Notes from a first course on complex analysis

Adam Coffman

These notes supplement the textbook [CB], which I’ve used when teaching MA 525, a first course on complex variables for upper-level undergraduates or graduate students in the M.S. program at Purdue Fort Wayne. Some of these miscellaneous topics appeared on class handouts and this compilation is not intended to be a self-contained reference.

Development of an online version of this course was supported by a 2021–2022 Online Course Development Grant from the Purdue Fort Wayne Division of Continuing Studies.

1 Complex differentiability

Let z_{0} = x_{0} + iy_{0} be a point in ℂ, and suppose z_{0} is in the domain of a complex-valued function f(z) = u(x, y) + iv(x, y).

These first two Propositions are sufficient conditions for ℂ-differentiability of f at the point z_{0}.

Proposition 1.1 ([CB], §22, p. 66). If u_{x}, u_{y}, v_{x}, v_{y} are continuous at z_{0}, and satisfy the Cauchy-Riemann equations at that point: u_{x}(x_{0}, y_{0}) = v_{y}(x_{0}, y_{0}) and u_{y}(x_{0}, y_{0}) = −v_{x}(x_{0}, y_{0}), then f is ℂ-differentiable at z_{0}.

Remark. In particular, the definition of two-variable continuity requires that u_{x}, u_{y}, v_{x}, v_{y} must exist in some neighborhood of z_{0}, not just at z_{0}. The idea is that the continuity of the partial derivatives (called the “C^{1}” property) implies the real differentiability property of f at z_{0}, so the next Proposition applies.

Proposition 1.2. If u and v are ℝ-differentiable at (x_{0}, y_{0}) and the partial derivatives u_{x}, u_{y}, v_{x}, v_{y} satisfy the Cauchy-Riemann equations at that point: u_{x}(x_{0}, y_{0}) = v_{y}(x_{0}, y_{0}) and u_{y}(x_{0}, y_{0}) = −v_{x}(x_{0}, y_{0}), then f is ℂ-differentiable at z_{0}.

Remark. Recall a real-valued two-variable function u(x, y) is ℝ-differentiable at (x_{0}, y_{0}) means there exist real constants a, c so that

lim_{(x,y)→(x_{0},y_{0})} \frac{|u(x, y) − [u(x_{0}, y_{0}) + a(x − x_{0}) + c(y − y_{0})]|}{|(x, y) − (x_{0}, y_{0})|} = 0.

Again, the two-dimensional limit requires that (x_{0}, y_{0}) is an interior point of the domain of u. This is equivalent to the properties (2) – (4) from [CB] §22, p. 67.

The next two Propositions are sufficient conditions for f to be analytic on an open set (meaning, f is ℂ-differentiable at every point in the set).

Proposition 1.3 (The Looman-Menchoff Theorem). If f(z) is continuous on an open set D, and the partial derivatives satisfy the Cauchy-Riemann equations at every point of D: u_{x}(x, y) = v_{y}(x, y) and u_{y}(x, y) = −v_{x}(x, y), then f is analytic on D.

Proposition 1.4 (Montel, Tolstoff). If f(z) is locally bounded on an open set D, and the partial derivatives satisfy the Cauchy-Riemann equations at every point of D: u_{x}(x, y) = v_{y}(x, y) and u_{y}(x, y) = −v_{x}(x, y), then f is analytic on D.

Remark. A function is \underline{locally bounded} on a set D means that for each w ∈ D, there is some neighborhood N_{w}, w ∈ N_{w} ⊆ D, and some bound M_{w}, so that |f(z)| ≤ M_{w} for all z ∈ N_{w}. Every continuous function is locally bounded (for each w, |f(z)| < |f(w)| + ϵ for all z within some δ of w), so this result improves the Looman-Menchoff Theorem by requiring less in the hypothesis. For more about these Propositions, see [GM].

Exercise 1.5. In any of the above Propositions, ℂ-differentiability of f at a point does not follow from checking only the Cauchy-Riemann equations at one point, without any further hypothesis. For example, let

f(z) = \{ \begin{matrix}z^{5}/|z|^{4} & if z ≠ 0 \\ 0 & if z = 0 \end{matrix} .

Show that f = u + iv has the following properties:

• On the set {z ≠ 0}, expand u and v as rational functions of x, y (so f is continuous for z ≠ 0).

• Calculate a limit as z → 0 to show that f is continuous at 0. (Hint: Use [CB] Exercise #18.9., p. 56.)

• Using the limit definition of real partial derivatives at (0, 0), show that u and v satisfy the Cauchy-Riemann equations at (0, 0).

• Show that u and v do not satisfy the Cauchy-Riemann equations at any point other than (0, 0).

• Using the limit definition of complex derivative, show that f is not ℂ-differentiable at z_{0} = 0. (Hint: this is related to [CB] Exercise #20.9., p. 63.)

Exercise 1.6. In any of the above Propositions, ℂ-differentiability of f at a point does not follow only from checking the Cauchy-Riemann equations on an open set, even on all of ℂ, without any further hypothesis. For example, let

f(z) = \{ \begin{matrix}e^{−1/z^{4}} & if z ≠ 0 \\ 0 & if z = 0 \end{matrix} .

Show that f = u + iv has the following properties:

• On the set {z ≠ 0}, show that f is ℂ-differentiable by using the rules for derivatives to find f′(z). (So, the C-R equations are satisfied at every point z ≠ 0 by the Theorem from [CB] §20, p. 65.)

• Using the limit definition of real partial derivatives at (0, 0), show that u and v satisfy the Cauchy-Riemann equations at (0, 0). (Hint: one step in a limit calculation could involve this substitution:)

lim_{x→0^{+}} \frac{e^{−1/x^{4}}}{x} = lim_{X→+∞} \frac{e^{−1/(1/X)^{4}}}{(1/X)} .

• Show that f is not continuous at z = 0, by showing that lim f(z) = +∞ along some direction (and so f is not locally bounded, and f is not ℂ-differentiable at z = 0 by the remark in [CB] §15, p. 59).

2 Bonus exercises

The following exercise is similar to [CB] Exercise #26.1, p. 81.

Exercise 2.1. For the following functions u(x, y) with domain D ⊆ ℂ, check that u is harmonic on D, and find a “harmonic conjugate” v(x, y) on D, using the method from [CB] Example 26.5, page 81.

1. u = e^{x} sin(y)

2. u = 2x^{3} − 3x^{2}y − 6xy^{2} + y^{3}

3. u = \frac{3x^{2} + 8xy − 3y^{2}}{(x^{2} + y^{2})^{2}} , domain D = \{(x, y) ≠ (0, 0)\}. (some computer algebra might help

on this one)

4. u = tan^{−1} (\frac{y}{x}), domain D = \{x > 0\} (this is related to, but not the same as, [CB] Exercise #26.6, p. 82).

Exercise 2.2. For the principal branch of the logarithm, Log(z) = ln |z|+iθ, for z = |z| exp(iθ), −π < θ < π, define the analytic function f(z) = \frac{z}{Log(z)}, on the domain D = \{|z − 1| < 1\}.

Find the derivatives f′(z) and f′′(z) on D, in terms of Log(z). Prove the following statements about limits as z approaches 0 but stays in the domain D:

lim_{z∈D, z→0} f(z) = 0.
lim_{z∈D, z→0} f′(z) = 0,
lim_{z∈D, z→0} |f′′(z)| = +∞.

