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One can see that \(\dfrac{x^{n+1}}{1-x}\) is strictly increasing on \([0,1)\) and thus \(\dfrac{x^{n}}{1-x}\to\infty\) as \(x\to 1\). Thus \(d_{n}\not\to 0\) and \(\sum z^{k}\) does not converge uniformly on \(D(0;1)\).
### Laurent Series
By a series of the form \(\sum_{n=-\infty}^{\infty}c_{n}(z-a)^{n}\) we mean
\[\sum_{n=-\infty}^{\infty}c_{n}(z-a)^{n} =\sum_{n=1}^{\infty}c_{-n}(z-a)^{-n}+\sum_{n=0}^{\infty}c_{n}(z- a)^{n}\] \[=\sum_{n=1}^{\infty}\dfrac{b_{n}}{(z-a)^{n}}+\sum_{n=0}^{\infty} c_{n}(z-a)^{n}\]
where we set \(c_{-n}=b_{n}\). By the convergence of \(\sum_{n=-\infty}^{\infty}c_{n}(z-a)^{n}\) we mean convergence of both series on the right. A power series converges inside a disc, and consequently a power series with negative powers alone should converge outside a disc. Those with positive and negative powers should therefore converge in a region between two concentric circles. Such a region is called an annulus.
**Theorem 145** (Laurent Series).: _Suppose \(f\) is holomorphic on \(A=\{z:R_{1}<|z-a| \[f(z)=\sum_{n=-\infty}^{\infty}c_{n}(z-a)^{n}\] _where both series converge absolutely on \(A\). Moreover,_ \[c_{n}=\dfrac{1}{2\pi i}\int_{\gamma}\dfrac{f(w)}{(w-a)^{n+1}}dw\] : Without loss of generality assume \(a=0\). As in the proof of Taylor’s Theorem, our starting point is Cauchy’s Integral Formula. Fix \(z\in A\) and choose \(r_{1}\) and \(r_{2}\) such that \(R_{1} Adding and noting that integrals along the line segments cancel, we have \[f(z)=\frac{1}{2\pi i}\int_{\gamma_{2}}\frac{f(w)}{w-z}dw-\frac{1}{2\pi i}\int_{ \gamma_{1}}\frac{f(w)}{w-z}dw. \tag{3.6} \]\] Now we can use Taylor’s theorem for \(w\in\gamma_{2}\), since \(z\) is fixed inside \(\gamma_{2}\); thus \(\left|\frac{z}{w}\right|<1\) and \[\frac{1}{w-z}=\frac{1}{w\left(1-\frac{z}{w}\right)}=\sum_{n=0}^{\infty}\frac{ z^{n}}{w^{n+1}}.\] We can use Taylor’s theorem again, this time with \(w\in\gamma_{2}\) and \(z\) fixed outside \(\gamma_{1}\). Thus \(\left|\frac{w}{z}\right|<1\) and \[\frac{1}{w-z}=\frac{-1}{-w+z}=-\frac{1}{z}\cdot\frac{1}{1-\left(\frac{w}{z} \right)}=-\sum_{m=0}^{\infty}\frac{w^{m}}{z^{m+1}}.\] Returning to equation (3.6), we have \[f(z)=\frac{1}{2\pi i}\int_{\gamma_{2}}f(w)\sum_{n=0}^{\infty}\frac{z^{n}}{w^{n +1}}dw-\frac{1}{2\pi i}\int_{\gamma_{1}}f(w)\sum_{m=0}^{\infty}\left(-\frac{w ^{m}}{z^{m+1}}\right)dw.\] Figure 3.41: Proof of Laurent’s theorem We can apply the Weierstrass M-test to obtain uniform convergence in order to interchange the summation and integration operators. This then leads to \[f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma_{2}}\frac{f(w)}{w^{n+1 }}dw\right)z^{n}+\sum_{m=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma_{1}}f(w) w^{m}dw\right)z^{-m-1}.\] If we then set \(m=-(n+1)\) in the second sum, and use the Deformation Theorem to replace \(\gamma_{1}\) and \(\gamma_{2}\) by \(\gamma\) in the statement of Laurent’s Theorem, we get \[f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma}\frac {f(w)}{w^{n+1}}dw\right)z^{n}+\sum_{n=-\infty}^{-1}\left(\frac{1}{2\pi i}\int _{\gamma}\frac{f(w)}{w^{n+1}}dw\right)z^{n}\] \[=\sum_{n=-\infty}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{w^{n+1}}dw\right)z^{n},\] which gives the desired result. **Remark 78**.: **Example 113**.: Let \(f(z)=\frac{1}{z(2-z)}\); suppose we want to find the Laurent expansion of \(f\) in 1. \(A=\{z:0<|z|<2\}\) and
2. \(A=\{z:|z|>2\}\). For part a), since \(\frac{|z|}{2}<1\), we have \[f(z)=\frac{1}{z(2-z)}=\frac{1}{2z}\cdot\frac{1}{\left(1-\frac{z}{2}\right)}= \frac{1}{2z}\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^{n}=\sum_{n=0}^{ \infty}\frac{z^{n-1}}{2^{n+1}}.