Power Series

By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

September 30, 2019

Contents

Convergence of Power Series

Definition & Theorem – Radius of Convergence.

‌ Let \(c \in \mathbb{C}\) and \(a_n \in \mathbb{C}\) for every \(n \in \mathbb{N}\). The radius of convergence of the power series \[ \sum_{n = 0}^{+\infty} a_{n} (z-c)^{n} \] is the unique \(r \in [0,+\infty]\) such that the series converges if \(|z - c|< r\) and diverges if \(|z - c| > r.\) The disk \(D(c,r)\) – the largest open disk centered on \(c\) where the series converges – is the open disk of convergence of the series.

The radius of convergence \(r\) is the inverse of the growth ratio of the sequence \(a_n,\) defined as the infimum in \([0,+\infty]\) of the set of values \(\sigma \in [0,+\infty)\) such that \(a_n\) is eventually dominated by \(\sigma^n\): \[ \exists \, m \in \mathbb{N}, \; \forall \, n \in \mathbb{N}, \; (n \geq m) \, \Rightarrow \, |a_{n}| \leq \sigma^{n}. \] (or equivalently, such that \(\exists \, \kappa > 0,\) \(\forall \, n \in \mathbb{N},\) \(|a_{n}| \leq \kappa \sigma^{n}.\)) This growth ratio is equal to \(\limsup_{{n} \to +\infty} |a_{n}|^{1/{n}},\) which leads to the Cauchy-Hadamard formula1: \[ r = \frac{1}{\displaystyle \limsup_{n \to +\infty} |a_{n}|^{1/{n}}}. \] By convention here, \(1/0 = +\infty\) and \(1/(+\infty) = 0.\)

Proof.

‌Let \(\rho\) be the growth ratio of the sequence \(a_n.\) If a complex number \(z\) satisfies \(|z - c| < \rho^{-1},\) \(\rho\) is finite and there is a \(\sigma > \rho\) such that \(|z-c| < \sigma^{-1}.\) Eventually, we have \(|a_n| \leq \sigma^n\) and thus \[ |a_n (z - c)^n| \leq (\sigma |z-c|)^n. \] As \(\sigma |z-c| < 1,\) the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is convergent. Conversely, if \(|z - c| > \rho^{-1},\) \(\rho > 0\) and there is a \(\sigma < \rho\) such that \(|z-c| > \sigma^{-1}.\) As \(\sigma < \rho,\) there is a strictly increasing sequence of \(n \in \mathbb{N}\) such that \(|a_n| > \sigma^n\) and thus \(|a_n (z - c)^n| > (\sigma \sigma^{-1})^n = 1.\) Since its terms do not converge to zero, the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is divergent.

We now prove that the growth ratio of \(|a_n|\) is equal \(\limsup_n |a_n|^{1/n}.\) Indeed, for any \(\sigma\) greater than the growth ratio \(\rho,\) eventually \(|a_n| \leq \sigma^n,\) hence \(|a_n|^{1/n} \leq \sigma\) and \(\limsup_n |a_n|^{1/n} \leq \sigma,\) therefore \(\limsup_n |a_n|^{1/n} \leq \rho.\) Conversely, if \(\sigma\) is smaller than the growth ratio, there is a strictly increasing sequence of \(n \in \mathbb{N}\) such that \(|a_n| > \sigma^n,\) hence \(|a_n|^{1/n} > \sigma\) and \(\limsup_n |a_n|^{1/n} \geq \sigma,\) thus \(\limsup_n |a_n|^{1/n} \geq \rho.\) \(\blacksquare\)

Example – A Geometric Series.

‌ Consider the power series \[ \sum_{n=0}^{+\infty} (-1/2)^n z^n. \] Since \(|(-1/2)^n| = 1/2^n \leq \sigma^n\) eventually if and only if \(\sigma \geq 1/2,\) the growth bound of the geometric sequence \((-1/2)^n\) is \(1/2.\) Thus the open disk of convergence of this power series is \(D(0,2).\)

Example – A Lacunary Series.

‌ Consider the power series: \[ \sum_{n=0}^{+\infty} z^{2^n} = z + z^2 + z^4 + z^8 + \cdots. \] The “lacunary” adjective refers to the large gaps between nonzero coefficients; These coefficients are defined by \[ a_n = \left| \begin{array}{ll} 1 & \mbox{ if } \; \exists \, p \in \mathbb{N}, \; n = 2^p, \\ 0 & \mbox{ otherwise.} \end{array} \right. \] It is plain that \(|a_n| \leq \sigma^n\) eventually if and only if \(\sigma \geq 1.\) Hence the growth bound of the sequence if \(1\) and the open disk of convergence of the power series is \(D(0,1).\)

Lemma – Multiplication of Power Series Coefficients.