Exercise 2.3. Using the exponential formula (1) from [CB] §34, p. 105, find infinitely many different complex solutions of the equation sin(z) = 5. (You do not have to find all the solutions.)

3 Review of parametric curve calculus

The following result on real curves in ℝ^{n} states that for a sufficiently smooth curve r⃗ with finite arclength, there is a change of parameter so that the composite has the same image but constant speed 1.

Theorem 3.1. Given r⃗(t) which has finite arclength L on [a, b], and such that \frac{d}{dt}r⃗(t) is continuous and nonvanishing on (a, b), there exists a function f(s) on the domain [0, L] so that (r⃗ ∘ f)(0) = r⃗(a), (r⃗ ∘ f)(L) = r⃗(b), and r⃗ ∘ f has constant speed 1 on (0, L).

Proof. Let s = ℓ(t) be the arclength from the start point r⃗(a) to the point on the curve at time t, r⃗(t). By the formula for arclength,

s = ℓ(t) = ∫_{a}^{t} |\frac{d}{dx}r⃗(x)| dx,  (3.1)

and assuming the total arclength on the interval [a, b] exists, we can conclude that the integral on the subinterval [a, t] exists for every t ∈ [a, b].

The Fundamental Theorem of Calculus applies for t in (a, b):

\frac{d}{dt}ℓ(t) = \frac{d}{dt} ∫_{a}^{t} |\frac{d}{dx}r⃗(x)| dx = |\frac{d}{dx}r⃗(x)|]_{x=t} = |r ′(t)| .

From the assumption that r ′ ≠ 0 on (a, b), we can conclude that |r ′(t)| > 0 on (a, b), so ℓ(t) is the integral from a to t of a positive, continuous function, and therefore s = ℓ(t) is an increasing function on [a, b]. It follows that ℓ is invertible: there exists an inverse function t = ℓ^{−1}(s), so that if s is the arclength, then t is the unique time at which the plot r⃗ gets to length s.

From (ℓ ∘ ℓ^{−1})(s) = s, we can \frac{d}{ds} both sides to show that the derivative of the composite is constant: \frac{d}{ds}(ℓ ∘ ℓ^{−1})(s) = 1. Applying the Chain rule,

1 = \frac{d}{ds}((ℓ ∘ ℓ^{−1})(s)) = ℓ′(ℓ^{−1}(s)) · (\frac{d}{ds}(ℓ^{−1}(s)))
⟹ \frac{d}{ds}(ℓ^{−1}(s)) = \frac{1}{ℓ′(ℓ^{−1}(s))} = \frac{1}{ℓ′(t)}.

Combining the above two equations gives:

\frac{d}{ds}(ℓ^{−1}(s)) = \frac{1}{ℓ′(t)} = \frac{1}{|r ′(t)|} > 0.

Let s be the input parameter, 0 ≤ s ≤ L, and consider the composition (r⃗ ∘ (ℓ^{−1}))(s) = r⃗(ℓ^{−1}(s)). This composite takes input s, gives ℓ^{−1}(s), which is the time at which r⃗ plots an arc of length s, and plugs this time into the function r⃗. So, r⃗(ℓ^{−1}(s)) is the position on the curve at which the arclength is s. This change of parameter is called a \underline{parametrization by arclength}, and the claim is that the function f from the statement of the Theorem can be chosen to be the function ℓ^{−1} that we’ve constructed.

Returning to the calculations, we want to show that r⃗ ∘ (ℓ^{−1}) has constant speed 1 with respect to the parameter s, 0 < s < L. The velocity of the composite is given by the Chain Rule:

\frac{d}{ds} ((r⃗ ∘ (ℓ^{−1}))(s)) = r ′(ℓ^{−1}(s)) · \frac{d}{ds}(ℓ^{−1}(s)) = r ′(t) · \frac{d}{ds}(ℓ^{−1}(s)),

and the speed is the magnitude:

|\frac{d}{ds} ((r⃗ ∘ (ℓ^{−1}))(s))| = |r ′(t) · \frac{d}{ds}(ℓ^{−1}(s))| = |r ′(t)| · |\frac{d}{ds}(ℓ^{−1}(s))| ,

and from the above equation \frac{d}{ds}(ℓ^{−1}(s)) = \frac{1}{|r ′(t)|}, this product cancels to exactly 1.

Lemma 3.2. Given r⃗(t) : [a, b] → ℝ^{n} which satisfies lim_{t→a^{+}} r⃗(t) = r⃗(a), and lim_{t→a^{+}} (\frac{d}{dt}r⃗(t)) = V⃗,

the following limit also exists: lim_{t→a^{+}} \frac{r⃗(t) − r⃗(a)}{t − a} = V⃗.

Proof. The existence of the one-sided derivative follows from the Mean Value Theorem (applied to the components r⃗(t) = (r_{1}(t), …, r_{n}(t))).

So, if r⃗ is continuous on [a, b] and the derivative extends continuously to the value V⃗ at endpoint a, then V⃗ is also equal to the one-sided derivative at a⃗.

Theorem 3.3. Given r⃗(t) which is continuous on [a, b], and such that \frac{d}{dt}r⃗(t) extends to a continuous and nonvanishing function on the closed interval [a, b], there exists a function f(s) on the domain [0, L] so that (r⃗ ∘ f)(0) = r⃗(a), (r⃗ ∘ f)(L) = r⃗(b), r⃗ ∘ f is continuous on [0, L], has constant speed 1 on (0, L), and the one-sided derivatives are also unit vectors, so that the velocity \frac{d}{ds}(r ∘ f) also extends continuously to [0, L]:

lim_{s→0^{+}} \frac{(r⃗ ∘ f)(s) − (r⃗ ∘ f)(0)}{s} = lim_{s→0^{+}} (\frac{d}{ds}(r ∘ f)) ,

and similarly for the other endpoint.

Proof. The finiteness of the arclength, L, is a consequence of the continuity of \frac{d}{dt}r⃗(t) on the closed interval [a, b]. Use the same f = ℓ^{−1} constructed in the Proof of Theorem 3.1; since ℓ is continuous on [a, b], f is continuous on [0, L]. Let lim_{t→a^{+}} r ′(t) = V⃗ ≠ 0. The following calculation,

using the Composite Limit Theorem, establishes the existence of the limit.

lim_{s→0^{+}} (\frac{d}{ds}(r ∘ ℓ^{−1})) = lim_{s→0^{+}} (r ′(ℓ^{−1}(s)) · \frac{d}{ds}(ℓ^{−1}(s)))
= lim_{s→0^{+}} (r ′(ℓ^{−1}(s))) · lim_{s→0^{+}} (\frac{d}{ds}(ℓ^{−1}(s)))
= V⃗ · lim_{s→0^{+}} \frac{1}{|r ′(ℓ^{−1}(s))|} = V⃗ · \frac{1}{|V⃗ |}.

The equality of this unit vector with the one-sided derivative is Lemma 3.2, using the continuity of r⃗. The other endpoint is considered similarly.

