{3}\right)}^2}\right)}^n\\ & ={\left(\sqrt{\frac{13}{36}}\right)}^n.\end{array}$$ Since $$\left|\sqrt{\frac{13}{36}}\right|<1,$$ by the Geometric Sequence Test for real sequences, we conclude that $$\left|a_n\right|\to 0.$$ 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

Question. Consider the sequence $$a_n:=\frac{2i\cos (3n)n+(7-i)n^2}{3n^2+2in+\sin (2n)}.$$ Prove that $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\frac{7}{3}-\frac{1}{3}i.$$

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

7.3 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 $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$$

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

Question. Prove that $a_n\to 0$, where $$a_n=\frac{(-1+4i{)}^n}{(7+3i{)}^n}.$$

Solution. We first rewrite $$a_n=\frac{(-1+4i{)}^n}{(7+3i{)}^n}={\left(\frac{-1+4i}{7+3i}\right)}^n$$ Then, we compute $$\begin{array}{rl}\left|\frac{-1+4i}{7+3i}\right|& =\frac{|-1+4i|}{|7+3i|}\\ & =\frac{\sqrt{(-1{)}^2+4^2}}{\sqrt{7^2+3^2}}\\ & =\frac{\sqrt{17}}{\sqrt{58}}\\ & =\sqrt{\frac{17}{58}}\\ & <1\end{array}$$ By the Geometric Sequence Test $a_n\to 0$.

Example 15

Question. Prove that $a_n$ diverges, where $$a_n=\frac{(-5+12i{)}^n}{(3-4i{)}^n}.$$

Solution. We first rewrite $$a_n=\frac{(-5+12i{)}^n}{(3-4i{)}^n}={\left(\frac{-5+12i}{3-4i}\right)}^n.$$ We compute $$\begin{array}{rl}\left|\frac{-5+12i}{3-4i}\right|& =\frac{|-5+12i|}{|3-4i|}\\ & =\frac{\sqrt{5^2+(-12{)}^2}}{\sqrt{3^2+(-4{)}^2}}\\ & =\frac{13}{5}\\ & >1.\end{array}$$ By the Geometric Sequence Test, the sequence $a_n$ diverges.

Example 16

Question. Prove that $a_n$ diverges, where $$a_n=\exp \left(\frac{i\pi }{2}n\right).$$

Solution. We have $$\left|a_n\right|=\left|e^{\frac{i\pi }{2}n}\right|=1,$$ and hence the Geometric Sequence Test cannot be applied. However, we can see that $$a_n=(i,-1,-i,1,i,-1,-i,1,\dots ),$$ that is, $a_n$ assumes only the values $\{i,-1,-i,1\}$, and each of them is assumed infinitely many times. Therefore $a_n$ is oscillating, and thus divergent.

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

Theorem 17: Ratio Test in $\mathbb{C}$

Let $\left(a_n\right)$ be a sequence in $\mathbb{C}$ such that $$a_n\ne 0,\mathrm{\forall }n\in \mathbb{N}.$$

1. Suppose that the following limit exists: $$L:=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|.$$ Then,

   • If $L<1$ we have $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$$

   • 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 $$\left|\frac{a_{n+1}}{a_n}\right|\ge L,\mathrm{\forall }n\ge N.$$ Then the sequence $a_n$ is unbounded, and hence does not converge.

The proof of Theorem 17 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 18

Question. Study the convergence / divergence of the sequence $$a_n=\frac{(4-3i{)}^n}{(2n)!}.$$

Solution. We compute $$\begin{array}{rl}\left|\frac{a_{n+1}}{a_n}\right|& =\left|\frac{(4-3i{)}^{n+1}}{(2(n+1))!}\frac{(2n)!}{(4-3i{)}^n}\right|\\ & =\frac{|4-3i{|}^{n+1}}{|4-3i{|}^n}\cdot \frac{(2n)!}{(2n+2)!}\\ & =\frac{|4-3i|}{(2n+2)(2n+1)}\\ & =\frac{\sqrt{4^2+(-3{)}^2}}{(2n+2)(2n+1)}\\ & =\frac{5}{(2n+2)(2n+1)}\\ & =\frac{{\displaystyle \frac{5}{n^2}}}{(2+{\displaystyle \frac{2}{n}})(2+{\displaystyle \frac{1}{n}})}⟶L=0.\end{array}$$ Since $L=0<1$, by the Ratio Test in $\mathbb{C}$ we infer $a_n\to 0$.