\] For part b), since \(\frac{2}{|z|}<1\), we have \[f(z)=\frac{1}{z(2-z)}=\frac{1}{z^{2}\left(\frac{2}{z}-1\right)}=-\frac{1}{z^{ 2}}\sum_{n=0}^{\infty}\left(\frac{2}{z}\right)^{n}=-\sum_{n=0}^{\infty}\frac{2 ^{n}}{z^{n+2}}.\] ### Classification of Singularities Consider the Laurent expansion of \(f\) when \(R_{1}=0\). In this case \(f\) is holomorphic in the region \(\{z:0<|z-a| \[f(z)=\cdots+\frac{bn}{(z-a)^{n}}+\cdots+\frac{b_{1}}{(z-a)}+a_{0}+a_{1}(z-a)+a_ {2}(z-a)^{2}+\cdots\] is valid for \(0<|z-a| **Definition 112**.: **Example 114**.: **Theorem 146**.: _Suppose \(z=a\) is an isolated singularity of the function \(f(z)\). Then_ 1. \(a\) _is a removable singularity_ \(\iff\lim_{z\to a}(z-a)f(z)=0\)_._ **Example 115**.: ### Calculation of Residues **Notation.** Suppose \(f\) is holomorphic in the punctured disc \(D^{\prime}(a;r)=\{z\in\mathbb{C}:0<|z-a| \[\text{Res}(f(z),a)=b_{1}.\] **Theorem 147**.: : \[\frac{d^{k-1}}{dz^{k-1}}\left[b_{k}+b_{k-1}(z-a)+b_{k-2}(z-a)^{2}+ \cdots+b_{1}(z-a)^{k-1}+\sum_{n=0}^{\infty}a_{n}(z-a)^{n+k}\right]\] \[\qquad=0+0+0+\cdots+b_{1}(k-1)!+\sum_{n=0}^{\infty}(n+k)(n+k-1) \cdots(n+2)a_{n}(z-a)^{n+1}.\] Finally, as we take \(z\to a\), we see that \[\lim_{z\to a}\left[b_{1}(k-1)!+\sum_{n=0}^{\infty}(n+k)(n+k-1)\cdots(n+2)a_{n} (z-a)^{n+1}\right]=b_{1}(k-1)!.\] That is, the power series vanishes. If we finally divide by \((k-1)!\), we’ll obtain \(b_{1}\). Overall, the way we obtain \(b_{1}\) is by multiplying through by \((z-a)^{k}\), differentiating the result \((k-1)\) times and taking the limit as \(z\to a\) to obtain
Proof
1. Note that we cannot set \(c_{n}=\frac{f^{n}(a)}{n!}\) as we did in Taylor’s theorem. Indeed \(f^{(n)}(a)\) is not even defined since \(a\not\in A\).
2. (Uniqueness of the coefficients). Let \(f\in H(A)\) where \(A=\{z:R_{1}<|z-a|
* \(a\) is called a **removable singularity** if \(b_{n}=0\) for all \(n\geq 0\).
* \(a\) is called a **pole of order \(m\)** if \(b_{m}\neq 0\) but \(b_{n}=0\) for all \(n>m\).
* \(a\) is called an **essential singularity** if \(b_{n}\neq 0\) for infinitely many \(n\).
* \(\frac{\sin z}{z}\) has a removable singularity at \(z=0\) because \[\frac{\sin z}{z}=\frac{z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\cdots}{z}=1-\frac {z^{2}}{3!}+\frac{z^{4}}{5!}-\cdots\] for \(|z|>0\).
* \(\frac{1-\cos z}{z^{3}}\) has a simple pole (a pole of order one) at \(z=0\) (\(a=0\)) because \[\frac{1-\cos z}{z^{3}}=\frac{1-(1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-\cdots)}{ z^{3}}=\frac{1}{2!}\cdot\frac{1}{z}-\frac{z}{4!}+\frac{z^{3}}{6!}-\cdots\] for \(|z|>0\).
* \(e^{1/z^{2}}\) has an essential singularity at \(z=0\) because \[e^{1/z^{2}} =1+\frac{1}{z^{2}}+\frac{\left(\frac{1}{z^{2}}\right)^{2}}{2!}+ \frac{\left(\frac{1}{z^{2}}\right)^{3}}{3!}+\cdots+\frac{\left(\frac{1}{z^{2} }\right)^{n}}{n!}+\cdots\] \[=1+\frac{1}{z^{2}}+\frac{1}{2!}\cdot\frac{1}{z^{4}}+\frac{1}{3!} \cdot\frac{1}{z^{6}}+\cdots+\frac{1}{n!}\cdot\frac{1}{z^{2n}}+\cdots.\]Now we give a theorem of identification, for more details consult [54].