Lemma 3.4. Given r⃗ : [a, b] → ℝ^{n}, if r⃗(t) is continuous on [a, b], and \frac{d}{dt}r⃗(t) exists on (a, b) and extends to a continuous and nonvanishing function on [a, b), with

lim_{t→a^{+}} (\frac{d}{dt}r⃗(t)) = V⃗ ,

then for any c, 0 < c < |V⃗ |, there exists some δ > 0 so that for a ≤ t < a + δ,

c · (t − a) ≤ |r⃗(t) − r⃗(a)| ≤ (2|V⃗ | − c) · (t − a).

Proof. By Lemma 3.2,

lim_{t→a^{+}} \frac{r⃗(t) − r⃗(a)}{t − a} = V⃗ .

Corresponding to ϵ = |V⃗ | − c > 0, there is some δ > 0 so that for 0 < t − a < δ,

|\frac{r⃗(t) − r⃗(a)}{t − a} − V⃗ | < |V⃗ | − c.

By the triangle inequality,

|r⃗(t) − r⃗(a)| ≥ |(t − a)V⃗ | − |r⃗(t) − r⃗(a) − (t − a)V⃗ |
> (t − a)|V⃗ | − (t − a)(|V⃗ | − c)
= c · (t − a),

and

|r⃗(t) − r⃗(a)| = |\frac{r⃗(t) − r⃗(a)}{t − a} − V⃗ + V⃗ | · |t − a|
< (|V⃗ | − c + |V⃗ |) · |t − a|.

Theorem 3.5. Given r⃗(t) which is continuous on [a, b], and such that \frac{d}{dt}r⃗(t) extends to a continuous and nonvanishing function on [a, b) and which has arclength function ℓ(t) as in (3.1), then for any c, 0 < c < 1, there exists some δ > 0 so that for a ≤ t < a + δ,

c · ℓ(t) ≤ |r⃗(t) − r⃗(a)| ≤ ℓ(t).

Proof. By Theorem 3.3 (possibly applied to some shorter interval [a, b_{0}], b_{0} ≤ b), there exists a function f(s) on the domain [0, L] so that (r⃗ ∘ f)(0) = r⃗(a), r⃗ ∘ f is continuous on [0, L], has constant speed 1 on (0, L), and the one-sided derivative at a is also a unit vector. Lemma 3.4 applies to any c, 0 < c < 1, to give a lower bound, and there is a better upper bound:

c · s ≤ |r⃗(f(s)) − r⃗(f(0))| ≤ s

for s in some interval [0, δ_{1}), depending on 0 < c < 1. From the Proof of Theorem 3.1, f = ℓ^{−1}, so if 0 < s = ℓ(t) < δ_{1}, then

c · ℓ(t) ≤ |r⃗(t) − r⃗(a)| ≤ ℓ(t)

for 0 < t < ℓ^{−1}(δ_{1}).

The conclusion from the Theorem is that for some initial interval, the magnitude of the displacement is comparable to the arclength.

Given a continuous function r⃗(t) : [a, b] → ℝ^{n}, the composite |r⃗(t) − r⃗(a)| is a continuous function [a, b] → ℝ. If there is some interval (a, c) on which |r⃗(t) − r⃗(a)| is nonvanishing and \frac{d}{dt}r⃗(t) exists (for example, (a, a + δ) from Lemma 3.4), then:

\frac{d}{dt}(|r⃗(t) − r⃗(a)|) = \frac{d}{dt}\sqrt{∑_{k=1}^{n}(r_{k}(t) − r_{k}(a))^{2}}

= \frac{1}{2} · \frac{∑_{k=1}^{n} 2(r_{k}(t) − r_{k}(a)) · \frac{dr_{k}(t)}{dt}}{\sqrt{∑_{k=1}^{n}(r_{k}(t) − r_{k}(a))^{2}}}
= \frac{1}{|r⃗(t) − r⃗(a)|}(r⃗(t) − r⃗(a)) · \frac{dr⃗}{dt}
= cos(α(t)) |\frac{dr⃗}{dt} | .

The cosine appears from the dot product formula, where α(t) is the angle between the direction vector r⃗(t) − r⃗(a) and the velocity vector \frac{dr⃗}{dt}. If r⃗ happens to have unit speed for a < t < c, then \frac{d}{dt}(|r⃗(t) − r⃗(a)|) = cos(α(t)).

Exercise 3.6. Let F be an analytic function on a domain containing the closed unit disk {|z| ≤ 1}, with derivative f(z) = F′(z). If F (1) is real and f(1) = 1, then the squared modulus of the values of F on the unit circle, given by the real function

g(θ) = |F (e^{iθ})|^{2},

has a critical point at θ = 0.

4 Cauchy integrals

Notation 4.1. For r > 0 and z_{0} ∈ ℂ, let D(z_{0}, r) denote the Euclidean disk with center z_{0} and radius r, and as the special case with z_{0} = 0, abbreviate D(0, r) = D_{r}.

Notation 4.2. By a \underline{smooth arc}, we mean a continuous parametric map z : [0, 1] → ℂ with image Γ, which is one-to-one on [0, 1] with the possible exception of z(0) = z(1), and differentiable on (0, 1) with \frac{dz}{dt} extending to a continuous, non-vanishing function on [0, 1]. By a \underline{piecewise smooth arc}, we mean a continuous parametric map z : [0, 1] → ℂ with image Γ, which is one-to-one on [0, 1], with the possible exception of z(0) = z(1) (in this special case we say \underline{piecewise smooth contour}), so that the domain has a partition 0 = t_{0} < t_{1} < … < t_{N} = 1, where z restricted to each [t_{i}, t_{i+1}] is, after a suitable re-scaling of the domain, a smooth arc. In any case, the notation can be abused by referring only to Γ, with the parametrization (and the induced orientation) understood.

By Lemma 3.2, the one-sided derivatives are defined at the endpoints of a smooth arc.

Notation 4.3. For φ integrable on a piecewise smooth arc Γ (meaning, as in the previous Notation, (φ ∘ z)(t) · z′(t) is integrable on [0, 1]), define a function Φ : ℂ \ Γ → ℂ by the formula:

Φ(z) = \frac{1}{2πi} ∫_{Γ} \frac{φ(ζ)}{ζ − z} dζ.

More generally, for n = 1, 2, 3, 4, …, define

Φ_{n}(z) = \frac{1}{2πi} ∫_{Γ} \frac{φ(ζ)}{(ζ − z)^{n}} dζ,

so Φ_{1} = Φ.

Theorem 4.4. For a piecewise smooth arc Γ and φ continuous on Γ, Φ_{n} is complex differentiable on the complement of Γ, with \frac{d}{dz} Φ_{n} = n · Φ_{n+1}.

Proof. Note that the derivative formula is just a special case of interchanging derivative and integral signs: formally, \frac{d}{dz} ∫_{Γ} F (z, w)dw = ∫_{Γ} \frac{d}{dz} F (z, w)dw. We will prove only the special case and not make any more general claim; the outline of the proof follows [A].

Step 1: a factoring trick.

(w − a)^{n} − (w − z)^{n}
= (w − a)^{n} − (w − z)(w − a)^{n−1} + (w − z)(w − a)^{n−1} − (w − z)^{n}
= (w − a)^{n−1}((w − a) − (w − z)) + (w − z)(w − a)^{n−1} − (w − z)^{n}  (4.1)
= (w − a)^{n−1}(z − a) + (z − a)Q_{n−1}(z, w, a),  (4.2)

where the (w − z)(w − a)^{n−1} − (w − z)^{n} quantity in (4.1) is a polynomial in z of degree n, which has value 0 at z = a, so it factors as in (4.2). When n = 1, Q_{0} is identically zero.