7.4 Convergence of real and imaginary part

A complex number $z\in \mathbb{C}$ can be written as $$z=a+bi$$ 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 19

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 $z=a+bi$, with $a,b\in \mathbb{R}$. Then $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=z⟺\underset{n\to \mathrm{\infty }}{lim}{a}_n=a,\underset{n\to \mathrm{\infty }}{lim}{b}_n=b\text{. }$$

Proof

Part 1. Suppose that $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=z.$$ To prove that $a_n\to a$ we need to show that $$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N}\text{ s.t. }\mathrm{\forall }n\ge N,|a_n-a|<\epsilon .$$

Let $\epsilon >0$. Since $z_n\to z$, there exists $N\in \mathbb{N}$ such that $$|z_n-z|<\epsilon ,\mathrm{\forall }n\ge N.$$ Let $n\ge N$. Then $$\begin{array}{rl}|a_n-a|& =\sqrt{{(a_n-a)}^2}\\ & \le \sqrt{{(a_n-a)}^2+{(b_n-b)}^2}\\ & =|(a_n-a)+(b_n-b)i|\\ & =|(a_n+b_{n}i)-(a+bi)|\\ & =|z_n-z|\\ & <\epsilon .\end{array}$$ The proof for $b_n\to b$ is similar.

Part 2. Suppose that $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a,\underset{n\to \mathrm{\infty }}{lim}{b}_n=b.$$ To prove that $z_n\to z$ we need to show that $$\mathrm{\forall }\epsilon >0,\mathrm{\exists }N\in \mathbb{N}\text{ s.t. }\mathrm{\forall }n\ge N,|z_n-z|<\epsilon .$$ Let $\epsilon >0$. Since $a_n\to a$, there exists $N_1\in \mathbb{N}$ such that $$|a_n-a|<\frac{\epsilon }{2},\mathrm{\forall }n\ge N_1.$$ Since $b_n\to b$, there exists $N_2\in \mathbb{N}$ such that $$|b_n-b|<\frac{\epsilon }{2},\mathrm{\forall }n\ge N_2.$$ Let $$N:=max\{N_1,N_2\}.$$ Let $n\ge N$. By the triangle inequality, $$\begin{array}{rl}|z_n-z|& =|(a_n+b_{n}i)-(a+bi)|\\ & =|(a_n-a)+(b_n-b)i|\\ & \le |a_n-a|+|b_n-b|\cdot |i|\\ & =|a_n-a|+|b_n-b|\\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}\\ & =\epsilon .\end{array}$$

Example 20

Question. Consider the complex sequence $$z_n:=\frac{(4n+3n^{2}i)(2n^2+i)}{5n^4}.$$ Show that $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=\frac{6}{5}i.$$

Solution. We find the real and imaginary parts of $z_n$ $$\begin{array}{rl}{z}_n& =\frac{(4n+3n^{2}i)(2n^2+i)}{5n^4}\\ & =\frac{8n^3+4ni+6n^{4}i+3n^2{i}^2}{5n^4}\\ & =\frac{8n^3-3n^2}{5n^4}+\frac{6n^4+4n}{5n^4}i\\ & =a_n+b_{n}i.\end{array}$$ Using the Algebra of Limits for real sequences we have that $$\begin{array}{rl}& a_n=\frac{8n^3-3n^2}{5n^4}=\frac{{\displaystyle \frac{8}{n}}-{\displaystyle \frac{3}{n^2}}}{5}⟶\frac{0-0}{5}=0,\\ & b_n=\frac{6n^4+4n}{5n^4}=\frac{6+{\displaystyle \frac{4}{n^3}}}{5}⟶\frac{6+0}{5}=\frac{6}{5}.\end{array}$$ By Theorem 19 we conclude $$\underset{n\to \mathrm{\infty }}{lim}{z}_n=\underset{n\to \mathrm{\infty }}{lim}{a}_n+i\underset{n\to \mathrm{\infty }}{lim}{b}_n=0+\frac{6}{5}i=\frac{6}{5}i.$$