Sigularities and Residue Theorem

Singularity

\(z_0\) is an isolated singularity of \(f\) if \(f(z)\) is not differentiable or is undefined at \(z_0\), but \(f(z)\) is analytic on \(A_{0, r_0}(z_0)\) for some \(r_0 > 0\)

Classes of Singularities

Suppose \(f\) has a isolated singularity at \(z_0\), and its Larent expansion is \(f(z) = \sum_{-\infty}^{\infty} c_n (z-z_0)^n\) on \(A_{0, r_0}(z_0)\) for some \(r_0 > 0\), then - \(z_0\) is a removable singularity if \(c_n = 0\) for all \(n < 0\). - \(z_0\) is a pole of order \(N\) if \(c_{-N}\neq 0\) abd \(c_n = 0\) for all \(n < -N\). - \(z_0\) is a essential singularity if \(c_n\neq 0\) for all \(n < 0\).

Another definition - \(z_0\) is a removable singularity if \(\exists s, 0 < s < r\) s.t. \(f(z)\) is bounded on \(A_{0, s}(z_0)\), i.e. \(\lim_{z\rightarrow z_0} f(z)\) exist. - \(z_0\) is a pole of order \(N\) if \(\exists s, 0 < s < r\) s.t. \((z-z_0)^N f(z)\) is bounded on \(A_{0, s}(z_0)\), i.e. \(\lim_{z\rightarrow z_0} (z-z_0)^N f(z)\) exist but \(\lim_{z\rightarrow z_0} (z-z_0)^{N-1} f(z)\) DNE. - \(z_0\) is a essential singularity if \(\forall N, \forall s, 0 < s < r\) \((z-z_0)^N f(z)\) is unbounded, or \(\lim_{z\rightarrow z_0} (z-z_0)^N f(z)\) DNE.

The two definitions are equivalent

Claim 1

Let \(f(z)\) have a removable singularity at \(z_0\), and analytic on \(A_{0, r}(z_0)\) for some \(r > 0\). Then, \(\exists g(z)\) analytic on \(B_r(z_0)\) and \(f=g\) on \(A_{0,r}(z_0)\).

proof. Note that for removable sinularity, we have the Laurent series \(f(z) = \sum_{n=0}^\infty c_n(z-z_0)^n\) since for \(\forall m \leq -1, c_m = 0\) on the anulus.
Then, simply write as the power series on the ball.

Example 1

consider \(\frac{\sin z}{z}\) on \(A_{0,r}(0)\). its Laurent series is given by

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

Therefore, take \(z=0\), we have the series equals to \(1\). we can write \(g(z) = \begin{cases}\sin z / z &z\neq 0\\ 1&z=0\end{cases}\)

Residue

Note that using Cauchy Theorem we can do contour deformation to solve singularity in \(C_{int}\), and each integral around singularity can be related to CIF. Therefore, we can unify them together.

For \(f\) analytic, for some singularity \(z_0\) on \(f\) s.t. \(f\) is analytic on \(A_{o,r}(z_0)\) for \(r>0\). The residue is defined as

\[Res(f, z_0) = c_{-1} = \frac{1}{2\pi i} \oint_{C}\frac{f(w)}{(w - z_0)^{-1+1} }dw = \frac{1}{2\pi i} \oint_{C_s}f(w)dw\]

for some circle \(C_s\) of radius \(0 < s < r\) centered at \(z_0\). Note that \(c_{-1}\) is the coefficient of Larent series expansion on \(A_{0,r}(z_0)\).

Theorem 1. Residue Theorem

Let \(D\) e a domain, Let \(C\subset D\) be a Jordon contour, let the set of singularities \(Z = \{z_1,..., z_N\} \subset C_{int}, f(z)\) analytic on \(D - Z\). Then,

\[\oint_C f(z)dz = 2\pi i \sum_{k=1}^N Res(f, z_k)\]

proof. First, by contour deformation, we only need to show that

\[\int_{C_k} f(z)dz = 2\pi i Res(f, z_k)\]

where \(C_k \subset A_{0, r_k}(z_k)\) s.t. \(f\) is analytic on each anulus. From the definition of residue above,

\[2\pi i Res(f, z_k) = 2\pi i \frac{1}{2\pi i} \oint_{C_k} f(w)dw = \oint_{C_k} f(z)dz\]

Claim 2

\(f(z)\) has a pole of order 1 at \(z_0\) IFF \(Res(f, z_0) = \lim_{z\rightarrow z_0} (z-z_0)f(z)\)

proof. \(f\) has pole of order 1 IFF the Larent series starts from \(-1\), so that

\[(z-z_0)f(z) = \sum_{n=0}^{\infty} c_{n-1} (z-z_0)^n\]

Therefore, consider the limit \(\lim_{z\rightarrow z_0} (z-z_0)f(z)\), if the limit exists, all terms with \((z-z_0)\) must approach \(0\) as \(z\rightarrow z_0\), the only term left can be the 0th term, i.e. \(c_{-1}\)

