# Series: Exercises

## Analytic Properties of Series

Show that any power series converges uniformly within its radius of convergence.

Write $$S_N(z) \coloneqq\sum_{0\leq k\leq N} c_k (z-z_0)^k$$ and $$S \coloneqq\lim_{N\to\infty} S_N$$. Suppose $$R\coloneqq\qty{\limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} }^{-1}$$ is the radius of convergence and let $$r\leq R$$, we’ll show $$S_N\to S$$ uniformly on any disc $${\left\lvert {z-z_0} \right\rvert}< r$$.

Use the $$M{\hbox{-}}$$test: $$\sum f_k$$ converges if $${\left\lVert {f_k} \right\rVert}_\infty\leq M_k$$ where $$\left\{{M_k}\right\}\in \ell^1({\mathbb{N}})$$. Define $$f_k \coloneqq c_k (z-z_0)^k$$, then \begin{align*} {\left\lVert {f_k} \right\rVert}_\infty =\sup_{{\left\lvert {z-z_0} \right\rvert}\leq r} {\left\lvert {c_k(z-z_0)^k} \right\rvert} \leq {\left\lvert {c_k} \right\rvert} r^k \coloneqq M_k .\end{align*} Then \begin{align*} \sum_{k\geq 0} M_k = \sum_{k\geq 0} {\left\lvert {c_k} \right\rvert} r^k ,\end{align*} and the claim is that this converges.

Note that since $$r\leq R$$, we have convergence of \begin{align*} \sum_{k\geq 0} c_k r^k .\end{align*} Recall that root test: \begin{align*} \sum_k a_k \text{ converges absolutely if } \limsup_k {\left\lvert {a_k} \right\rvert}^{1\over k} < 1 .\end{align*} Here we take $$a_k \coloneqq c_k r^k$$, then \begin{align*} \limsup_k {\left\lvert {a_k} \right\rvert}^{1\over k} &\coloneqq\limsup_k {\left\lvert {c_k r^k} \right\rvert}^{1\over k} \\ &= \limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} r \\ &< \limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} R\\ &\coloneqq\limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} \qty{\limsup_{k} {\left\lvert {c_k} \right\rvert}^{1\over k} }^{-1}\\ &= 1 ,\end{align*} so $$\sum_k {\left\lvert {c_k r^k} \right\rvert} < \infty$$. Thus $$\left\{{M_k}\right\}\in \ell^1({\mathbb{N}})$$, and so $$\sum_k f_k$$ converges uniformly and absolutely on $${\left\lvert {z-z_0} \right\rvert} = r < R$$.

Show that any power series is continuous on its domain of convergence.

Let $$f(z) = \lim_{N\to\infty} \sum_{k\leq N} c_k (z-z_0)^k$$. Use that power series converge uniformly and absolutely within their disc of convergence, each term is a continuous function, and finite sums of continuous functions are again continuous. So the partial sums $$S_N$$ are continuous, and since $$S_N\to f$$ uniformly, $$f$$ is continuous by the uniform limit theorem.

Suppose $$f_k: \Omega\to {\mathbf{C}}$$ is a sequence of differentiable functions converging locally uniformly to $$f:\Omega\to {\mathbf{C}}$$. Show that

• $$f$$ is continuous,
• $$f$$ is differentiable,
• $$\left\{{f_k'}\right\}\to f'$$ locally uniformly.

Thus if $$f(z) = \sum{k\geq 0} c_k (z-z_0)^k$$ is a power series, since $$S_N\to f$$ locally uniformly, $$f$$ can be differentiated term-by-term within its radius of convergence.

That $$f$$ is continuous is a local question: fixing a point $$z_0$$, take a closed disc $${\mathbb{D}}+z_0$$ about $$z_0$$. By local uniform convergence $$f_k\to f$$ uniformly on $${\mathbb{D}}+z_0$$, and differentiable $$\implies$$ continuous. So each $$f_k$$ is continuous, making $$f$$ continuous on $${\mathbb{D}}+z_0$$ by the uniform limit theorem.

That $$f$$ is differentiable is again a local question: fix $$z$$ and write $$\gamma \coloneqq\overline{{\mathbb{D}}+ z}$$ as the boundary of the disc about $$z$$. Define $$g_k(\xi) \coloneqq{f_k\over \xi-z}$$, so $$g_k \to {f \over \xi-a}$$ locally uniformly. Now apply Cauchy’s integral formula at $$z$$: \begin{align*} f(z) &= \lim_k f_k(z) \\ &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over \xi - z}\,d\xi\\ &= \lim_k {1\over 2\pi i}\int_\gamma g_k(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma \lim_k g_k(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma g(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over \xi - z} \,d\xi ,\end{align*} where we’ve used uniform convergence on $$\gamma$$ to commute the limit and integral. So $$f$$ has an integral representation, making it differentiable.

