Power Series and Larent Series

Series

For definitions of series, check real analysis notes.

Power Series (Taylor Series)

Let \(f(z) = \sum^\infty_{n=0} a_n(z-z_0)^n\) for all \(z\in\mathbb C\), if the RHS converges, then we call such \(f(z)\) a power series centered at \(z_0\).

Commonly used Power Series

The power series for complex is almost the same as for reals (same proof).

\[S_n = \frac{1-c^{n+1}}{1-c} = \sum_{k=0}^n c^k, \forall |c| < 1. S_\infty = \frac{1}{1-c} = \sum_{k=0}^\infty c^k\]

\[e^z = \sum_{k=0}^\infty \frac{z^k}{k!}\]

\[\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{1}{2}(\sum_{k=0}^\infty \frac{i^kz^k}{k!} + \sum_{k=0}^\infty \frac{(-i)^kz^k}{k!}) = \sum_{k=0}^\infty \frac{(-1)^kz^{2k}}{(2k)!}\]

\[\sin z = \sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)!}\]

Theorem 1

Let \(f(z) = \sum^\infty a_n z^n\), if \(f(z_1)\) converges absolutely for some \(z_1\in \mathbb C\), then \(f(z)\) converges on \(B_{|z_1|}(0)\) and converges uniformly on \(\overline{B_r(0)}\) for all \(r \in (0, |z_1|)\).

proof. First note that \(\lim_\infty |a_n z_1^n| = 0\) since the series converges absolutely at \(z_1\). Let \(r \in (0, z_1), z \in \overline{B_r(0)}\).
take \(N > 0\) s.t. \(\forall n\geq N. |a_n z_1^n| < 1\). Therefore,

\[|a_nz^n| = |a_nz_1^n||\frac z{z_1}|^n < |\frac z{z_1}|^n \leq (\frac r{|z_1|})^n\]

Therefore, by M-test, \(f(z)\) is uniformly convergent on \(\overline{B_r(0)}\).

Example 1

\(f(z) = \sum^\infty \frac{z^n}{n!}\) converges on \(\mathbb C\) and is continuous.

proof. Let \(z\in\mathbb C\), take \(r > |z|\), since \(r\in\mathbb R\), we know that \(f(r) = e^r\). Thus, by the above theorem, \(f(z)\) converges for all \(z\in\mathbb C\). In addition, we have uniform converges of \(f(z)\) on any closed ball, hence it is also continuous.

Existence of Power Series Expansion

Theorem If \(f(z)\) is analytic on \(B_r(z_0)\), then on \(B_r(z_0)\) we have

\[f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n\]

and the series is absolutely convergent and uniformly convergent on \(\overline{B_\delta(z_0)}\) for all \(\delta \in (0,r)\), also convergent on \(B_r(z_0)\).

proof. Let \(z\in B_r(0)\), let \(\delta\) where \(|z| < \delta < r_1 < r\). Take \(C\) be the circle of radius \(r_1\) centered at 0. Then

\[\begin{align*} f(z) &= \frac1{2\pi i}\oint_C\frac{f(w)}{w-z}dw &\text{CIF}\\ &= \frac1{2\pi i}\oint_C\frac{f(w)}{w}\frac1{1-z/w}dw \\ &= \frac1{2\pi i}\oint_C\frac{f(w)}w\sum^\infty (\frac zw)^ndw \\ &= \frac1{2\pi i}\oint_C\sum^\infty \frac{f(w)}{w^{n+1}}z^ndw \\ &= \sum^\infty \frac1{2\pi i}\oint_C \frac{f(w)}{w^{n+1}}z^ndw\\ &= \sum^\infty \frac{f^{(n)}(0)}{n!}z^n &\text{CIF} \end{align*}\]

Therefore, we can use M-test to prove the claim.

Radius of Convergence

Theorem (existence) Let \(f(z) = \sum^\infty a_nz^n\), then \(\exists R \in [0, \infty)\) s.t. \(f(z)\rightarrow z\in B_r(0)\), does not converge for \(z\in \mathbb C - \overline{B_R(0)}\) and converge uniformly on \(\overline{B_r(0)}\) for all \(r\in(0,R)\).

And we have 1. ratio test If \(\lim_{n\rightarrow\infty} \frac{|a_{n+1}|}{|a_n|} = S, R = S^{-1}\) 2. root test. If \(\lim_\infty |a_n|^{1/n} = S, R = S^{-1}\) 3. Hadamard's Theorem. \(S = \lim_\infty\sup |a_n|^{1/n} = S, R = S^{-1}\)

Lemma \(\lim_\infty |a_n|^{1/n} = \lim_\infty\sup \big((n+1)|a_{n+1}|\big)^{1/n}\).
proof. Part of Hadamard's Theorem.

Theorem Let \(f(z) = \sum^\infty a_nz^n\) with radius of convergence \(R>0\), then \(f(z)\) is analytic on \(B_R(0)\).

proof. Let \(g(z) = \sum^\infty (n+1)a_{n+1}z^n\), by lemma, \(g\) also has radius of convergence \(R\).

