Complex Sequences

Convergence of a Complex Sequence

The previous tutorial introduced the metric \(d(z,w)=|z-w|\) on \(\mathbb{C}\). A sequence of complex numbers converges when its terms eventually lie as close as desired to one fixed complex number in this metric. This definition looks just like the metric definition for real sequences, but complex numbers have two coordinates. A useful first result shows that convergence in the complex plane is exactly convergence of both real coordinates.

Writing \(z_n=x_n+iy_n\) and \(z=x+iy\), the distance to the proposed limit is \(|z_n-z|=\sqrt{(x_n-x)^2+(y_n-y)^2}\). Each coordinate difference is no larger than this distance. Conversely, if both coordinate differences are small, then the distance is small as well. The following theorem makes this equivalence precise.

Proof. Suppose first that \(z_n\to z\). For every \(n\), the coordinate formula for the modulus gives

Given \(\varepsilon>0\), convergence of \(z_n\) gives an \(N\) such that \(|z_n-z|<\varepsilon\) for every \(n\geq N\). The displayed inequalities then imply both \(|x_n-x|<\varepsilon\) and \(|y_n-y|<\varepsilon\) for those \(n\). Thus \(x_n\to x\) and \(y_n\to y\).

Conversely, suppose \(x_n\to x\) and \(y_n\to y\). Fix \(\varepsilon>0\). There are positive integers \(N_1,N_2\) such that for \(n\geq N_1\), \(|x_n-x|<\varepsilon/2\), and for \(n\geq N_2\), \(|y_n-y|<\varepsilon/2\). Set \(N=\max(N_1,N_2)\). For \(n\geq N\), the triangle inequality for the complex metric, or the coordinate formula, gives

Therefore \(z_n\to z\). Both directions hold, proving the criterion. \(\square\)

The two coordinates must approach their respective limits along the same sequence of indices. It is not enough for each coordinate to be close to its target at different, incompatible sets of indices. The common choice \(N=\max(N_1,N_2)\) in the proof ensures that both estimates hold simultaneously.

Limit Laws for Complex Sequences

Once convergence is understood through the metric, the familiar algebraic operations can be handled by limit laws. These laws let us calculate limits without repeatedly returning to the definition for every expression. We prove them using the triangle inequality and the fact that a convergent sequence is bounded.

Proof. By convergence, there is an \(N\) such that \(|z_n-z|<1\) for \(n\geq N\). The triangle inequality gives \(|z_n|\leq |z_n-z|+|z|<1+|z|\) for these indices. There are only finitely many terms \(z_1,\ldots,z_{N-1}\). Choose \(M=\max\{1+|z|,|z_1|,\ldots,|z_{N-1}|\}\), omitting the finite list if \(N=1\). Then \(|z_n|\leq M\) for every \(n\), as required. \(\square\)

Proof. For sums, the triangle inequality gives \(|(z_n+w_n)-(z+w)|\leq|z_n-z|+|w_n-w|\). Given \(\varepsilon>0\), each term on the right is less than \(\varepsilon/2\) for all sufficiently large \(n\), so the sum tends to zero. For scalar multiples, \(|cz_n-cz|=|c||z_n-z|\). If \(c=0\), this is always zero; if \(c\neq0\), choose the error in \(z_n\) smaller than \(\varepsilon/|c|\). Thus \(cz_n\to cz\).

For products, write \(z_nw_n-zw=z_n(w_n-w)+w(z_n-z)\). The lemma shows that \((z_n)\) is bounded, so there is an \(M\geq0\) with \(|z_n|\leq M\) for every \(n\). Hence

Both differences on the right tend to zero, so the whole expression tends to zero. More explicitly, if \(M>0\), choose the eventual bounds \(|w_n-w|<\varepsilon/(2M)\) and \(|z_n-z|<\varepsilon/(2(|w|+1))\). Then the right side is less than \(\varepsilon/2+\varepsilon|w|/(2(|w|+1))<\varepsilon\). If \(M=0\), every \(z_n=0\), so \(z=0\) and the products are all zero.

For the quotient, since \(w_n\to w\neq0\), there is an \(N_0\) such that \(|w_n-w|<|w|/2\) for \(n\geq N_0\). The reverse triangle inequality gives \(|w_n|\geq |w|-|w_n-w|>|w|/2\). For all sufficiently large \(n\), the quotient is defined, and

The right-hand side tends to zero. This proves the quotient limit and completes the proof. \(\square\)

Oscillation, Subsequences, and Nonconvergence

The coordinate criterion is also an efficient way to prove that a complex sequence does not converge: it is enough to show that one coordinate fails to converge. A basic obstruction is persistent oscillation. A subsequence, formed by retaining terms at a strictly increasing sequence of indices, provides another useful test: if the original sequence converges, every subsequence must have the same limit.

Proof. Fix \(\varepsilon>0\). Since \(z_n\to z\), there is an \(N\) such that \(|z_n-z|<\varepsilon\) whenever \(n\geq N\). Because the indices \(n_k\) are strictly increasing positive integers, \(n_k\geq k\). Thus \(k\geq N\) implies \(n_k\geq N\), and consequently \(|z_{n_k}-z|<\varepsilon\). This is precisely \(z_{n_k}\to z\). \(\square\)

Why Complex Sequence Limits Matter

Complex sequences will soon be used to define and analyze complex series. The limit laws permit termwise algebra in many familiar situations, while the coordinate criterion reduces convergence questions to real sequences. The metric definition remains the foundation: coordinate calculations are a practical equivalent, not a different notion of convergence.

A common pitfall is to infer convergence from boundedness. The sequence \(i^n\) is bounded, because \(|i^n|=1\) for every \(n\), but it does not converge. Boundedness only says that all terms remain within some fixed distance of the origin; it does not say that the terms approach a single point. Likewise, convergence of only the real parts is insufficient: both real and imaginary parts must converge to obtain a complex limit.