That $$f_k'\to f'$$: \begin{align*} \lim_k f_k'(z) &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over (\xi - z)^2 }\,d\xi\\ &= {1\over 2\pi i}\int_\gamma \lim_k {f_k(\xi) \over (\xi - z)^2}\,d\xi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over (\xi - z)^2}\,d\xi\\ &= f'(z) .\end{align*}

That the convergence is locally uniform: first consider what happens on an closed discs $$K = D$$ with $$\gamma \coloneqq{{\partial}}{D}$$. Then for $$z\in D$$, \begin{align*} {\left\lvert {f'(z) - f_k'(z) } \right\rvert} &= {\left\lvert {{1\over 2\pi i} \int_{\gamma} {f(\xi) - f_k(\xi) \over (\xi - z)^2}\,d\xi} \right\rvert}\\ &\leq {1\over 2\pi}\int_{\gamma} {{\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^2} \,d\xi\\ &\leq {1\over 2\pi}\int_{\gamma } { \sup_{\xi \in \gamma} {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over r^2 } \,d\xi\\ &= {1\over 2\pi} { \sup_{\xi \in \gamma } {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over r^2} \cdot {2\pi r} \\ &= { \sup_{\xi \in \gamma } {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert}}/r .\end{align*} Since $$\gamma$$ is compact, using locally uniform convergence of $$f_k\to f$$, there exists an $$n_0$$ such that $$n\geq n_0$$ bounds this $$\sup$$ by $${\varepsilon}$$. For $$K$$ arbitrary, cover $$K$$ by discs $$D_z$$ for every $$z\in K$$ and extract a finite cover $$\left\{{D_{z_k}}\right\}_{k\leq N}$$. Produce $$n_0, n_1,\cdots, n_N$$ as in the above argument, and take $$n\coloneqq\max\left\{{n_k}\right\}_{k\leq N}$$ to obtain uniform convergence on every $$D_{z_k}$$ and thus on $$K$$.

Find the radius of convergence of

• $$\sum a^k z^k$$ for $$a$$ a constant.
• $$\sum a^{k^2}z^k$$

• $$R = {1 \over \limsup {\left\lvert {a^k} \right\rvert}^{1\over k}} = {1\over {\left\lvert {a} \right\rvert}}$$
• $$R = {1 \over \limsup {\left\lvert {a^{k^2}} \right\rvert}^{1\over k}} = {1\over \limsup {\left\lvert {a} \right\rvert}^k}$$, so $$R=\infty$$ if $${\left\lvert {a} \right\rvert}< 1$$, $$R=0$$ if $${\left\lvert {a} \right\rvert}<1$$, and $$R=1$$ if $${\left\lvert {a} \right\rvert} = 1$$.

Find the radius of convergences for the power series expansion of $$\sqrt{z}$$ about $$z_0 = 4 +3i$$. Repeat with $$z_1=-4+3i$$.

Choose the principal branch of $$\log$$, so take a branch cut at $${\mathbf{R}}_{\leq 0}$$, to define $$z^{1\over 2} = e^{{1\over 2}\log(z)}$$. The radius of convergence is the distance to the nearest singularity or branch cut, so note that $$f(z) = z^{1\over 2}$$ is singular at $$z=0$$, so we compute $${\left\lvert {z_0 - 0} \right\rvert} = {\left\lvert {4+3i} \right\rvert} = 5$$. The distance to the branch is also 5, so $$R=5$$.

For $$z_1$$, the distance to zero is $${\left\lvert {4+3i - 0} \right\rvert} = 5$$ but the distance to the branch is 4, so $$R=4$$.

Note the subtle distinction: the series converges to $$f$$ in a disc $${\left\lvert {z-z_0} \right\rvert}<1$$, but the series itself converges in larger discs.