Dividing by (w − z)^{n}(w − a)^{n},

\frac{1}{(w − z)^{n}} − \frac{1}{(w − a)^{n}}

= \frac{z − a}{(w − z)^{n}(w − a)} + \frac{1}{(w − z)^{n−1}(w − a)} − \frac{1}{(w − a)^{n}}  (4.3)
= (z − a) · \frac{(w − a)^{n−1} + Q_{n−1}(z, w, a)}{(w − z)^{n}(w − a)^{n}}.  (4.4)

Step 2: Φ_{n} is continuous on the complement of Γ. To show this, fix a ∉ Γ, and we want lim_{z→a} Φ_{n}(z) = Φ_{n}(a). Since the complement is open, we can find some radius ρ > 0 so that D(a, ρ) is contained in the complement, and for all w ∈ Γ, |w − a| > ρ and |w − z| > ρ/2 for z ∈ D(a, ρ/2). Using (4.4),

Φ_{n}(z) − Φ_{n}(a) = \frac{1}{2πi} ∫_{Γ} \frac{φ(w)}{(w − z)^{n}} dw − \frac{1}{2πi} ∫_{Γ} \frac{φ(w)}{(w − a)^{n}} dw
= \frac{1}{2πi} ∫_{Γ} φ(w) · (z − a) · \frac{(w − a)^{n−1} + Q_{n−1}(z, w, a)}{(w − z)^{n}(w − a)^{n}}dw.  (4.5)

Taking absolute value and using L for the arclength of Γ,

|Φ_{n}(z) − Φ_{n}(a)|
= \frac{|z − a|}{|2πi|} · |∫_{Γ} φ(w) · \frac{(w − a)^{n−1} + Q_{n−1}(z, w, a)}{(w − z)^{n}(w − a)^{n}}dw|
≤ \frac{|z − a|}{2π} \frac{1}{(ρ/2)^{n}} \frac{1}{ρ^{n}} max_{w∈Γ}\{|φ(w)|\} max\{|(w − a)^{n−1} + Q_{n−1}(z, w, a)|\} · L,

where the second max is over the compact product space {(z, w) : w ∈ Γ, |z − a| ≤ ρ/2}. Every factor in the last line except |z − a| depends only on Γ, φ, ρ, a, and n, but not on z, and this is enough to establish the claimed continuity.

Step 3: Φ_{1} is complex differentiable on ℂ \ Γ. Again, fix a ∉ Γ. Using the n = 1 case of (4.5), where Q_{0} ≡ 0,

\frac{Φ_{1}(z) − Φ_{1}(a)}{z − a} = \frac{1}{2πi} ∫_{Γ} \frac{φ(w)}{(w − z)(w − a)}dw  (4.6)

= \frac{1}{2πi} ∫_{Γ} \frac{(\frac{φ(w)}{(w−a)})}{(w − z)}dw.  (4.7)

In (4.7), the numerator \frac{φ(w)}{w−a} is continuous on Γ, so the n = 1 case of Step 2 applies with this fraction substituted for φ(w), and quantity (4.7) is, as a function of z, continuous at a. Taking the z → a limit of (4.6)=(4.7) gives:

Φ′_{1}(a) = \frac{1}{2πi} ∫_{Γ} \frac{(\frac{φ(w)}{(w−a)})}{(w − a)}dw = Φ_{2}(a).

The equation Φ′_{1} = 1 · Φ_{2} is the n = 1 case of the claimed derivative formula, and also the start of an induction on n.

Step 4: The inductive step. Assume, for any continuous φ on Γ, Φ′_{n−1} = (n − 1) · Φ_{n} on ℂ \ Γ.

\frac{Φ_{n}(z) − Φ_{n}(a)}{z − a}  (4.8)

= \frac{∫_{Γ} \frac{φ(w)}{(w−z)^{n}} dw − ∫_{Γ} \frac{φ(w)}{(w−a)^{n}} dw}{2πi(z − a)}

= \frac{∫_{Γ} φ(w) (\frac{z−a}{(w−z)^{n}(w−a)} + \frac{1}{(w−z)^{n−1}(w−a)} − \frac{1}{(w−a)^{n}} ) dw}{2πi(z − a)}  (4.9)

= \frac{1}{2πi} ∫_{Γ} \frac{(\frac{φ(w)}{w−a})}{(w − z)^{n}} dw + \frac{∫_{Γ} \frac{(\frac{φ(w)}{w−a})}{(w−z)^{n−1}} dw − ∫_{Γ} \frac{(\frac{φ(w)}{w−a})}{(w−a)^{n−1}} dw}{2πi(z − a)}
= Φ̃_{n}(z) + \frac{Φ̃_{n−1}(z) − Φ̃_{n−1}(a)}{z − a},  (4.10)

where (4.9) uses (4.3), and Φ̃_{n} in (4.10) denotes the Φ_{n} expression with the continuous function \frac{φ(w)}{w−a} substituted for φ(w), as in the previous step. Taking the z → a limit, quantity (4.8) has limit Φ′_{n}(a), and in (4.10), the first term Φ̃_{n}(z) is continuous at a by Step 2, and the limit of the second term is Φ̃′_{n−1}(a) = (n − 1) · Φ̃_{n}(a), by the inductive hypothesis. The conclusion is

Φ′_{n}(a) = Φ̃_{n}(a) + (n − 1)Φ̃_{n}(a)
= nΦ̃_{n}(a) = \frac{n}{2πi} ∫_{Γ} \frac{(\frac{φ(w)}{w−a})}{(w − a)^{n}} dw
= n · Φ_{n+1}(a).

Corollary 4.5. If f(z) is complex differentiable on an open set D, then f′(z) is also complex differentiable on D.

Proof. Given any a ∈ D, there is some disk D(a, r) ⊆ D; let Γ be the circle {|z − a| = r/2}, and let φ be the (continuous) restriction of f to Γ. Theorem 4.4 applies to z in D(a, r/2), where

Φ_{1}(z) = \frac{1}{2πi} ∫_{Γ} \frac{f(w)}{(w − z)^{1}} dw = f(z)

by the Cauchy Integral Formula, and the conclusion of the Theorem is that the derivative of f = Φ_{1} on D(a, r/2) is:

f′(z) = Φ_{2}(z) = \frac{1}{2πi} ∫_{Γ} \frac{f(w)}{(w − z)^{2}} dw,

which is complex differentiable on D(a, r/2).

Repeating the above construction shows that all higher derivatives f^{(n)} exist, with the formula

f^{(n)}(z) = \frac{n!}{2πi} ∫_{Γ} \frac{f(w)}{(w − z)^{n+1}} dw.

5 Principal values

Definition 5.1. For Γ as in Notation 4.3, a point τ ∈ Γ, and ε > 0, let Γ_{ε} denote the complement Γ\D(τ, ε), which for small ε is a pair of piecewise smooth arcs (or possibly one arc), parametrized by restricting the parametrization of Γ. If, for φ as in Notation 4.3, the limit

lim_{ε→0^{+}} \frac{1}{2πi} ∫_{Γ_{ε}} \frac{φ(ζ)}{ζ − τ} dζ  (5.1)

exists, then it is called the \underline{Principal Value} of the Cauchy integral Φ at τ, P.V.Φ(τ).

