Calculating Residues of Complex Functions

If you hear a voice within you say ‘you cannot paint’, then by all means paint and that voice will be silenced, Vincent van Gogh

Recall

Laurent’s theorem. Let A = {$z \in \mathbb{C}: R \lt |z -a| \lt S$} be an annular region ($0 \le R \lt S \le \infty$) and let f be analytic on A. Then, $f(z) = \sum_{n=-\infty}^\infty c_n(z -a)^n, \forall z \in A$ where $c_n = \frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$ and $\gamma$ is a positively oriented circle of radius r centered at a and R < r < S ($\gamma$ is inside the annular region).

Corollary (aka Rosetta Stone of Complex Analysis). Let f be analytic in $\mathbb{B'}(a; r)$. (i) f has a removable singularity at a if and only if in the Laurent's series expansion of f around a ($\sum_{n = -\infty}^\infty (z-a)^n$) all negative coefficients are zero $c_n = 0, \forall n \lt 0$. (ii) f has a pole of order m ($m \ge 1$) at a if and only if the series stop at the negative power -m, $c_{-m} \ne 0, c_n = 0, \forall n \lt -m$; (iii) f has an isolated essential singularity at a if and only if the series has infinitely many negative terms (it never stops), i.e., there is no m such that $c_n = 0, \forall n \lt -m$

The Argument Principle. Suppose f is meromorphic on and inside a simple closed curve $\gamma$ with a finite number of zeroes $a_j, 1 \le j \le l_1$ and poles $b_k, 1 \le k \le l_2$. Suppose none of those zeros and poles lie on $\gamma$. Then, $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz = M - N$ where M is the sum of order of zeroes at $a_j, 1 \le j \le l_1$ and N is the sum of order of poles at $b_k, 1 \le k \le l_2$

• Find the residue of $f(z) = \frac{z^2-2z}{(z+1)^2(z^2+4)}$

First, we look at the denominator of $f(z) = \frac{z^2-2z}{(z+1)^2(z^2+4)}$:

1. $(z+1)^2 = 0$: This gives $z = -1$. Since it is squared, this is a pole of order 2.
2. $(z^2+4) = 0$: This factors into $(z-2i)(z+2i) = 0$. This gives two simple poles at $z = 2i$ and $z = -2i$.

Residue at z = -1. For a pole of order 2, we must multiply by $(z+1)^2$, take the derivative, and then plug in the limit: $\text{Res}(f, -1) = \lim_{z \to -1} \frac{d}{dz} \left[ (z+1)^2 f(z) \right]$

$(z+1)^2 f(z) = (z+1)^2 \frac{z^2-2z}{(z+1)^2(z^2+4)} = \frac{z^2-2z}{z^2+4}$

$\frac{d}{dz}\frac{z^2-2z}{z^2+4} =[\text{Using the Quotient Rule: } \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}] \frac{(2z-2)(z^2+4) - (z^2-2z)(2z)}{(z^2+4)^2}$

$\frac{d}{dz}|_{z=-1} = \frac{(-4)(5) - (3)(-2)}{(5)^2} = \frac{-20 - (-6)}{25} = \frac{-14}{25}$.

Residue at $z = 2i$. For a simple pole, remove the singular term $(z-2i)$ and plug in the value $\text{Res}(f, 2i) = \lim_{z \to 2i} \left[ (z-2i) f(z) \right] = \lim_{z \to 2i} \frac{z^2-2z}{(z+1)^2(z+2i)} = \frac{(2i)^2 - 2(2i)}{(2i+1)^2(2i+2i)} = \frac{-4-4i}{-16-12i} = \frac{1+i}{4+3i} = \frac{(1+i)(4-3i)}{(4+3i)(4-3i)} = \frac{4 - 3i + 4i - 3i^2}{16 + 9} = \frac{4 + i + 3}{25} = \frac{7+i}{25}$

Residue at $z = -2i$, $\text{Res}(f, -2i) = \lim_{z \to -2i} \left[ (z+2i) f(z) \right] = \lim_{z \to -2i}\frac{z^2-2z}{(z+1)^2(z-2i)} = \frac{(-2i)^2 - 2(-2i)}{(-2i+1)^2(-2i-2i)} = \frac{1-i}{4-3i} = \frac{(1-i)(4+3i)}{(4-3i)(4+3i)} = \frac{4+3i-4i-3i^2}{25} = \frac{7-i}{25}$.

• Find the residue of $f(z) = \frac{z-\sinh(z)}{z^2\sinh(z)}$ at $z = \pi i$.

Solution.

Let’s check the numerator and denominator separately:

• Numerator: $P(z) = z - \sinh(z)$. At $z = \pi i$, $\sinh(\pi i) = i\sin(\pi) = 0$. So, $P(\pi i) = \pi i - 0 = \mathbf{\pi i}$ (Non-zero).
• Denominator: $Q(z) = z^2 \sinh(z)$. At $z = \pi i$, since $\sinh(\pi i) = 0$, the whole denominator is $\mathbf{0}$.

