Questions: Sequences and Convergence in the Complex Plane

Question 1

Consider the sequence zₙ = (n/(n+1)) + i·(1/n). What is the limit as n → ∞?

• A. The sequence diverges because complex sequences behave differently from real sequences
• B. 1 + 0i = 1, because both real and imaginary parts converge separately
• C. 1 + i, because the modulus of the terms converges to 1
• D. The limit cannot be determined without knowing the spiral behavior in the plane

Answer: (B)

Complex convergence reduces to simultaneous real convergence: Re(zₙ) = n/(n+1) → 1 and Im(zₙ) = 1/n → 0. Since both parts converge, zₙ → 1 + 0i = 1 in ℂ. Option A is wrong — complex sequences use the same epsilon-modulus definition as real sequences. Option C confuses the modulus of the terms (which approaches 1) with the limit of the sequence. The key tool: |zₙ − 1| = √((n/(n+1)−1)² + (1/n)²) → 0, confirmed by checking each component.

Question 2

A sequence of complex numbers {zₙ} satisfies |zₙ| ≤ 5 for all n but does not converge. Which is the most that the Bolzano-Weierstrass theorem guarantees?

• A. The sequence converges — boundedness in ℂ implies convergence
• B. There exists a subsequence {zₙₖ} that converges to some complex number w
• C. The sequence is Cauchy even though it does not converge
• D. The sequence has at most finitely many distinct values

Answer: (B)

Bolzano-Weierstrass guarantees a convergent subsequence from any bounded sequence, not convergence of the sequence itself. The sequence zₙ = (−1)ⁿ is bounded (|zₙ| = 1) but does not converge — it oscillates between 1 and −1. However, it has convergent subsequences: {z₂ₙ} → 1 and {z₂ₙ₊₁} → −1. Option A is the common misconception that boundedness implies convergence (it guarantees only a convergent subsequence). Option C is false — a non-convergent sequence need not be Cauchy.

Question 3

A sequence {zₙ} in ℂ converges to w if and only if Re(zₙ) → Re(w) and Im(zₙ) → Im(w) as real sequences.

• T. True
• F. False

Answer: True

This equivalence is the central practical tool. It follows from the identity |zₙ − w|² = (Re(zₙ) − Re(w))² + (Im(zₙ) − Im(w))². The modulus is small exactly when both squared real differences are small — so complex convergence is equivalent to simultaneous convergence of real and imaginary parts. This lets you reduce all questions about complex sequences to the real analysis you already know.

Question 4

Since ℂ contains square roots of negative numbers and behaves differently from ℝ algebraically, Cauchy sequences in ℂ may fail to converge.

• T. True
• F. False

Answer: False

ℂ is complete: every Cauchy sequence in ℂ converges to a limit in ℂ. This follows because ℂ inherits completeness from ℝ via the isomorphism ℂ ≅ ℝ × ℝ. A Cauchy sequence {zₙ} in ℂ has Cauchy sequences of real and imaginary parts (since |Re(zₙ) − Re(zₘ)| ≤ |zₙ − zₘ|), and since ℝ is complete, those converge. The algebraic 'exoticness' of i has no bearing on completeness, which is a metric property.

Question 5

Why does extending convergence from ℝ to ℂ require replacing the absolute value with the complex modulus, and what geometric change does this represent?

Model answer: In ℝ, |xₙ − L| measures distance on the real line — a one-dimensional notion of closeness. In ℂ, convergence must capture closeness in the complex plane, which is two-dimensional. The complex modulus |zₙ − w| = √((xₙ−u)² + (yₙ−v)²) is the Euclidean distance between two points in ℝ², measuring closeness in all directions simultaneously. This matters for limits of complex functions, where approaching a limit point means approaching from any direction in the plane — a richer geometry than the real case where approach is only from left or right.

The modulus is the bridge between algebraic structure (complex numbers as a+bi) and metric structure (complex numbers as points in the plane). All of convergence theory transfers because the modulus satisfies the same properties as the absolute value: positivity, symmetry, and the triangle inequality.