Definition 5.2. For any constant 0 < α < 1 and any set B ⊆ ℂ, a function φ : B → ℂ is \underline{Hölder continuous} with exponent α on B means: there is a constant C so that for any z_{1}, z_{2} ∈ B, |φ(z_{1}) − φ(z_{2})| < C|z_{1} − z_{2}|^{α}. This property is abbreviated φ ∈ C^{0,α}(B).

Theorem 5.3. For 0 < α < 1, and a piecewise smooth contour Γ, if φ ∈ C^{0,α}(Γ), then:

• The Principal Value of Φ exists at any point τ ∈ Γ;

• For any smooth point τ ∈ Γ, the Principal Value of Φ is equal to

\frac{1}{2πi} ∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ + \frac{1}{2}φ(τ).

Proof. The first step is to show that for a piecewise smooth arc Γ,

∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ

exists, as a complex valued improper integral, in the following sense. On Γ_{ϵ}, the quantity being integrated is continuous, so the only parts requiring attention are the one or two arcs with endpoint τ in the disk D(τ, ε). It is enough to show that there is some smooth arc reparametrizing a piece of Γ, z(t) ∈ Γ ∩ D(τ, ε), t ∈ [0, 1], with z(0) = τ, with the property: for any η > 0 there is some δ > 0 so that for all 0 < a < δ:

|∫_{z(a)}^{z(δ)} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ| < η.

By Lemma 3.4, we can choose a small enough sub-arc of Γ ∩ D(τ, ε), parametrized by z(t) on [0, B], z(0) = τ, |z′(t)| = 1 on (0, B), and |z(t) − z(0)| ≥ ct for some c > 0. (This is not yet related to η.)

Then, for any 0 < b < B,

∫_{b}^{B} \frac{|φ(z(t)) − φ(z(0))|}{|z(t) − z(0)|}dt

≤ ∫_{b}^{B} \frac{C|z(t) − z(0)|^{α}}{|z(t) − z(0)|} dt
= ∫_{b}^{B} \frac{C}{|z(t) − z(0)|^{1−α}} dt
≤ C ∫_{b}^{B} \frac{1}{(ct)^{1−α}} dt

= \frac{C}{c^{1−α}} \frac{t^{α}}{α} ]_{b}^{B} < \frac{CB^{α}}{αc^{1−α}} .

Since the quantity being integrated is nonnegative and has bounded integral for all b, as b decreases to 0, ∫_{b}^{B} weakly increases to some finite least upper bound 0 ≤ U ≤ \frac{CB^{α}}{αc^{1−α}}. In particular, for any η, there is some small δ > 0 so that for all 0 < a ≤ δ, U − η < ∫_{a}^{B} ≤ U. So, for all 0 < a < δ,

|∫_{z(a)}^{z(δ)} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ| = |∫_{t=a}^{t=δ} \frac{φ(z(t)) − φ(z(0))}{z(t) − z(0)}z′(t)dt|

≤ ∫_{a}^{δ} \frac{|φ(z(t)) − φ(z(0))|}{|z(t) − z(0)|}dt
= ∫_{δ}^{B} − ∫_{a}^{B} < U − (U − η) = η.

and the claim follows.

The previous part only required that Γ is a piecewise smooth arc; the next step uses the assumption that it is a contour, with global parametrization z(t). We make the simplifying assumption that τ = z(t_{0}) for some t_{0} ≠ 0, 1; otherwise, the argument can be easily modified.

For a smooth point τ, and sufficiently small ε > 0, there are numbers t_{1} < t_{2} so that Γ_{ε} is a connected arc, that is parametrized by z(t) in two sub-arcs, one for the restricted domain [0, t_{1}], and the other for [t_{2}, 1]. As a function of ζ, the expression \frac{1}{ζ−τ} has an antiderivative log(ζ − τ), away from some branch cut starting at τ and avoiding Γ_{ε} by staying in the connected exterior of Γ. Considering the quantity

∫_{Γ_{ε}} \frac{dζ}{ζ − τ} = ∫_{z(0)}^{z(t_{1})} \frac{dζ}{ζ − z(t_{0})} + ∫_{z(t_{2})}^{z(1)} \frac{dζ}{ζ − z(t_{0})}
= (log(z(t_{1}) − τ) − log(z(0) − τ))
+ (log(z(1) − τ) − log(z(t_{2}) − τ))
= log (\frac{z(t_{1}) − τ}{z(t_{2}) − τ} )
= ln |\frac{z(t_{1}) − τ}{z(t_{2}) − τ} | + i arg (\frac{z(t_{1}) − τ}{z(t_{2}) − τ} ) ,

as ε → 0^{+}, the integral approaches 0 + iπ. If τ is a \underline{corner} point (a shared endpoint of arcs in the contour) with angle ρ, then the above difference in arguments approaches ρ and the integral approaches iρ. The claimed formula for the smooth point follows from the add-and-subtract trick:

lim_{ε→0^{+}} \frac{1}{2πi} ∫_{Γ_{ε}} \frac{φ(ζ)}{ζ − τ} dζ = lim_{ε→0^{+}} (\frac{1}{2πi} ∫_{Γ_{ε}} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ + \frac{φ(τ)}{2πi} ∫_{Γ_{ε}} \frac{dζ}{ζ − τ} )
= \frac{1}{2πi} ∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ + \frac{1}{2}φ(τ).

Notation 5.4. Given a C^{0,α} function φ on a piecewise smooth arc Γ, and any point τ ∈ Γ, define Ψ_{τ} by

Ψ_{τ} (z) = ∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − z}dζ.

It was checked in the above proof that Ψ_{τ} (τ) is well-defined as an improper integral, and Ψ_{τ} (z) is analytic on ℂ \ Γ by Theorem 4.4.

Theorem 5.5. For any limit with z approaching τ non-tangentially,

lim_{z→τ} Ψ_{τ} (z) = Ψ_{τ} (τ).

Proof. The arc Γ and the point τ are given, and we also fix an arc on which z will approach τ, that has the following property: There is some δ_{1} > 0 and some c > 0 so that if z is on the approach arc and 0 < |z − τ| < δ_{1}, then for all ζ ∈ Γ, |z − ζ| > c|z − τ|. We take this as the definition of \underline{non-tangential}, although if we choose some other approach arc, the constants δ_{1} and c may be different.

Given η > 0, we want to show that there is some δ > 0 so that if z is on the approach arc and 0 < |z − τ| < δ, then |Ψ_{τ} (z) − Ψ_{τ} (τ)| < η. By definition,

Ψ_{τ} (z) − Ψ_{τ} (τ) = ∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − z}dζ − ∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ
= ∫_{Γ} \frac{(φ(ζ) − φ(τ))(z − τ)}{(ζ − z)(ζ − τ)}dζ.

From the Proof of Theorem 5.3, there is some ε > 0 and some parametrization ζ(t) of Γ near τ so that the improper integral satisfies:

|∫_{Γ\Γ_{ε}} \frac{φ(ζ) − φ(τ)}{ζ − τ}dζ| ≤ ∫_{t_{0}−δ_{2}}^{t_{0}+δ_{3}} \frac{|φ(ζ(t)) − φ(τ)|}{|ζ(t) − τ|}dt < \frac{ηc}{2}.