2. _i)_ \(a\) _is a pole of order_ \(m\)__\(\iff\lim_{z\to a}(z-a)^{m}f(z)=l\neq 0\)_._ _ii)_ \(a\) _is a pole of order 1 (or_ \(a\) _is a simple pole)_ \(\iff\lim_{z\to a}f(z)=\infty\)_._
3. \(a\) _is an essential singularity_ \(\iff\lim_{z\to a}f(z)\) _does not exist (i.e.,_ \(f(z)\) _tends to no limit, finite or infinite)._
1. \(\frac{\sin z}{z}\) has a removable singularity at \(z=0\) because \[\lim_{z\to 0}(z-0)\frac{\sin z}{z}=\sin(0)=0\ (\text{or}\ \lim_{z\to 0}\frac{\sin z}{z}=1).\]
2. \(\frac{1-\cos z}{z^{3}}\) has a pole of order 1 at \(z=0\) since \[\lim_{z\to 0}\frac{1-\cos z}{z^{3}}=\lim_{z\to 0}\frac{-\sin z}{3z^{2}}=\lim_{z \to 0}\frac{\cos z}{6z}\to\infty.\]
3. \(f(z)=\frac{5z+3}{(1-z)^{3}\sin^{2}z}\) where \(0<|z|<1\) has a pole of order 3 at \(z=1\) and a pole of order 2 at \(z=0\) because \[\lim_{z\to 1}(z-1)^{3}\frac{5z+3}{(1-z)^{3}\sin^{2}z}=\frac{-8}{\sin^{2}(1)}\neq 0\] and \[\lim_{z\to 0}(z-0)^{2}f(z)=\lim_{z\to 0}z^{2}\cdot\frac{5z+3}{(1-z)^{3}\sin^{2}z}=3.\]
4. \(e^{1/z^{2}}\) has an essential singularity at \(z=0\) because \(\lim_{z\to 0}e^{1/z^{2}}\) does not exist. If \(z=x\), then \[\lim_{z\to 0}e^{1/z^{2}}=\lim_{x\to 0}e^{1/x^{2}}\to\infty\] while if \(z=iy\), then \[\lim_{z\to 0}e^{1/z^{2}}=\lim_{y\to 0}e^{1/(iy)^{2}}=\lim_{y\to 0}\frac{1}{e^{1/y^{2}}}=0.\]
1. _If_ \(f(z)\) _has a removable singularity at_ \(a\)_, then_ \[\text{Res}(f(z),a)=0.\]
2. _If_ \(a\) _is a simple pole of_ \(f(z)\)_, then_ \[\text{Res}(f(z),a)=\lim_{z\to a}(z-a)f(z)\]
3. _If_ \(a\) _is a pole of order_ \(k\) _(_\(k\geq 2\)_) for_ \(f(z)\)_, then_ \[\text{Res}(f(z),a)=\frac{1}{(k-1)!}\lim_{z\to a}\left[\frac{d^{k-1}}{dz^{k-1}} [(z-a)^{k}f(z)]\right].\]Proof
1. If \(a\) is a removable singularity of \(f(z)\), then \(f(z)=\sum_{n=0}^{\infty}a_{n}(z-a)^{n}\). Thus there is no coefficient \(b_{1}\), so \(\text{Res}(f(z),a)=0\).
2. If \(a\) is a simple pole, then \(f(z)=\frac{b_{1}}{z-a}+\sum_{n=0}^{\infty}a_{n}(z-a)^{n}\). Therefore \[(z-a)f(z)=b_{1}+\sum_{n=0}^{\infty}a_{n}(z-a)^{n+1}\] and \[\lim_{z\to a}(z-a)f(z)=b_{1}.\]
3. If \(a\) is a pole of order \(k\), for \(f(z)\), then \[f(z)=\frac{b_{k}}{(z-a)^{k}}+\frac{b_{k-1}}{(z-a)^{k-1}}+\cdots+\frac{b_{1}}{ (z-a)}+\sum_{n=0}^{\infty}a_{n}(z-a)^{n}\quad\text{ with }b_{k}\neq 0.\] The goal here is to obtain the coefficient \(b_{1}\). To show this formula does in fact yield \(b_{1}\), observe that \[(z-a)^{k}f(z)=b_{k}+b_{k-1}(z-a)+b_{k-2}(z-a)^{2}+\cdots+b_{1}(z-a)^{k-1}+\sum _{n=0}^{\infty}a_{n}(z-a)^{n+k}.\]Now if we differentiate this \(k-1\) times, only the terms with an order \(k-1\) or higher survive. That is, we get
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