Claim 3

For \(z_0\) is a pole of order \(k\) singularity, or a removable singularity, or an analytic point, (i.e. \(z_0\) has at most order \(k\))take any \(M\geq k\), we have

\[Res(f, z_0) = \frac{1}{(m-1)!} \lim_{z\rightarrow z_0} \frac{d^{M-1}}{dz^{M-1}} (z-z_0)^M f(z)\]

proof.

\[\begin{align*} (z-z_0)^M f(z) &= \sum_{n=0}^\infty c_{n-M} (z-z_0)^n\\ \frac{d^{M-1}}{dz^{m-1}}(z-z_0)^M f(z) &= \sum_{n=M-1}^\infty (\prod_{k=0}^{M-2} (n-k))c_{n-M} (z-z_0)^{n-M-1} \end{align*}\]

Therefore, the coefficient at \((z-z_0)^0\) is when \(n = M-1\)

\[(\prod_{k=0}^{M-2} (M-1 - k))c_{-1} = (M-1)! c_{-1}\]

Residue At Infinity

If \(f(z)\) analytic on \(A_{s,\infty}(0)\) for some \(s>0\), define the residue at inifinity as

\[Res(f,\infty) = \lim_{R\rightarrow\infty} \frac{1}{2\pi i} \oint_{C_R} f(z)dz\]

Note that this definition makes sense because on \(A_{s,\infty}(0)\), consider Laurent series expansion \(f(z) = \sum_{-\infty}^\infty c_nz^n\) we have

\[\begin{align*} \lim_{R\rightarrow\infty} \frac{1}{2\pi i} \oint_{C_R} f(z)dz &= \frac{1}{2\pi i}\lim_{R\rightarrow\infty} \oint_{C_R} \sum_{n=-\infty}^\infty c_n z^ndz\\ &\rightarrow^{u.c} \frac{1}{2\pi i}\lim_{R\rightarrow\infty} \sum_{n=-\infty}^\infty \oint_{C_R} c_n z^ndz\\ &= \frac{1}{2\pi i} \oint_{C_R} c_{-1}z^{-1}dz&\text{CT}\\ &= c_{-1} \end{align*}\]

Claim 1

If \(f\) has finitely many singularities \(z_1,...,z_n \in \mathbb C\), then

\[Res(f,\infty) = \sum_{k=1}^n Res(f, z_k)\]

proof.

\[\begin{align*} Res(f,\infty) &= \lim_{R\rightarrow\infty} \frac{1}{2\pi i} \oint_{C_R} f(z)dz\\ &= \sum_{k=1}^n Res(f, z_k) &\text{Residue Thrm} \end{align*}\]

Claim 2

\(Res(f,\infty) = -Res(\frac{1}{z^2} f(\frac{1}{z}), 0)\)

proof. By our definition

\[\begin{align*} Res(f,\infty) &= \lim_{R\rightarrow\infty} \frac{1}{2\pi i} \oint_{C_R} f(z)dz\\ &= \lim_{R\rightarrow\infty} \frac{1}{2\pi i} \oint_{C_{R^{-1}}} f(u^{-1})(-u^{-2})du &z=u^{-1}, dz=-u^{-2}du\\ &= \lim_{\epsilon\rightarrow0} -\frac{1}{2\pi i} \oint_{C_{\epsilon}} \frac{1}{u^2}f(\frac{1}{u})du\\ &= -Res(\frac{1}{z^2} f(\frac{1}{z}), 0) \end{align*}\]

Claim 3

For \(p, q\) polynomials, if some \(C\) large enough circle to contain all the roots of polynomial \(q(z)\) and \(deg(q)\geq deg(p) + 2\), then

\[\oint_C \frac{p(z)}{q(z)} dz = 0\]

proof. Consider \(\lim_{z\rightarrow\infty} z\frac{p(z)}{q(z)}, deg(zp(z)) = deg(p(z)) + 2\leq deg(q)\) so that the limit exists and \(\lim_{z\rightarrow\infty} z\frac{p(z)}{q(z)} = 0\). Therefore,

\[Res(p/q,\infty) = \frac{1}{2\pi i}\oint_C \frac{p(z)}{q(z)} dz = 0\]

Winding Number

For closed curve \(C\), define for the winding number of \(C\) at \(z_0\) (not on \(C\)) as

\[w(C, z_0) = \frac{1}{2\pi i}\int_C \frac{1}{z-z_0} dz\]

Graphically, winding number is the number of circles of \(C\) oriented c.c.w. around \(z_0\).

Theorem For curves \(C\) that does not go across \(0\), any parameterization \(c(t)\) of \(C\) can be decomposed into

\[c(t) = r(t)e^{i\theta(t)}\]

Theorem \(w(C, 0) = \frac{\theta(b) - \theta(a)}{2\pi i}\)

Generalized Residue Theorem

Let \(D\) be a simply connected domain, \(\mathbf z = \{z_1,...,z_n\} \in D\) be singularities. \(C \subset D - \mathbf z\) be a closed curve and \(f\) is analytic on \(D-\mathbf z\). Then,

\[\int_C f(z)dz = 2\pi i \sum_{i=1}^n w(C, z_k) Res(f, z_k)\]