‌ The radius of convergence of the power series \(\sum_{n=0}^{+\infty} a_n b_n (z-c)^n\) is at least the product of the radii of convergence of the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) and \(\sum_{n=0}^{+\infty} b_n (z-c)^n\). In particular, for any nonzero polynomial sequence \[ a_n = \alpha_0 + \alpha_1 n + \dots + \alpha_p n^p, \] the radii of convergence of \(\sum_{n=0}^{+\infty} a_n b_n (z-c)^n\) and \(\sum_{n=0}^{+\infty} b_n (z-c)^n\) are identical.

Proof.

‌ Denote by \(\rho_a\) and \(\rho_b\) the respective growth bounds of the sequences \(a_n\) and \(b_n\); the growth bound of the product sequence \(a_n b_n\) is at most \(\rho_a \rho_b\): for any \(\sigma > \rho_a \rho_b,\) we may find some \(\sigma_a > \rho_a\) and \(\sigma_b > \rho_b\) such that \(\sigma= \sigma_a \sigma_b.\) Since \(|a_n| \leq (\sigma_a)^n\) and \(|b_n| \leq (\sigma_b)^n\) eventually, \(|a_n b_n| \leq \sigma^n\) eventually.

The growth bound of any polynomial sequence \(a_n\) is at most \(1\): the inequality \[|\alpha_0 + \alpha_1 n + \dots + \alpha_p n^p| \leq \rho^n\] holds for any \(\rho > 1\) eventually. Now, for any nonzero polynomial sequence \(a_n\) and any sequence \(b_n\), eventually \(|b_n|\) is dominated by a multiple of \(|a_n b_n|\), thus the growth bound of \(|b_n|\) is at most the growth bound of \(|a_n b_n|\). Reciprocally, the growth bound of \(|a_n b_n|\) is at most the product of the growth bound of \(|a_n|\) – at most one – and the growth bound of \(|b_n|\) and thus at most the growth bound of \(|b_n|\). \(\blacksquare\)

Theorem – Locally Normal Convergence.

‌ The convergence of the power series \(\sum_{n=0}^{+\infty} a_n(z-c)^n\) in its open disk of convergence \(D(c,r)\) is locally normal: for any \(z \in D(c,r)\), there is an open neighbourghood \(U\) of \(z\) in \(D(c,r)\) such that \[ \exists \, \kappa > 0, \; \forall \, z \in U, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \kappa \] or equivalently, for every compact subset \(K\) of \(D(c,r),\) \[ \exists \, \kappa > 0, \; \forall \, z \in K, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \kappa. \]

Proof.

‌ If \(K\) is compact subset of \(D(c,r)\) and \(\rho = \sup \, \{|z-c| \; | \; z \in K\},\) \[ \forall \, z \in K, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \sum_{n=0}^{+\infty} |a_n| \rho^n. \] Since the growth bound of the sequence \(a_n\) and \(|a_n|\) are identical, the radius of convergence of the series \(\sum_{n=0}^{+\infty} |a_n| (z-c)^n\) is \(r.\) Given that \(\rho < r,\) the series \(\sum_{n=0}^{+\infty} |a_n| \rho^n\) is convergent; all its terms are non-negative real numbers, thus the sum is finite: there is a \(\kappa>0\) such that \(\sum_{n=0}^{+\infty} |a_n| \rho^n \leq \kappa.\) \(\blacksquare\)

Remark – Other Types of Convergence.

‌ The locally normal convergence implies the absolute convergence: \[ \forall \, z \in D(c, r), \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| < + \infty. \] It also provides the locally uniform convergence: on any compact subset \(K\) of \(D(c,r),\) the partial sums \(\sum_{n=0}^p a_n(z-c)^n\) converge uniformly to the sum \(\sum_{n=0}^{+\infty} a_n(z-c)^n\): \[ \lim_{p\to +\infty} \sup_{z \in K} \left|\sum_{n=0}^p a_n(z-c)^n - \sum_{n=0}^{+\infty} a_n(z-c)^n \right| = 0. \]

Power Series and Holomorphic Functions

Theorem – Power Series Derivative.