Find the radius of convergence for \begin{align*} f(z) \coloneqq\sum_{k\in {\mathbf{Z}}} 2^{-{\left\lvert {k} \right\rvert}}z^k .\end{align*}

Break this up into a principal part at $$z=0$$ and a holomorphic part: \begin{align*} f(z) = f_1(z) + f_2(z) \coloneqq\sum_{k\geq 1} 2^{-k}z^{-k} + \sum_{k\geq 0} 2^{-k}z^k .\end{align*}

Using the ratio test: \begin{align*} f_1(z) < \infty &\impliedby \limsup_k {\left\lvert {2^{-k}z^{-k}} \right\rvert}^{1\over k} < 1 \iff \limsup_k {\left\lvert {1\over 2z} \right\rvert} < 1 \iff {1\over 2}< {\left\lvert {z} \right\rvert} \\ f_2(z) < \infty &\impliedby \limsup_k {\left\lvert {2^{-k}z^{k}} \right\rvert}^{1\over k} < 1 \iff \limsup_k {\left\lvert {z\over 2} \right\rvert}< 1 \iff {\left\lvert {z} \right\rvert} < 2 .\end{align*}

So $$f$$ converges on $${1\over 2}< {\left\lvert {z} \right\rvert} < 2$$.

## Finding Laurent Expansions

Expand $$f(z) = {1\over z(z-1)}$$ in both

• $${\left\lvert {z} \right\rvert} < 1$$
• $${\left\lvert {z} \right\rvert} > 1$$

\begin{align*} {1\over z(z-1)} = -{1\over z}{1 \over 1-z} = -{1\over z}\sum z^k .\end{align*} and \begin{align*} {1\over z(z-1)} = {1\over z^2(1 - {1\over z})} = {1\over z^2} \sum \qty{1\over z}^k .\end{align*}

Find the Laurent expansion about $$z=0$$ and $$z=1$$ respectively of the following function: \begin{align*} f(z) \coloneqq{z+1 \over z(z-1)} .\end{align*}

Note: once you see that everything is in terms of powers of $$(z-z_0)$$, you’re essentially done. For $$z=0$$: \begin{align*} {z+1 \over z(z-1)} &= {1\over z} {z+1 \over z-1} \\ &= -{z+1\over z} {1\over 1-z} \\ &= -\qty{1 + {1\over z}}\sum_{k\geq 0} z^k .\end{align*}

For $$z=1$$: \begin{align*} {z+1 \over z(z-1)} &= {1\over z-1}\qty{1 + {1\over z} } \\ &= {1\over z-1}\qty{1 + {1\over 1 - (1-z)} } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (1-z)^k } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (-1)^k (z-1)^k } .\end{align*}

Find a Laurent expansion for $$f(z) \coloneqq{1\over (z-3)(z-1)}$$ on the 3 annular regions centered at $$0$$ where $$f$$ is holomorphic.

The three regions are

• $$0 \leq {\left\lvert {z} \right\rvert} < 1$$
• $$1 < {\left\lvert {z} \right\rvert} < 3$$
• $$3 < {\left\lvert {z} \right\rvert} < \infty$$

Write $$f$$ in terms of its principal parts at $$z=1$$ and $$z=3$$ by computing the residues:

• $$\mathop{\mathrm{Res}}_{z=1}f(z) = (z-1)f(z)\Big|_{z=1} = {1\over z-3}\Big|_{z=1} = -{1\over 2}$$
• $$\mathop{\mathrm{Res}}_{z=3}f(z) = (z-3)f(z)\Big|_{z=3} = {1\over z-1}\Big|_{z=3} = {1\over 2}$$

Thus \begin{align*} f(z) = {-1/2 \over z-1} + {1/2 \over z-3} .\end{align*}

Now find the two expansions for each term:

\begin{align*} {-1/2 \over z-1} &= {1/2 \over 1-z} = {1\over 2}\sum_{k\geq 0} z^k && 0 < {\left\lvert {z} \right\rvert} < 1 \\ {-1/2 \over z-1} &= -{1\over 2}{z^{-1}\over z^{-1}- 1} = -{1\over 2z}{1\over 1-z^{-1}} = -{1\over 2}\sum_{k\geq 0}z^{-k-1} && 1 < {\left\lvert {z} \right\rvert} < \infty \\ {1/2\over z-3} &= -{1\over 2}{1\over 3-z} = -{1\over 6}{1\over 1-{z\over 3}} = -{1\over 6}\sum_{k\geq 0}3^{-k} z^k && 0 < {\left\lvert {z} \right\rvert} < 3 \\ {1/2\over z-3} &= {1\over 2z}{1\over 1-3z^{-1}} = {1\over 2z} \sum_{k\geq 0}3^kz^{-k} = {1\over 2}\sum_{k\geq 0}3^k z^{-k-1} && 3 < {\left\lvert {z} \right\rvert} < \infty .\end{align*}