The ε depends on the constant c from the fixed approach arc, but not on the point z. For z on the approach arc such that 0 < |z − τ| < δ_{1}, \frac{|z−τ|}{|ζ−z|} < \frac{1}{c}, so

|∫_{Γ\Γ_{ε}} \frac{(φ(ζ) − φ(τ))(z − τ)}{(ζ − z)(ζ − τ)}dζ| < \frac{η}{2} .

Suppose 0 < |z − τ| < ε/2, so |ζ − z| > ε/2 for all ζ ∈ Γ_{ε}. Let L_{ε} be the arclength of Γ_{ε}. Then

|∫_{Γ_{ε}} \frac{φ(ζ) − φ(τ)}{(ζ − z)(ζ − τ)}dζ| ≤ max |\frac{φ(ζ) − φ(t)}{(ζ − z)(ζ − τ)}| L_{ε}
< \frac{2}{ε} max |\frac{φ(ζ) − φ(τ)}{ζ − τ}| L_{ε},

where the max is over ζ ∈ Γ_{ε}.

If max |\frac{φ(ζ)−φ(τ)}{ζ−τ} | L_{ε} = 0, then the claim is established, with δ = min\{δ_{1}, ε/2\}. Otherwise, if

0 < |z − τ| < \frac{ηε}{4 max |\frac{φ(ζ)−φ(τ)}{ζ−τ} | L_{ε}} ,

then

|∫_{Γ_{ε}} \frac{(φ(ζ) − φ(τ))(z − τ)}{(ζ − z)(ζ − τ)}dζ| = |z − τ| |∫_{Γ_{ε}} \frac{φ(ζ) − φ(τ)}{(ζ − z)(ζ − τ)}dζ|
< \frac{η}{2} .

For a piecewise smooth contour Γ as in Theorem 5.3, let D^{+} denote the interior region of Γ and let D^{−} denote the exterior. For φ and Φ as in Notation 4.3, define Φ^{±} : Γ → ℂ as a non-tangential limit, if it exists:

Φ^{±}(τ) = lim_{z→τ, z∈D^{±}} Φ(z).

Theorem 5.6 (Plemelj Jump). Given a C^{0,α} function φ on Γ, the functions Φ^{±} are well-defined at every smooth point τ, and satisfy

Φ^{±}(τ) = P.V.Φ(τ) ± \frac{1}{2}φ(τ).

Proof. At any smooth point, there is a non-tangential approach arc on either side. Use the add-and-subtract trick again.

Φ^{±}(τ) = lim_{z→τ, z∈D^{±}} \frac{1}{2πi} ∫_{Γ} \frac{φ(ζ)}{ζ − z} dζ
= \frac{1}{2πi} lim_{z→τ, z∈D^{±}} (∫_{Γ} \frac{φ(ζ) − φ(τ)}{ζ − z}dζ + φ(τ) ∫_{Γ} \frac{dζ}{ζ − z} ) .

So, using the Cauchy Integral Formula,

Φ^{+}(τ) = \frac{1}{2πi} lim_{z→τ, z∈D^{+}} Ψ_{τ} (z) + φ(τ),

and

Φ^{−}(τ) = \frac{1}{2πi} lim_{z→τ, z∈D^{−}} Ψ_{τ} (z).

The claimed result follows from the limit in Theorem 5.5 and the formula for the Principal Value in Theorem 5.3.

6 Power series

Definition 6.1. An infinite series of the form ∑_{n=0}^{∞} c_{n}(z − a)^{n} is called a \underline{power series}. The c_{n}

are called the \underline{coefficients}, and a is called the \underline{center} of the power series.

The coefficients c_{n}, the center a, and the variable z can all be complex numbers. The index n usually starts at 0 (the constant term is c_{0}z^{0} = c_{0}), or, if the first few coefficients are 0, n may start at any positive integer. (This definition of power series excludes negative or non-integer exponents n.)

Definition 6.2. The \underline{set of convergence} of a power series ∑_{n=0}^{∞} c_{n}(z − a)^{n} is the set of all (com-

plex) numbers z so that the series is convergent.

Note that the set of convergence cannot be the empty set, since any power series always

converges at its center, z = a: ∑_{n=0}^{∞} c_{n}(a − a)^{n} = c_{0} + c_{1}0^{1} + c_{2}0^{2} + ⋯ = c_{0}.

Proposition 6.3 (“Abel’s Lemma”). For a power series centered at a, ∑_{n=0}^{∞} c_{n}(z − a)^{n}, exactly

one of the following holds:

• ∑_{n=0}^{∞} c_{n}(z − a)^{n} converges to c_{0} at z = a, and diverges for all other z.

• ∑_{n=0}^{∞} c_{n}(z − a)^{n} is absolutely convergent for all z ∈ ℂ.

• There is a real number R > 0 so that ∑_{n=0}^{∞} c_{n}(z − a)^{n} is absolutely convergent for |z−a| < R,

and the series is divergent for |z − a| > R.

Proof. For a proof, see [C].

Definition 6.4. The number R is the \underline{radius of convergence} of the power series, and it must be nonnegative. The first two cases are referred to as R = 0 and R = ∞.

Note that Proposition 6.3 is inconclusive when both 0 < R < ∞ and |z − a| = R. (Geometrically, this is the case where the set of convergence is a disk in ℂ with positive radius, and z is on the circular boundary of the disk.) The power series could be divergent, absolutely convergent, or conditionally convergent for z on the boundary. In the case where the center a is on the real number line, then the real values of z for which the series is convergent form an interval centered at a (the intersection of the disk and the real axis), and the points on the boundary are the two endpoints, a − R and a + R.

Proposition 6.3 shows that in the case R > 0, the set of convergence is a “region” in the technical sense of [CB] Section 11, page 32, because it is the non-empty union of the open disk {|z − a| < R} and possibly some of its boundary points on the “circle of convergence” {|z − a| = R}, as on [CB] page 209, Section 63.

Proposition 6.5. If ∑_{n=0}^{∞} c_{n}(z − a)^{n} has radius of convergence R, then ∑_{n=1}^{∞} c_{n}n(z − a)^{n−1} also

has radius of convergence R.

Proof. For a proof, see [C].

Theorem 6.6. The function defined by f(z) = ∑_{n=0}^{∞} c_{n}(z − a)^{n} is analytic on the disk {|z − a| <

R}, with f′(z) = ∑_{n=1}^{∞} c_{n}n(z − a)^{n−1}.

Proof. The following steps give an elementary ϵ/δ argument, based on a proof from [BC]; unlike [CB] Chapter 5, we do not use integration and we do not use the general theory of uniform convergence, only what is specifically needed here.

First pick a specific point z in the disk {|z − a| < R}. Then choose some ρ so that |z − a| < ρ < R.

We want to check the definition of limit appearing in the definition of derivative at the point z, f′(z), so we want to show:

lim_{w→z} \frac{f(w) − f(z)}{w − z} = ∑_{n=1}^{∞} c_{n}n(z − a)^{n−1}.

The absolute convergence of the RHS series is Proposition 6.5. So, given ϵ > 0, we want to find δ > 0 so that

0 < |w − z| < δ ⟹ |\frac{f(w) − f(z)}{w − z} − ∑_{n=1}^{∞} c_{n}n(z − a)^{n−1}| < ϵ.  (6.1)

The δ may depend on ϵ, a, c_{n}, R, ρ, and z, but not on w.