‌ A power series and its formal derivative \[ \sum_{n = 0}^{+\infty} a_{n} (z-c)^{n} \; \mbox{ and } \; \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] have the same radius of convergence \(r.\) The sum \[f: z \in D(c,r) \mapsto \sum_{n=0}^{+\infty} a_n (z-c)^n\] is holomorphic; its derivative is the sum of the formal derivative: \[ \forall \, z \in D(c,r), \; f'(z) = \sum_{n=1}^{+\infty} n a_{n} (z-c)^{n-1}. \] More generally, the \(p\)-th order derivative of \(f\) is defined for any \(p \in \mathbb{N}\) and \[ \forall \, z \in D(c,r), \; f^{(p)}(z) = \sum_{n=p}^{+\infty} n(n-1) \cdots (n-p+1) a_n (z-c)^{n-p}. \]

Lemma.

‌For any \(z \in \mathbb{C},\) \(h \in \mathbb{C}^*\) and \(n \geq 2,\) \[ \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| \leq \frac{n(n-1)}{2} (|z| +|h|)^{n-2} |h|. \]

Proof – Lemma.

‌Using the identity \(a^n - b^n = (a-b) \sum_{m=0}^{n-1} a^m b^{n-1-m}\) yields \[ (z+h)^n - z^n = h \sum_{m=0}^{n-1} (z+h)^m z^{n-1-m}, \] hence \begin{eqnarray*} \frac{(z+h)^n - z^n}{h} - n z^{n-1} &=& \sum_{m=0}^{n-1} (z+h)^m z^{n-1-m} - \sum_{m=0}^{n-1} z^m z^{n-1-m} \\ &=& \sum_{m=0}^{n-1} \left[(z+h)^m - z^m\right] z^{n-1-m}. \end{eqnarray*} By the same identity, we also have \[ \left| (z+h)^m - z^m \right| = \left| h \sum_{l=0}^{m-1} (z+h)^{l} z^{m-1-l} \right| \leq m (|z| + |h|)^{m - 1} |h|. \] Therefore \begin{eqnarray*} \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| &\leq& \left[ \sum_{m=0}^{n-1} m \ (|z| + |h|)^{m - 1} (|z| + |h|)^{n - 1 - m} \right] |h| \\ &\leq& \frac{n(n-1)}{2} (|z| + |h|)^{n - 2} |h| \end{eqnarray*}

as expected. \(\blacksquare\)

Proof – Power Series Derivative.

‌ Let \(D(c,r)\) be the open disk of convergence of the series \[ f(z) = \sum_{n = 0}^{+\infty} a_n (z-c)^{n}. \] The radii of convergence of the series \[ \sum_{n = 1}^{+\infty} na_n (z-c)^{n-1} \; \mbox{ and } \; \sum_{n = 0}^{+\infty} na_n (z-c)^{n} \] are equal. Since the coefficient sequence of the latter series is the product of \(a_n\) and a nonzero polynomial sequence, the open radius of convergence of \(f\) and of its the formal derivative are identical. For any \(z \in D(c,r)\) and any complex number \(h\) such that \(0 < |h| < r\), define \(e(z, h)\) as \[ e(z, h) = \frac{f(z+h) - f(z)}{h} - \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] A straightforward calculation leads to \[ e(z, h) = \sum_{n=1}^{+\infty} a_{n} \left[\frac{(z + h - c)^{n} - (z - c)^n}{h} - n(z-c)^{n-1} \right], \] hence, using the lemma, we obtain \[ \left| e(z,h) \right| \leq \left[\sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| (|z - c| + |h|)^{n-2}\right] \times |h|. \] The power series \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| z^{n-2} \] has the same radius of convergence than \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} a_{n} (z - c)^{n-2} \] which is the the formal derivative of order 2 of the original series, hence the three series have the same radius of convergence \(r.\) Consequently, for any \(h\) such that \(|z-c| + |h| < r,\) \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| (|z - c| + |h|)^{n-2} < + \infty \] and therefore \[ \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] The statement about the \(p\)-th order derivative of \(f\) can be obtained by a simple induction on \(p.\) \(\blacksquare\)

Theorem & Definition – Taylor Series.

‌ If the complex-valued function \(f\) has a power series expansion centered at \(c\) inside the non-empty open disk \(D(c,r),\) it is the Taylor series of \(f\): \[ \forall \, z \in D(c,r), \; f(z) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(c)}{n!} (z-c)^n. \]

Proof.

‌If \(f(z) = \sum_{n=0}^{+\infty} a_n (z - c)^n,\) then for any \(p \in \mathbb{N},\) the \(p\)-th order derivative of \(f\) inside \(D(c, r)\) is given by \[ f^{(p)}(z) = \sum_{n=p}^{+\infty} n(n-1) \dots (n-p+1) a_n (z - c)^{n-p} \] and consequently, \(f^{(p)}(c) = p ! a_p.\) \(\blacksquare\)

Note that the above theorem is only a uniqueness result; it says nothing about the existence of the power series expansion. This is the role of the following theorem.

