5.2 Residue

1. A singular point is a point where the function $f(z)$ is not regular. It is the point where the denominator is 0 in a rational function.

Basic formula

$$\int_{\vert z\vert = r}\frac{1}{(z - z)^{n}} dz = \left\{\begin{array}{ll} 2\pi i, & n = 1\\ 0, & n \neq 1 \end{array}\right.$$

As you can see from this formula, the integral of $\frac{1}{z-a}$ does not become 0 when Laurent expanded, but everything else becomes 0. From this, the coefficient of $\frac{1}{z-a}$ is called a residue, which means that it does not become 0 when integrated, and is expressed as $Res[a]$.

Residue formula For $a$ is the singularity of $f(z)$ with pole of order $m$,

$$Res[a] = \frac{1}{(m-1)!}\lim_{z \to a}\frac{d^{m-1}}{dz^{m-1}}(z -a)^{m}f(z)$$

(a) The points where the denominator is 0 are $z =0, 1$. Then we find the residue of $z =0, 1$. Expand $\frac{1}{z(z-1)^{2}}$ using partial fraction expansion. Then

$$\frac{1}{z(z-1)^{2}} = \frac{A}{z} + \frac{B}{(z-1)} + \frac{C}{(z-1)^{2}}$$

Clear the denominators.

$$1 = A(z-1)^2 + Bz(z-1) + Cz$$

Let $z=0$. Then

$$1 = A$$

Let $z=1$. Then

$$1 = C$$

Match the coefficients of $z^2$. Then

$$0 = A + B \Rightarrow B = -A =-1$$

Thus

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

Therefore,

$$Res[0] =$$   coefficient of $$\frac{1}{z} = 1$$

$$Res[1] =$$   coefficient of $$\frac{1}{z-1} = -1$$

(b) The points where the denominator is 0 are $z =-\frac{1}{2}, 2$. Then we find the residue of $z =-\frac{1}{2}, 2$. Expand $\frac{z}{(2z+1)(z-2)}$ using partial fraction expansion.

$$\frac{z}{(2z+1)(z-2)} = \frac{A}{2z+1} + \frac{B}{z-2}$$

Clear the denominators. Then

$$z = A(z-2) + B(2z+1)$$

Let $z = 2$. Then we have

$$2 = 5B \Rightarrow B = \frac{2}{5}$$

Let $z=-\frac{1}{2}$. Then we have

$$-\frac{1}{2} = A(-\frac{5}{2} \Rightarrow A = \frac{1}{5}$$

Thus.

$$\frac{z}{(2z+1)(z-2)} = \frac{1/5}{2z+1} + \frac{2/5}{z-2}$$

But the residue is the coefficient of $\frac{1}{z-a}$. So, we rewrite

$$\frac{z}{(2z+1)(z-2)} = \frac{1/5}{2(z+\frac{1}{2}} + \frac{2/5}{z-2}$$

Thus,

$$Res[2] =$$   coefficient of $$\frac{1}{z-2} = \frac{2}{5}$$

$$Res[-\frac{1}{2}] =$$   coefficient of $$\frac{1}{z+\frac{1}{2}} = \frac{1}{10}$$

(c) The points where the denominator is 0 are $z = n\pi (n=0,\pm1,\pm2,\ldots)$. Then find the residue of $n\pi$. We can not use the partial fraction expansion on $\frac{1}{\sin{z}}$. So, we use Taylor expansion on $\sin{z}$ at $z=n\pi$. Then divide. Let $t = z - n\pi$. Then

$$\sin{z} = \sin(t + n\pi) = \left\{\begin{array}{ll} \sin{t}, & n \mbox{蛛ｶ謨ｰ}\\ -\sin{t}, & n \mbox{螂\x87謨ｰ} \end{array}\right.$$

First for $n$ is even.

$$\frac{1}{\sin{z}} = \frac{1}{\sin{t}}$$

$$= \frac{1}{t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots} = \frac{1}{t} + \frac{t}{3!} + \cdots$$

$$= \frac{1}{z - n\pi} + \frac{z- n\pi}{3!} + \cdots$$

Thus,

$$Res[n\pi] =$$   coefficient of $$\frac{1}{z-n\pi} = 1$$

Next for $n$ is odd.