Then, we need to show that there exists some anti-derivative of \(g\) and g is continuous on \(B_R(0)\), so we need to show that all Jordon contour \(C \subset B_R(0)\), \(\oint_C g(z)dz = 0\). Note that this holds since there is some \(0 <\delta < R\) s.t. \(C\subset \overline{B_\delta(0)}\) and g is uniformly convergent on \(\overline{B_\delta(0)}\), and thus on \(C\). Therefore,

\[\oint_C g(z)dz= \oint_C \sum^\infty (n+1)a_{n+1} z^n dz = \sum^\infty \oint_C (n+1)a_{n+1} z^n dz = \sum^\infty 0\]

Therefore, take \(G\) be some anti-derivative s.t.

\[\begin{align*} G(z_1) &= \int_0^{z_1} g(z)dz \\ &= \int_0^{z_1} \sum^\infty_{n=0} (n+1)a_{n+1} z^n dz\\ &= \sum^\infty_{n=0} \int_0^{z_1} (n+1)a_{n+1} z^n dz&\text{u.c. on} B_{|z_1|}(0)\\ &= \sum^\infty_{n=1} \frac{na_n}{n}z_1^n\\ &= \sum^\infty_{n=1} a_nz_1^n\\ &= f(z_1) - a_0 \end{align*}\]

Therefore, \(f(z) = G(z) + a_0\) is analytic.

Laurent Series

Define an annulus \(A_{r_1, r_2}(z_0) = \{z: r_1 < |z-z_0| < r_2\}\) and \(A_{r_1, \infty} (z_0) = \{z : |z-z_0| > r_1\}\).

Theorem Let \(f(z)\) be analytic on \(A_{r_1,r_2}(0)\), then \(\forall z \in A_{r_1,r_2}(0)\),

\[f(z) = \sum_{n=-\infty}^{\infty}\bigg(\frac1{2\pi i}\oint_C \frac{f(w)}{w^{n+1}}dw\bigg) z^n\]

where \(C \subset A_{r_1,r_2}(0)\) is a Jordon contour s.t. \(0\in C_{int}\). Moreover, \(f\) converges uniformly on \(\overline{A_{s_1,s_2}(0)}\) for all \(r_1<s_1<s_2<r_2\).

proof. Let \(z \in A_{r_1,r_2}(0)\), let \(s_1, s_2, r_1', r_2'\) s.t. \(r_1< r_1' < s_1 \leq |z| \leq s_2 < r_2' < r_2\). Then, CIF and CT gives

\[f(z) = \frac1{2\pi i}\oint_{C_{r_2'}}\frac{f(w)}{w-z}dw - \frac1{2\pi i}\oint_{C_{r_1'}}\frac{f(w)}{w-z}dw = I_2 - I_1\]

As from power series proof.

\[I_2 =\sum_{n=0}^\infty\frac{1}{2\pi i}\oint_C \frac{f(w)}{w^{n+1}} dw z^n\]

And similarly,

\[-I_1 =\frac1{2\pi i} \oint_C\frac{f(w)}{z}\frac1{1-w/z} dw = \sum_{n=-\infty}^{-1} \frac1{2\pi i}\oint_C \frac{f(w)}{w^{n+1}} dw z^n\]

Lemma 1

If \(f(z) = \sum_0^\infty a_n z^n\) and \(g(z) = \sum_0^\infty b_nz^n\) on \(B_r(0)\), then either \(f=g\) or \(\exists r > \delta > 0\) s.t. \(f\neq g\) on \(A_{0, \delta}(0)\).

proof. If \(f(0)\neq g(0)\), then by continuity, \(\exists \delta > 0\) s.t. \(f(z)\neq g(z)\) on \(B_\delta(0)\). Now suppose that \(f(0)= g(0)\) i.e. If \(f\neq g\), then there must exist some \(n\) s.t. \(a_n \neq b_n\). Take \(n_0\) be the least \(n\) s.t. \(a_{n_0}\neq b_{n_0}\). Therefore,

\[\begin{align*} f(z) &= &\sum_{n=0}^{n_0-1} a_nz^n + &z^{n_0} \sum_{n=0}^\infty a_{n - n_0}z^n\\ &= &p(z) + &z^{n_0}\tilde f(z)\\ g(z) &= &\sum_{n=0}^{n_0-1} a_nz^n + &z^{n_0} \sum_{n=0}^\infty b_{n - n_0}z^n\\ &= &p(z) + &z^{n_0}\tilde g(z) \end{align*}\]

Notablly, \(\tilde f(0) = a_{n_0} \neq b_{n_0} = \tilde g(0)\), so exists \(\delta > 0\) s.t. \(\tilde f\neq \tilde g\) on \(B_\delta(0)\). Therefore, on \(z^{n_0}\tilde f\neq z^{n_0}\tilde g\) on \(A_{0, \delta}(0)\)

Theorem 2

Let \(f,g\) be analytic on a domain \(D\). Let \(A\subset D\) s.t. \(f(z) = g(z)\) for all \(z\in A\). If \(A\) has a limit point contained in \(D\) then \(f=g\) on \(D\).

proof. Let \(z_0\) be a limit point of \(A \subseteq D\). Write the functions as their power series expansion \(f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n\) and \(g(z)=\sum_{n=0}^\infty b_n(z-z_0)^n\) on \(B_r(z_0)\subset D\). We want to show \(f=g\) on such \(B_r(z_0)\).

By the lemma above, - Either \(f=g\) on \(B_r(z_0)\) - Or exists \(\delta > 0\) s.t. \(f\neq g\) on \(A_{0, \delta}(z_0)\). Take such \(\delta\), however, \(z_0\) is a limit point of \(A\) so that this is a contradiction.

Now, take \(D'\subseteq A \subseteq D\) be the largest open set s.t. \(f=g\). If \(D'\neq D\), let \(z_1 \in \partial D'\cap D\), since \(z_1\) is a limit point, we can expand \(D'\) with \(B_r(z_1)\). \(D'\cup B_r(z_1)\) is larger than \(D'\) and \(f=g\), hence contradiction.