For integer N = 1, 2, 3, …, define:

S_{N}(z) = ∑_{n=0}^{N} c_{n}(z − a)^{n}

R_{N}(z) = ∑_{n=N+1}^{∞} c_{n}(z − a)^{n}.

So, for any N, f(z) + S_{N}(z) + R_{N}(z), and lim_{N→∞} S_{N}(z) = f(z) and lim_{N→∞} R_{N}(z) = 0. We will choose a specific N later, in a way that may depend on ϵ, a, c_{n}, R, ρ, and z, but not on w and not on the δ we have not yet found. The quantity from (6.1) satisfies the following inequality for any N:

|\frac{f(w) − f(z)}{w − z} − ∑_{n=1}^{∞} c_{n}n(z − a)^{n−1}|

= |\frac{(S_{N}(w) + R_{N}(w)) − (S_{N}(z) + R_{N}(z))}{z − w}

− ((∑_{n=1}^{N} c_{n}n(z − a)^{n−1}) + ( ∑_{n=N+1}^{∞} c_{n}n(z − a)^{n−1}))|

≤ |\frac{S_{N}(w) − S_{N}(z)}{w − z} − ∑_{n=1}^{N} c_{n}n(z − a)^{n−1}|  (6.2)

+ | ∑_{n=N+1}^{∞} c_{n}n(z − a)^{n−1}|  (6.3)

+ |\frac{R_{N}(w) − R_{N}(z)}{w − z}| .  (6.4)

The term (6.3) is the tail end of the convergent series from Proposition 6.5: there’s some N_{1} so that if N > N_{1}, then the term (6.3) is less than ϵ/3. This cutoff N_{1} depends on ϵ, a, c_{n}, and z, but not on w.

The term (6.4) can be re-arranged:

\frac{R_{N}(w) − R_{N}(z)}{w − z} = \frac{( ∑_{n=N+1}^{∞} c_{n}(w − a)^{n}) − ( ∑_{n=N+1}^{∞} c_{n}(z − a)^{n})}{w − z}
= ∑_{n=N+1}^{∞} c_{n} \frac{(w − a)^{n} − (z − a)^{n}}{(w − a) − (z − a)}

= ∑_{n=N+1}^{∞} c_{n} ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k}  (6.5)

The step (6.5) follows from the polynomial identity

(w − a)^{n} − (z − a)^{n} = ((w − a) − (z − a)) ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k},

a telescoping sum similar to the geometric series formula, which holds for all z, w, and a. Assuming |w − a| < ρ, the following estimate holds:

|c_{n} ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k}| ≤ |c_{n}| ∑_{k=0}^{n−1} |z − a|^{k}|w − a|^{k}

< |c_{n}| ∑_{k=0}^{n−1} ρ^{k}ρ^{n−1−k} = |c_{n}|nρ^{n−1}.

The series ∑_{n=0}^{∞} c_{n} ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k} is absolutely convergent, by the Comparison Test

([C]) applied to

∑_{n=0}^{∞} |c_{n} ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k}| ≤ ∑_{n=0}^{∞} |c_{n}|nρ^{n−1},

which is absolutely convergent by Proposition 6.5 (with z = ρ + a). So (6.5) is the tail end of a convergent series, and there is some there is some N_{2} so that if N > N_{2} then

| ∑_{n=N+1}^{∞} c_{n} ∑_{k=0}^{n−1}(z − a)^{k}(w − a)^{n−1−k}| ≤ ∑_{n=N+1}^{∞} |c_{n}|nρ^{n−1} < ϵ/3.

This cutoff N_{2} depends on ϵ, c_{n}, and ρ, but not on the specific values of z or w, only that they both are in the disk D(a, ρ), and in particular w does not have to be close to z. It follows that if N > N_{2}, w ≠ z, |z − a| < ρ, and |w − a| < ρ, then |\frac{R_{N} (w)−R_{N} (z)}{w−z} | < ϵ/3.

Now, fix a number N > max{N_{1}, N_{2}}, and consider the term (6.2). S_{N}(z) = ∑_{n=0}^{N} c_{n}(z − a)^{n}

is a polynomial of degree N, with derivative

S′_{N}(z) = lim_{w→z} \frac{S_{N}(w) − S_{N}(z)}{w − z} = ∑_{n=1}^{N} c_{n}n(z − a)^{n−1},

so corresponding to ϵ/3 > 0, there is some δ_{1} > 0 so that if 0 < |w − z| < δ_{1}, then the (6.2) quantity is less than ϵ/3.

To bound all three terms at the same time, let δ = min{δ_{1}, ρ − |z − a|} > 0. Then |w − a| = |w − a + z − z| ≤ |w − z| + |z − a|, and if 0 < |w − z| < δ, then |w − a| ≤ ρ, which, together with N > N_{2}, is all we need for the ϵ/3 bound on the (6.4) term. The (6.3) term is bounded by ϵ/3 because N > N_{1}, and the (6.2) term is bounded by ϵ/3 because 0 < |w − z| < δ ≤ δ_{1}.

The following property of analytic functions may be useful in answering [CB] Exercises #59.12.a., p. 195, and #66.4 and #66.5, p. 219.

Lemma 6.7. Given an open set D ⊆ ℂ, a point z_{0} ∈ D, and a number n = 1, 2, 3, …, if h(z) is analytic on D and there is some disk {z : |z − z_{0}| < r} where h has a series expansion:

h(z) = ∑_{k=n}^{∞} c_{k}(z − z_{0})^{k},

then this function f is also analytic on D:

f(z) = \{ \begin{matrix}\frac{h(z)}{(z − z_{0})^{n}} & for z ≠ z_{0} \\ c_{n} & for z = z_{0} \end{matrix} \} .

Proof. The series for h has a positive radius of convergence r > 0 by Taylor’s Theorem from [CB] §57, p. 189 (possibly r = ∞); the hypothesis of the Lemma is that the coefficients c_{0}, …, c_{n−1} are all = 0, so the series starts with coefficient c_{n} (which may or may not be = 0).

The function f(z) is complex differentiable at every point in D \ {z_{0}}, by the quotient rule for derivatives. We only need to check that f is complex differentiable at z_{0} to get the claimed conclusion. Consider the series:

g(z) = ∑_{k=n}^{∞} c_{k}(z − z_{0})^{k−n}.

This is a power series with all non-negative exponents, and defines some function g so that g(z_{0}) = c_{n} = f(z_{0}). For any point z with 0 < |z − z_{0}| < r, the series is convergent, because it is equal to a scalar multiple of the convergent series for h:

g(z) = ∑_{k=n}^{∞} (c_{k}(z − z_{0})^{k} · \frac{1}{(z − z_{0})^{n}} )

= \frac{1}{(z − z_{0})^{n}} · ∑_{k=n}^{∞} c_{k}(z − z_{0})^{k} = \frac{1}{(z − z_{0})^{n}} h(z).

The scalar multiple rule applies because \frac{1}{(z−z_{0})^{n}} does not depend on the summation index k. We can conclude that the series for g(z) converges for all z with |z − z_{0}| < r, so g(z) is complex differentiable at z_{0} by Theorem 6.6 (or the Corollary from [CB] §65, p. 215). The above construction shows g(z) = f(z) for all z with |z − z_{0}| < r, so f is also complex differentiable at z_{0} and has a series expansion centered at z_{0} given by the g(z) series.