Theorem – Power Series Expansion.

‌ Let \(\Omega\) be an open subset of \(\mathbb{C},\) let \(c \in \Omega\) and \(r \in \left]0,+\infty\right]\) such that the open disk \(D(c, r)\) is included in \(\Omega.\) For any holomorphic function \(f: \Omega \to \mathbb{C},\) there is a power series with coefficients \(a_n\) such that \[ \forall \, z\in D(c,r), \; f(z) = \sum_{n=0}^{+\infty} a_n (z-c)^n. \] Its coefficients are given by \[ \forall \, \rho \in \left]0,r\right[, \; a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}} \, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft]. \]

Proof – Power Series Expansion.

‌ For any \(n \in \mathbb{N},\) the complex number \[ a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}}\, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft] \] is independent of \(\rho\) as long as \(0 <\rho <r.\) Indeed, if \(\rho_1\) and \(\rho_2\) are two such numbers, denote \(\gamma_1 = c + \rho_1 [\circlearrowleft]\) and \(\gamma_2 = c + \rho_2 [\circlearrowleft].\) The interior of the sequence of paths \(\mu = \gamma_{1} \, | \, \gamma_2^{\leftarrow}\) is included in \(D(c, r) \setminus \{c\}\) where the function \(z \mapsto {f(z)/}{(z-c)^{n+1}}\) is holomorphic. Hence, by Cauchy’s integral theorem, \[ \int_{\mu} \frac{f(z)}{(z-c)^{n+1}}\, dz = \int_{\gamma_1} \frac{f(z)}{(z-c)^{n+1}}\, dz - \int_{\gamma_2} \frac{f(z)}{(z-c)^{n+1}}\, dz = 0. \]

Now, let \(z \in D(c, r)\) and let \(\rho \in \left]0,r \right[\) such that \(|z-c| < \rho.\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} \, dw. \] For any \(w \in \gamma([0,1]),\) we have \[ \frac{1}{w-z} = \frac{1}{(w-c) - (z-c)} = \frac{1}{w-c}\frac{1}{1 - \frac{z-c}{w-c}}. \] Since \[ \left|\frac{z-c}{w-c}\right| = \frac{|z-c|}{\rho} < 1, \] we may expand \(f(w)/(w-z)\) into \[ \frac{f(w)}{w-z} = \frac{f(w)}{w-c}\frac{1}{1 - \frac{z-c}{w-c}} = \sum_{n=0}^{+\infty} \frac{f(w)}{w-c} \left(\frac{z-c}{w-c}\right)^n. \] The term of this series is dominated by \[ \frac{\sup_{|w-c|=\rho} |f(w)|}{\rho} \left(\frac{|z-c|}{\rho}\right)^n; \] the convergence of the series is normal – and thus uniform – with respect to the variable \(w.\) Finally \[ \begin{split} f(z) &= \frac{1}{i2\pi} \int_{\gamma} \left[ \sum_{n=0}^{+\infty}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \right] dw \\ &= \sum_{n=0}^{+\infty} \left[ \frac{1}{i2\pi} \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \, dw \right]\\ &= \sum_{n=0}^{+\infty} \left[ \frac{1}{i2\pi} \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n \end{split} \] which is the desired expansion. \(\blacksquare\)

Laurent Series

Definition – Annulus.

‌ Let \(c \in \mathbb{C}\) and \(r_1, r_2 \in [0, +\infty].\) We denote by \[ A(c, r_1, r_2) = \{ z \in \mathbb{C} \; | \; r_1 < |z-c| < r_2 \} \] the open annulus with center \(c,\) inner radius \(r_1\) and outer radius \(r_2.\)
  1. The open annulus \(A(0, 0, +\infty),\) centered on the origin, with inner radius \(0\) and outer radius \(+\infty,\) is the set \(\mathbb{C}^*.\)

  2. The sets \(A(0,0,1),\) \(A(0,1,2)\) and \(A(0,2,+\infty)\) are three open annuli centered on the origin and included in the open set \(\Omega = \mathbb{C} \setminus \{i,2\}.\) They are maximal in \(\Omega\) – if we decrease their inner radius and/or increase their outer radius the resulting annulus is not a subset of \(\Omega\) anymore.

Definition – Laurent Series.