Now, just combinatorics to pick the various series that converge on the desired regions: \begin{align*} 0 \leq {\left\lvert {z} \right\rvert} < 1 \qquad & f(z) = {1\over 2}\sum_{k\geq 0}z^k - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 1 \leq {\left\lvert {z} \right\rvert} < 3 \qquad & f(z) = -{1\over 2}\sum_{k\geq 0}z^{-k-1} - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 3 \leq {\left\lvert {z} \right\rvert} < \infty \qquad & f(z) = - {1\over 2}\sum_{k\geq 0}z^{-k-1} + {1\over 2}\sum_{k\geq 0} 3^{k}z^{-k-1} .\end{align*}

Expand $$f(z) = z^2\cos\qty{z\over 3}$$ about $$z=0$$.

\begin{align*} f(z) = z^2\qty{ 1 + {1\over 2!}\qty{1\over 3z}^2 + {1\over 4!}\qty{1\over 3z}^4 } = z^2 + {1\over 2! \cdot 3^2} + {1\over 4! \cdot 3^4}z^{-2} + \cdots .\end{align*}

Find a power series expansion of \begin{align*} f(z) = {1\over e^z-1} .\end{align*}

One way: polynomial long division. \begin{align*} {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^2 + \cdots } \\ &= {1\over z}\qty{1 - {1\over 2}z + \qty{-{1\over 6} + {1\over 4} }z^2 + \cdots } \\ &= z^{-1}- {1\over 2} + {1\over 12}z + { \mathsf{O}}(z^3) .\end{align*} Alternatively, use geometric series. Note that something like $${1\over 1-e^z} = \sum_{k\geq 0} e^{kz}$$ won’t converge, and won’t even be calculable since each $$e^{kz}$$ contributes a constant term! \begin{align*} {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^3 + \cdots } \\ &= {1\over z(1 + {1\over 2}z + {1\over 6}z^2 + \cdots) } \\ &= z^{-1}{1\over 1 + q(z) } \qquad q(z) \coloneqq{1\over 2}z + {1\over 6}z^2 + \cdots \\ &= z^{-1}\sum_{k\geq 0}(-q(z))^k \\ &= z^{-1}\qty{1 - q(z) + q(z)^2 - \cdots } \\ &= z^{-1}\qty{1 - \qty{{1\over 2}z + {1\over 6}z^2 + \cdots } + \qty{{1\over 2}z + {1\over 6}z^2 + \cdots }^2 - \cdots } \\ &= z^{-1}\qty{ 1 - {1\over 2} z + \qty{-{1\over 6} + \qty{1\over 2}^2 } z^2 + { \mathsf{O}}(z^3) } \\ &= z^{-1}- {1\over 2} + {1\over 12}z + { \mathsf{O}}(z^2) .\end{align*}

Find a Laurent expansion at $$z=0$$ for \begin{align*} f(z) \coloneqq e^{1\over z}\cos\qty{1\over z} .\end{align*}

Note that a direct expansion won’t work, since there are infinitely many contributions to the constant term. Instead, a trick: consider $$g(z) \coloneqq e^z\cos(z)$$, so $$g(1/z ) = f(z)$$. Expanding $$g$$ is easier: \begin{align*} g(z) &= e^{z}\cos(z)\\ &= {1\over 2}e^z\qty{e^{iz} + e^{-iz}} \\ &= {1\over 2}\qty{e^{(1+i)z} + e^{(1-i)z}} \\ &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {z^k\over k!} \\ \implies f(z) &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {1 \over k!z^k } \\ .\end{align*}

Find a Laurent expansion about $$z=0$$ of \begin{align*} f(z) \coloneqq\cos\qty{1- {1\over z}} ,\end{align*} and compute the “residue” coefficient $$c_{-1}$$.

Write $$g(z) \coloneqq\cos(1-z)$$, so $$g(1/z) = f(z)$$, and expand: \begin{align*} g(z) &= \cos(1-z) \\ &= {1\over 2}\qty{e^{i(1-z)} + e^{-i(1-z)}} \\ &= {1\over 2}\qty{e^i e^{-iz} + e^{-i} e^{iz}}\\ &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {z^k \over k!} \\ \implies f(z) &= {1\over 2}\sum_{k\geq 0} \qty{ ( (-i)^k e^i + i^k e^{-i} } {1 \over k!z^k} .\end{align*}

Taking $$k=1$$ yields \begin{align*} c_{-1} = {-ie^i + ie^{-i} \over 2} = -i\cdot {e^i - e^{-i}\over 2} = {e^i - e^{-i}\over 2i} = \sin(1) .\end{align*}

Find a Laurent expansion that converges for $${\left\lvert {z} \right\rvert} > 1$$ of \begin{align*} f(z) \coloneqq{1 \over e^{1-z}} .\end{align*}

\begin{align*} f(z) = e^{-(1-z)} = e^{z-1} = e^{-1}e^z = e^{-1}\sum_{k\geq 0} {z^k\over k!} .\end{align*} Since $$e^z$$ is entire, this converges on $${\mathbf{C}}$$.