7 Rational functions

Definition 7.1. A \underline{rational function} of a complex variable z is defined by f(z) = P (z)/Q(z), where P and Q are polynomials and Q ≢ 0.

By the Fundamental Theorem of Algebra, there are finitely many points r where Q(r) = 0, so the domain of a rational function is {z ∈ ℂ : Q(z) ≠ 0}, the open, connected, non-empty complement of a finite set.

Lemma 7.2. Given any complex number w and polynomials P (z) and Q(z) so that the rational function f(z) = P (z)/Q(z) is non-constant on the domain D = {z ∈ ℂ : Q(z) ≠ 0}, the following are equivalent:

1. There exists a solution z = s ∈ D of the equation f(z) = w.

2. There exists s ∈ D and a polynomial F (z) ≢ 0 so that P (z) − wQ(z) = (z − s)F (z).

Proof. First, P (z) − wQ(z) is not the constant zero polynomial; otherwise, if P (z) − wQ(z) ≡ 0, then for any t ∈ D, P (t) = wQ(t) ⟺ P (t)/Q(t) = w = f(t), which contradicts the initial assumption that f is not a constant function in D.

For (2) ⟹ (1), plugging in z = s ∈ D gives P (s) − wQ(s) = (s − s)F (s) = 0 and Q(s) ≠ 0, so P (s) = wQ(s) and f(s) = P (s)/Q(s) = w.

Conversely, for (1) ⟹ (2), the hypothesis is that f(s) = P (s)/Q(s) = w, with Q(s) ≠ 0; this is equivalent to P (s) = wQ(s) ⟺ P (s) − wQ(s) = 0, so s is a root of the polynomial P (z) − wQ(z). P (z) − wQ(z) cannot be any constant polynomial: it was already shown that it can’t be ≡ 0, and it also cannot be any non-zero constant, because it has a root s. So P (z) − wQ(z) has degree ≥ 1 and root s, and it factors as (z − s)F (z) for some (possibly constant but not ≡ 0) polynomial F (z).

Theorem 7.3. For any non-constant rational function f(z) = P (z)/Q(z), the set of complex numbers w such that f(z) = w has no solution is a finite set.

Proof. Let P have degree M and let Q have degree N. Let f(z) = P (z)/Q(z) have domain D ⊆ ℂ as in Lemma 7.2.

Case 1. Q(z) is a constant function. Q ≢ 0 by definition, so f(z) = P (z)/Q(z) is a nonconstant polynomial and for any w, there is at least one root of the polynomial f(z) − w, by the Fundamental Theorem of Algebra. So the set of w with no solutions of f(z) = w is the empty set.

Case 2. Q(z) is a non-constant polynomial of degree N > 0 and, by the Fundamental Theorem of Algebra, factors as Q(z) = q_{N}(z − r_{1})(z − r_{2}) ⋯ (z − r_{N}), for leading coefficient q_{N} ≠ 0 and some possibly repeating list of roots (r_{1}, …, r_{N}). For any ordered list of nonnegative integers v⃗ = (v_{1}, v_{2}, …, v_{N}), define Rv⃗(z) = (z − r_{1})^{v_{1}}(z − r_{2})^{v_{2}} ⋯ (z − r_{N})^{v_{N}}, so Rv⃗ has leading coefficient 1 and degree v_{1} + ⋯ + v_{N}, and every root of Rv⃗ is one of the roots of Q.

If w is a number such that f(z) = w has no solution in D, then neither of the equivalent conditions in Lemma 7.2 holds. In particular, P (z) − wQ(z) cannot be equal to any polynomial of the form (z − s)F (z) for s ∈ D. We want to show that given P and Q, there are only finitely many such w.

Case 2.a. Consider the set of w such that P (z) − wQ(z) is a constant function. For any w in this set, applying the N^{th} derivative to both sides of P (z) − wQ(z) ≡ C gives

P^{(N)}(z) − wq_{N}N! ≡ 0,

so P^{(N)}(z) is a constant function and w = P^{(N)}(z)/(q_{N}N!) is uniquely determined by P and Q and does not depend on C (or z). The set of such w is either empty or has only this one element.

Case 2.b. Consider the set of w such that P (z) − wQ(z) is non-constant and not of the form (z − s)F (z) for any s ∈ D. So any root of P (z) − wQ(z) must be a root of Q(z), and we can conclude P (z) − wQ(z) must be equal to Cv⃗Rv⃗(z) for some non-zero leading coefficient Cv⃗ and some vector v⃗ with non-negative integer entries satisfying 0 < v_{1} + ⋯ + v_{N} ≤ max{M, N}. There are only finitely many such vectors v⃗; we will show that for each v⃗, there is at most one pair (w, Cv⃗) such that P (z) − wQ(z) = Cv⃗Rv⃗(z).

Pick any t in the non-empty set D, so Q(t) ≠ 0, Rv⃗(t) ≠ 0, and w and Cv⃗ satisfy P (t) − wQ(t) = Cv⃗Rv⃗(t) ≠ 0. Using this constant t, define a new polynomial

d(z) = det [ \begin{matrix}Q(t) & Rv⃗(t) \\ Q(z) & Rv⃗(z) \end{matrix} ] = Q(t)Rv⃗(z) − Rv⃗(t)Q(z).

If d(z) ≡ 0 then Rv⃗(z) = (\frac{Rv⃗(t)}{Q(t)}) Q(z), and P (z) − wQ(z) = Cv⃗ (\frac{Rv⃗(t)}{Q(t)}) Q(z), so P (z) −

(w − \frac{Cv⃗Rv⃗(t)}{Q(t)}) Q(z) ≡ 0. However, as in the Proof of Lemma 7.2, this implies f(z) = P (z)/Q(z) is constant on D, contradicting the assumption. So d(z) ≢ 0, and there is some u ∈ ℂ with d(u) ≠ 0. This means the pair (w, Cv⃗) satisfies P (u) − wQ(u) = Cv⃗Rv⃗(u) and also the linear system

[ \begin{matrix}Q(t) & Rv⃗(t) \\ Q(u) & Rv⃗(u) \end{matrix} ] [ \begin{matrix}w \\ Cv⃗ \end{matrix} ] = [ \begin{matrix}P (t) \\ P (u) \end{matrix} ] ,

where the constant coefficient matrix has non-zero determinant d(u), and there is exactly one solution (w, Cv⃗) satisfying this system. There may be other constraints on (w, Cv⃗) but checking P (z) − wQ(z) = Cv⃗Rv⃗(z) at these two points z = t and z = u already rules out the existence of more than one solution for (w, Cv⃗).

References

[A] L. Ahlfors, Complex Analysis, third ed., McGraw-Hill, New York, 1979.

[BC] J. Bruna and J. Cufí, Complex Analysis, Eur. Math. Soc., 2013.

[CB] R. Churchill and J. Brown, Complex Variables and Applications, eighth ed., McGraw-Hill, New York, 2009.

[C] A. Coffman, Notes on sequences and series in the calculus of one variable.

     https://users.pfw.edu/CoffmanA/

[GM] J. Gray and S. Morris, When is a function that satisfies the Cauchy–Riemann equations analytic?, Amer. Math. Monthly (4) 85 (1978), 246–256.