‌ The Laurent series centered on \(c\in \mathbb{C}\) with coefficients \(a_n \in \mathbb{C}\) for every \(n \in \mathbb{Z}\) is \[ \sum_{n=-\infty}^{+\infty} a_n (z-c)^n. \] It is convergent for some \(z \in \mathbb{C} \setminus \{c\}\) if the series \[ \sum_{n=0}^{+\infty} a_n(z-c)^n \; \mbox{ and } \; \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n} \] are both convergent – otherwise it is divergent. When the Laurent series is convergent its sum is defined as \[ \sum_{n=-\infty}^{+\infty} a_n (z-c)^n = \sum_{n=0}^{+\infty} a_n(z-c)^n + \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n}. \]

Theorem – Convergence of Laurent Series.

‌ Let \(c \in \mathbb{C}\) and let \(a_n \in \mathbb{C}\) for \(n \in \mathbb{Z}.\) The inner radius of convergence \(r_1 \in [0,+\infty]\) and outer radius of convergence \(r_2 \in [0,+\infty]\) of the Laurent series \(\sum_{n=-\infty}^{+\infty} a_n (z-c)^n\) defined by \[ r_1 = \limsup_{n\to+\infty} {|a_{-n}|^{1/n}} \; \mbox{ and } \; r_2 = \frac{1}{\displaystyle \limsup_{n\to+\infty}{|a_n|^{1/n}}}. \] are such that the series converges in \(A(c, r_1, r_2)\) and diverges if \(|z-c| < r_1\) or \(|z-c| > r_2.\) In this open annulus of convergence, the convergence is locally normal.

Proof – Convergence of Laurent Series.

‌ The first series converges if \(|z-c|\) is smaller than the radius of convergence \(r_2\) of this power series and diverges if it is greater. We may rewrite the second series as: \[ \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n} = \sum_{n=1}^{+\infty} a_{-n}\left(\frac{1}{z-c}\right)^{n}. \] Consequently, it converges if \(|1/(z-c)|\) is smaller than the radius of convergence \(1/r_1\) of the power series \(\sum_{n=1}^{+\infty} a_{-n} z^n,\) that is if \(|z-c| > r_1,\) and diverges if \(|1/(z-c)|\) is greater than \(1/r_1,\) that is \(|z-c|\) is smaller than \(r_1.\)

Now, for any \(z \in A(c,r_1,r_2),\) there is an open neighbourhood \(U\) of \(z\) where \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is normally convergent and an open neighbourhood \(V\) of \((z-c)^{-1}\) in \(\mathbb{C}^*\) where \(\sum_{n=1}^{+\infty} a_{-n}w^{n}\) is normally convergent. The Laurent series \(\sum_{n=-\infty}^{+\infty} a_n (z-c)^n\) is normally convergent in the open neighbourhood \(U \cap \{w^{-1} + c \; | \; w \in V\}\) of \(z.\) \(\blacksquare\)

Theorem – Laurent Series Expansion.

‌ Let \(\Omega\) be an open subset of \(\mathbb{C},\) let \(c \in \mathbb{C}\) and \(r_1, r_2 \in [0,+\infty]\) such that \(r_1 < r_2\) and the open annulus \(A(c, r_1,r_2)\) is included in \(\Omega.\) For any holomorphic function \(f: \Omega \to \mathbb{C},\) there is a Laurent series with coefficients \(a_n\) such that \[ \forall \, z\in A(c, r_1, r_2), \; f(z) = \sum_{n=-\infty}^{+\infty} a_n (z-c)^n. \] Its coefficients are given by \[ \forall \, \rho \in \left]r_1,r_2\right[, \; a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}} \, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft]. \]

Proof – Laurent Series Expansion.

‌ For any integer \(n,\) the coefficient \[ a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}}\, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft] \] is independent of \(\rho \in \left]r_1,r_2\right[\) – refer to the proof of “Power Series Expansion” for a detailled argument.

Let \(z \in A(c, r_1, r_2)\) and \(\rho_1,\rho_2 \in \left]r_1,r_2\right[\) such that \(\rho_1 <|z-c| < \rho_2.\) Let \(\gamma_1 = c + \rho_1 [\circlearrowleft]\) and \(\gamma_2 = c + \rho_2 [\circlearrowleft];\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw - \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw \] As in the proof of “Power Series Expansion”, we can establish that \[ \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw = \sum_{n=0}^{+\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_2}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] A similar argument, based on a series expansion of \[ \frac{1}{w-z} = -\frac{1}{(z-c) - (w-c)} = -\frac{1}{z-c}\frac{1}{1 - \frac{w-c}{z-c}} \] yields \[ \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw = -\sum_{n=-1}^{-\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_1}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] The combination of both expansions provides the expected result. \(\blacksquare\)

Notes


  1. to compute the limit superior of a sequence of (extended) real numbers, consider all subsequences that converge (as extended real numbers: in \([-\infty, +\infty]\)) and take the supremum of their limits.