Since we have a “Non-zero / Zero” situation, this is a Pole. Because the derivative of the denominator is not zero (as we will see), it is a Simple Pole.

The Derivative Rule: If $f(z) = \frac{P(z)}{Q(z)}$ has a simple pole at $a$ (where $P(a) \neq 0, Q(a) = 0, Q'(a) \neq 0$), then: $\text{Res}(f, a) = \frac{P(a)}{Q'(a)}$

$Q(z) = z^2 \sinh(z) \implies[\text{Product Rule}] Q'(z) = 2z\sinh(z) + z^2\cosh(z)$

$$ \begin{aligned} Q'(z)|_{z=\pi i} &=2(\pi i) \sinh(\pi i) + (\pi i)^2\cosh(\pi i) \\[2pt] &=[\text{Using the identities for hyperbolic functions with imaginary arguments: } \sinh(\pi i) = i\sin(\pi) = 0, \cosh(\pi i) = cos(\pi)= 1] \\[2pt] &=2(\pi i)0 + (-\pi^2)(-1)\\[2pt] &= 0 + \pi^2 = \pi^2 \end{aligned} $$

$\text{Res}(f, \pi i) = \frac{P(\pi i)}{Q'(\pi i)} = \frac{\pi i}{\pi^2} = \frac{i}{\pi}$

Let’s do it manually by peeling apart the function.

$f(z) = \frac{1}{\sinh(z)}\frac{z-\sinh(z)}{z^2}$ has a simple pole at $z = \pi i$ and $\frac{z-\sinh(z)}{z^2}$ is analytic at this point.

Let’s expand $\sinh(z)$ into a Taylor series at $z=\pi i$ to show exactly how it goes to zero.

$\sinh(z) = 0 + \frac{(\sinh(z))'|_{z=\pi i}}{1!}(z-\pi i) + \frac{(\sinh(z))''|_{z=\pi i}}{2!}(z-\pi i)^2 + \frac{(\sinh(z))'''|_{z=\pi i}}{3!}(z-\pi i)^3+ \cdots$

f(πi) = sinh(πi) = 0, f′(πi) = cosh(πi) = −1, f′′(πi) = sinh(πi) = 0, f′′′(πi) = cosh(πi) = −1, and the pattern repeats.

$\sinh(z) = \frac{(-1)}{1!}(z-\pi i) + \frac{0}{2!}(z-\pi i)^2 + \frac{(-1)}{3!}(z -\pi i)^3 + \cdots = (z-\pi i)(-1 + \frac{(-1)}{3!} (z -\pi i)^3 + \cdots )$

$\sinh(z) \approx -1(z - \pi i) + \dots$. This confirms that the zero is “simple” (power of 1).

$\sinh(z) = (z-\pi i)(-1 + \phi(z))$ where $(z-\pi i)$ is the part that causes the zero and $(-1 + \phi(z))$ is the “leftover” part, which is roughly equal to $-1$ when $z$ is close enough to $\pi i$.

$f(z) = \frac{z-\sinh(z)}{(z-\pi i)(-1 + \phi(z))z^2} = \frac{1}{z-\pi i}(\frac{z-\sinh(z)}{(-1 + \phi(z))z^2})$ in $\mathbb{B}(\pi i; r), r \gt 0$.

Recall. The definition of residue for a simple pole is $\lim_{z \to a} (z-a)f(z)$.

$Res(f(z); z = \pi i) =[\pi i \text{ is a simple pole}] \lim_{z \to \pi}(z-\pi i)f(z) = (\frac{z-\sinh(z)}{(-1 + \phi(z))z^2})|_{z=\pi i} = \frac{\pi i - 0}{(-1+0)(\pi i)^2} =\frac{\pi i}{\pi^2} = \frac{i}{\pi}$

Suppose that g is an analytic function in $\mathbb{B}(z_0; r)$ and that $g(z_0) = 0, g'(z_0) \ne 0, \text{ and } g(z) \ne 0, \forall z \in \mathbb{B'}(z_0; r)$ Define $f(z) := \frac{1}{(g(z))^2}, \forall z \in \mathbb{B'}(z_0; r)$, show that f has a pole of order 2 at $z_0$ and calculate $Res(f; z_0)$.

Solution.

Because g is analytic at $z_0, g(z_0) = 0, g'(z_0) \ne 0$, the Taylor series expansion of g at $z_0: g(z)= g(z_0) +g'(z_0)(z-z_0) + \frac{g''(z_0)}{2!}(z-z_0)^2+\cdots$.

Since $g(z_0)=0$, the constant term disappears: $g(z)=g'(z_0)(z-z_0)+\frac{g''(z_0)}{2!}(z-z_0)^2+\cdots$

Now factor out $(z-z_0): g(z)=(z-z_0)\left( g'(z_0)+\frac{g''(z_0)}{2!}(z-z_0)+\cdots \right)$. This is valid because $g'(z_0)\neq 0$, so the factor in parentheses is nonzero at $z_0$.

