Cauchy's theorem states that ∮_γ f(z) dz = 0 when f is holomorphic everywhere inside and on γ. Why does this theorem NOT immediately give ∮_γ f(z)/(z − z₀) dz = 0 when z₀ is inside γ?
ABecause f(z)/(z − z₀) fails to be holomorphic at z₀, which lies inside γ
BBecause the contour γ must be a circle for Cauchy's theorem to apply
CBecause f(z)/(z − z₀) is not bounded on γ when z₀ is close to the contour
DBecause Cauchy's theorem requires the function to be real-valued on the contour
Cauchy's theorem requires holomorphicity everywhere *inside and on* the contour. The function g(z) = f(z)/(z − z₀) has a singularity at z = z₀, which sits inside γ. So g is not holomorphic everywhere inside γ, and Cauchy's theorem cannot be applied. Instead of giving zero, the integral gives 2πi·f(z₀) — precisely Cauchy's integral formula. This is the key: the formula arises because we are forced to handle a function with an interior singularity.
Question 2 Multiple Choice
You want to compute f(i) for f(z) = z³ + 2z using Cauchy's integral formula. Which statement about the choice of contour is correct?
AYou must use a circle of radius 1 centered at i for the formula to be exact
BYou can use any simple closed contour that encloses i and stays within a simply connected region where f is holomorphic
CLarger contours give better approximations, so you should use the largest feasible contour
DThe contour must not enclose any other points besides i for the formula to apply
Cauchy's integral formula gives f(z₀) = (1/2πi) ∮_γ f(z)/(z − z₀) dz for *any* simple closed contour enclosing z₀, as long as f is holomorphic in the simply connected region bounded by γ. Since f(z) = z³ + 2z is holomorphic everywhere (it's a polynomial), any simple closed contour enclosing i works — circle, square, ellipse, or any other shape. The result is always f(i) = i³ + 2i = −i + 2i = i. Independence from contour shape (within a holomorphic region) is a direct consequence of Cauchy's theorem.
Question 3 True / False
Cauchy's integral formula implies that a holomorphic function is automatically infinitely differentiable — all derivatives of all orders exist.
TTrue
FFalse
Answer: True
By differentiating both sides of Cauchy's integral formula with respect to z₀ (differentiating under the integral sign), you obtain formulas for all higher derivatives: f^(n)(z₀) = (n!/2πi) ∮_γ f(z)/(z − z₀)^{n+1} dz. Since the right side is well-defined whenever f is holomorphic and z₀ is inside γ, every derivative exists automatically. This is a striking contrast to real analysis, where a function can be once differentiable without being twice differentiable. Complex differentiability once is a much stronger condition — it entails differentiability infinitely many times.
Question 4 True / False
A smooth real function f: ℝ → ℝ has the same 'boundary determines interior' property as a holomorphic function: knowing f on the boundary of an interval determines most its interior values.
TTrue
FFalse
Answer: False
For real smooth functions, this is completely false. You can freely change f on the interior of [0,1] without affecting its values at the endpoints. For example, f(x) = x and g(x) = x + sin(πx) both satisfy f(0) = g(0) = 0 and f(1) = g(1) = 1, but differ everywhere in between. Holomorphic functions have a rigidity with no real analogue: complex differentiability locks the function's values together globally. Cauchy's integral formula makes this explicit — boundary values on *any* surrounding contour completely reconstruct interior values.
Question 5 Short Answer
Why does Cauchy's integral formula represent something genuinely new about holomorphic functions that has no parallel in real analysis? What structural feature of complex differentiability makes it possible?
Think about your answer, then reveal below.
Model answer: In real analysis, a function's values in the interior of a domain are independent of its boundary values — you can modify f on (0,1) without touching f(0) or f(1). Complex differentiability (holomorphicity) imposes a rigid global constraint: the Cauchy-Riemann equations couple the real and imaginary parts of f so tightly that no isolated local modification is possible. This coupling is encoded in Cauchy's theorem (∮ f dz = 0), which forces the integral of f(z)/(z−z₀) to equal exactly 2πi·f(z₀). The interior value is not just influenced by boundary values — it is completely reconstructed from them.
This rigidity explains why holomorphic functions are so well-behaved (analytic, infinitely differentiable, expressible as convergent power series) compared to smooth real functions. The formula is not a computational trick — it is the quantitative expression of a deep structural rigidity that has no real-variable analogue.