Find a Laurent expansion for \begin{align*} f(z) = {1\over 1 + e^z} \end{align*} about $$z_0 = 0$$ and $$z_1 = i\pi$$.

At $$z=0$$, we can use a geometric series approach since $${\left\lvert {e^z} \right\rvert} = e^{\Re(z)} \leq 1$$ near $$0$$. However, we still have to get rid of the leading 1 in the expansion of $$e^z$$ in order to get a constant coefficient. \begin{align*} {1\over 1 + e^z} &= {1\over 1 + 1 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + { \mathsf{O}}(z^4)} \\ &= {1\over 2 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + { \mathsf{O}}(z^4) } \\ &= {1\over 2} {1\over 1 + {1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!} z^3 + { \mathsf{O}}(z^4) } \\ &= {1\over 2}{1\over 1 - (-p(z)) } \qquad p(z) \coloneqq{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}}(z^4) \\ &= {1\over 2} \sum_{k\geq 0} (-p(z))^k \\ &= {1\over 2}\Big[ 1 - \qty{{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}}\qty{z^4} } \\ &\qquad + \qty{ {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}}\qty{z^4} }^2 \\ &\qquad - \qty{{1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}}\qty{z^4} }^3 \\ &\qquad - { \mathsf{O}}(z^4) \Big]\\ &= {1\over 2} \Big[ 1 + z\qty{- {1\over 2}} + z^2\qty{- {1\over 2\cdot 2!} + \qty{1\over 2}^2}\\ &\qquad + z^3 \Big( -{1\over 2\cdot 3! } +\left[ \qty{1\over 2}^3 + {1\over 2}{1\over 2\cdot 2!}\right] - \qty{1\over 2}^3 \Big) \\ &\qquad + { \mathsf{O}}(z^4) \Big]\\ &= {1\over 2} - {1\over 4}z + 0z^2 + {1\over 48}z^3 + { \mathsf{O}}(z^4) .\end{align*}

Expanding at $$z-i\pi$$: quite a bit easier. Let $$\omega \coloneqq z-i\pi$$, then \begin{align*} {1\over 1 + e^z} &= {1\over 1 + e^{z-i\pi}e^{i\pi}} \\ &= {1\over 1 - e^{\omega} } \\ &= {1 \over -\omega - {1\over 2!}\omega^2 - {1\over 3!}\omega^3 - { \mathsf{O}}(\omega^4) } \\ &= -{1\over \omega} {1 \over 1 + {1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}}(\omega^3) } \\ &= -{1\over \omega}{1\over 1-(- p(z) ) } \qquad p(z) \coloneqq\sum_{k\geq 2}{\omega^{k-1}\over k!} \\ &= -{1\over \omega} \sum_{k\geq 0} (-p(z))^k \\ &= -{1\over \omega} \left[ 1 - \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}}(\omega^3)} + \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}}(\omega^3)}^2 - { \mathsf{O}}(\omega^3) \right] \\ &= -{1\over \omega} \left[ 1 + \omega\qty{-{1\over 2!}} + \omega^2\qty{-{1\over 3!} + \qty{1\over 2!}^2} + { \mathsf{O}}(\omega^3) \right] \\ &= -{1\over w} + {1\over 2} - {1\over 12}\omega + { \mathsf{O}}(\omega^2) .\end{align*}

## New Things in $${\mathbf{C}}$$

Show that if $$f(z) \sum_{k\in {\mathbf{Z}}} c_k (z-z_0)^k$$, then \begin{align*} c_k = {1\over 2\pi i}\int_\gamma {f(z) \over (z-z_0)^{n+1}}\,dz ,\end{align*} and that this converges in an annulus $$D_R(z_0)\setminus\overline{D_r(z_0)}$$ where \begin{align*} r=\limsup _{n \rightarrow \infty} \sqrt[n]{\left|a_{-n}\right|} \text { and } R=\frac{1}{\limsup _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|}} \text {. } .\end{align*}

Hint: start with \begin{align*} f(z)=\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=s_{2}} \frac{f(w)}{w-z} d w-\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=s_{1}} \frac{f(w)}{w-z} d w ,\end{align*} and try to obtain a geometric series to obtain \begin{align*} f(z)=\sum_{j=-\infty}^{\infty}\left(\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=r} \frac{f(w)}{\left(w-z_{0}\right)^{j+1}} d w\right)\left(z-z_{0}\right)^{j} .\end{align*}