$g(z) = (z-z_0)(g'(z) + \frac{g''(z_0)}{2!}(z-z_0)^2 + \cdots) = g(z)\rho(z), \forall z \in \mathbb{B}(z_0; r)$ where $\rho(z)$ is analytic in the neighborhood of $z_0$ and $\rho(z) \ne 0$. Besides, we know that $\forall z \in \mathbb{B}(z_0; \varepsilon)$ for some $\varepsilon \gt 0$ (because a continuous function that is nonzero at a point stays nonzero in a small neighborhood).

$g(z)=(z-z_0)\rho (z)$, where $\rho(z)$ is analytic and $\rho(z) \neq 0$. This is exactly the standard factorization for a simple zero of an analytic function. The function $\rho (z):=g'(z_0)+\frac{g''(z_0)}{2!}(z-z_0)+\cdots$ is analytic because it is a Taylor series with positive radius of convergence and $\rho(z_0)=g'(z_0) \neq 0$.

Substitute our factored form of g(z): $f(z) = \frac{1}{(z-z_0)^2\rho(z)^2}$ in $\mathbb{B}(z_0; r)$. This clearly shows that $f(z)$ has a Pole of Order 2 at $z_0$ because $(z-z_0)^2$ is in the denominator and $\rho(z)$ is analytic and never zero near $z_0$ (and obviously $\frac{1}{\rho (z)^2}$ is also analytic and never zero near $z_0$). In other words, f(z) is exactly of the form: $\frac{\text{analytic, nonvanishing}}{(z-z_0)^2}$

A function has a pole of order n at $z_0$ if it can be written as: $f(z)=\frac{h(z)}{(z-z_0)^n}$, where h(z) is analytic and nonzero at $z_0$.

To verify the residue, it is cleaner to rewrite $f(z)$ by separating the singular part from the analytic part: $f(z) = \frac{1}{(z-z_0)^2} \cdot \underbrace{\left[ \frac{1}{\rho(z)^2} \right]}_{\text{Let's call this } \phi(z)}$. Here, $\phi(z) = \frac{1}{\rho(z)^2}$ is analytic at $z_0$ because $\rho(z_0) \ne 0$.

For a function with a pole of order 2 looking like $\frac{\phi(z)}{(z-z_0)^2}$, the Residue is simply the first derivative of the analytic part $\phi(z)$ evaluated at the pole $\text{Res}(f; z_0) = \phi'(z_0)$.

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To find the residue, we need the coefficient of the $\frac{1}{z-z_0}$ term. Let’s just expand $\phi(z)$ using its standard Taylor Series: $\phi(z) = \phi(z_0) + \phi'(z_0)(z-z_0) + \frac{\phi''(z_0)}{2}(z-z_0)^2 + \dots$, so $f(z) = \frac{\phi(z_0) + \phi'(z_0)(z-z_0) + \dots}{(z-z_0)^2}$

Divide each term by $(z-z_0)^2$: $f(z) = \frac{\phi(z_0)}{(z-z_0)^2} + \mathbf{\frac{\phi'(z_0)}{z-z_0}} + \frac{\phi''(z_0)}{2} + \dots$. The coefficient of $(z-z_0)^{-1}$ is exactly $\phi'(z_0)$.

This is actually just a specific case of the general Residue formula for a pole of order m (m = 2, pole of order 2, $(z-z_0)^2 f(z)$ is just $\phi(z)$): $\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right] = \frac{d}{dz} [\phi(z)] \Big|_{z=z_0} = \phi'(z_0)$

Example, find the residue of $f(z) = \frac{e^z}{z^2}$ at z = 0. Here, $\phi(z) = e^z$. Take the derivative: $\phi'(z) = e^z$. Evaluate at 0: $\phi'(0) = 1$. Residue = 1.

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We need to differentiate $\phi(z) = (\rho(z))^{-2}$ using the Chain Rule:

$\phi'(z)|_{z=z_0} = (\frac{1}{\rho(z)^2})|_{z=z_0} = -2\rho(z)^{-3}(z)\rho'(z)|_{z=z_0} = -2(g'(z_0))^{-3}\frac{g''(z_0)}{2} = \frac{-g''(z_0)}{(g'(z_0))^{3}}$

$g(z) = (z-z_0) \left[ g'(z_0) + \frac{g''(z_0)}{2}(z-z_0) + \dots \right]$, $\rho(z) = g'(z_0) + \frac{g''(z_0)}{2}(z-z_0) + \dots, \rho(z_0) = g'(z_0)$ (the constant term) and $\rho'(z_0) = \frac{g''(z_0)}{2}$ (the coefficient of the linear term).

$Res(f; z_0) = \frac{-g''(z_0)}{(g'(z_0))^{-3}}$