A function f is defined on the closed disk S = {z ∈ ℂ : |z| ≤ 1}. A student claims f can be holomorphic on S because S contains every point f would ever need to evaluate. What is wrong with this claim?
ANothing — holomorphic functions can be defined on any set, open or closed
BThe closed disk is unbounded, so derivatives cannot be computed there
CBoundary points of S have no open disk contained entirely within S, so the complex derivative cannot be defined there
DClosed sets cannot be connected, which is required for holomorphic functions
Holomorphicity at a point requires the complex derivative to be defined in an entire neighborhood of that point — an open disk around it. At a boundary point of the closed disk (like z = 1), any open disk centered there extends outside S, so the derivative is not defined within the domain. The domain of a holomorphic function must be open precisely to guarantee every point has this necessary breathing room.
Question 2 Multiple Choice
Which of the following subsets of ℂ is an open set?
AThe closed disk {z : |z − 2| ≤ 1}
BThe real axis {z : Im(z) = 0}
CThe unit circle {z : |z| = 1}
DThe right half-plane {z : Re(z) > 0}
The right half-plane is open: for any z with Re(z) > 0, a sufficiently small open disk around z is entirely contained in the half-plane. The closed disk includes its boundary — boundary points have no open disk lying entirely within the set. The real axis and unit circle are both 'thin' sets (1-dimensional curves in ℝ²); no 2D open disk around any point on them is contained within them.
Question 3 True / False
The open disk D₁(0) = {z ∈ ℂ : |z| < 1} is both open and bounded.
TTrue
FFalse
Answer: True
Both properties hold independently. The disk is open because every point inside has an open neighborhood contained entirely within it. It is bounded because it is contained in the closed disk of radius 1 centered at the origin — no point in D₁(0) has modulus greater than 1. Openness and boundedness are logically independent properties; a set can be any combination of open/closed and bounded/unbounded.
Question 4 True / False
Most closed subset of ℂ is bounded.
TTrue
FFalse
Answer: False
Closed and bounded are independent properties. The entire real axis {z : Im(z) = 0} is closed — it contains all its limit points and its complement is open — but it is unbounded, extending infinitely in both directions. Similarly, the entire plane ℂ is both open and closed but is unbounded. The misconception often arises from conflating 'closed' with 'compact'; compactness requires both closed and bounded in ℝⁿ.
Question 5 Short Answer
Why must the domain of a holomorphic function be an open set rather than an arbitrary subset of ℂ?
Think about your answer, then reveal below.
Model answer: The complex derivative at a point z₀ is defined as the limit of (f(z) − f(z₀))/(z − z₀) as z → z₀ from any direction. This limit must be taken over points z in the domain, so z must be able to approach z₀ freely from all directions. An open set guarantees that every point has an open disk around it contained in the domain, providing the full two-dimensional neighborhood required for the directional limit to be well-defined. Without openness, the limit might only be defined along restricted paths, which is insufficient for complex differentiability.
Real differentiability only requires limits along the real line (one direction). Complex differentiability requires the limit to exist and be the same from every direction in the plane — horizontally, vertically, diagonally, and all others. This is what makes complex analysis so powerful (and restrictive). The open domain condition is what ensures every point has enough surrounding points for this multi-directional limit to be meaningful.