# 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: $r = \frac{1}\limsup_{n \to +\infty} |a_{n}|^{1/{n}}}.$ 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 $$\sum_{n=0}^{+\infty} |a_n| z^n$$ 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 $$h \in \mathbb{C},$$ 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 $\begin{split} f(z) &= \int_{\gamma} \left[ \sum_{n=0}^{+\infty}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \right] dw \\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \, dw \right]\\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n \end{split}$ 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 $$r_2.$$

### Examples – Annuli.

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 $r_1 = \limsup_{n\to+\infty} {|a_{-n}|^{1/n}} \; \mbox{ and } \; r_2 = \frac{1}\limsup_{n\to+\infty}{|a_n|^{1/n}}}.$ 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. $$\blacksquare$$