$$\frac{1}{\sin{z}} = -\frac{1}{\sin{t}}$$

$$= -\frac{1}{t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots} = \frac{1}{t} + \frac{t}{3!} + \cdots$$

$$= -\frac{1}{z - n\pi} - \frac{z- n\pi}{3!} - \cdots$$

Thus

$$Res[n\pi] =$$   coefficient of $$\frac{1}{z-n\pi} = -1$$

(d) The points where the denominator is 0 are $z =1, -2$. Then find the residue of $z =1, -2$. Expand $\frac{e^z}{(z-1)(z+2)^2}$ using the partial fraction expansion. Then

$$\frac{e^z}{(z-1)(z+2)^2} = \frac{A}{z-1} + \frac{B}{z+2} + \frac{C}{(z+2)^2}$$

Clear the denominators.

$$e^z = A(z+2)^2 + B(z-1)(z+2) + C(z-1)$$

Let $z=1$. Then

$$e = 9A \Rightarrow A = \frac{e}{9}$$

Let $z=-2$. Then

$$e^{-2} = -3C \Rightarrow C = \frac{-e^{-2}}{3}$$

Thus,

$$Res[1] =$$   coefficient of $$\frac{1}{z-1} = \frac{e}{9}$$

Lastly, we want to find the value of $B$. But we can not use the matching of coefficients. Because, $e^{z}$ is not polynomial. Then we use Heaviside's expansion theorem or residue formula. Here, we use residue formula. $-2$ is the pole of order 2, by residue formula, we have

$$Res[-2] = \frac{1}{(2-1)!}\lim_{z \to -2}\frac{d}{dz}(z +2)^{2} \frac{e^z}{(z-1)(z+2)^2} = -\frac{4e^{-2}}{9}$$

2. Residue formula

The function $f(z)$ is analytic on and inside the single closed curve $C$ and monovalent, except for the finite number of points $a_{1},a_{2},\ldots,a_{n}$ inside it. Then

$$\int_{C}f(z) dz = 2\pi i\{Res[a_{1}] + Res[a_{2}] + \cdots + Res[a_{n}]\}$$

holds.

(a) By the residue theorem,

$$\int_{\vert z\vert=2}\frac{dz}{z(z-1)^2} = 2\pi i\{Res[0] + Res[1]\} ( \vert z\vert = 2)$$

$$= 2\pi i (1 - 1) = 0 \ref{enshu:14-1-1a},$$

(b) By the residue theorem,

$$\int_{\vert z\vert=1}\frac{z}{(2z+1)(z-2)} dz = 2\pi i Res[-\frac{1}{2}] ( \vert z\vert = 1)$$

$$= 2\pi i \frac{1}{10} = \frac{\pi i}{5} by the result of \ref{enshu:14-1-1b}.$$

(c) By the residue theorem,

$$\int_{\vert z\vert=1}\frac{dz}{\sin{z}} = 2\pi i Res[0] ( \vert z\vert = 1)$$

$$= 2\pi i \ref{enshu:14-1-1c},$$

(d) By the residue theorem,

$$\int_{\vert z\vert=3}\frac{e^{z}}{(z-1)(z+2)^2} dz = 2\pi i (Res[1] + Res[-2]) ( \vert z\vert = 3)$$

$$= 2\pi i (\frac{e}{9} - \frac{e^{-2}}{9}) \ref{enshu:14-1-1d}$$

$$= \frac{2\pi i}{9}(e - e^{-2})$$

3.

(a) The circle $\vert z\vert = 1$ contains the singularity $z=0$. So find $Res[0]$. Then

$$Res[0] = \lim_{z \to 0}\frac{d}{dz}z^{2}(\frac{e^{2z}}{z^2(z^2 + 2z + 2)}$$

$$= \lim_{z \to 0}\frac{d}{dz}(\frac{e^{2z}}{z^2 + 2z +2})$$

$$= \lim_{z \to 0}(\frac{2e^{2z}(z^2 + 2z +2) - e^{2z}(2z+2)}{(z^2 + 2z +2)^2}$$

$$= \frac{4-2}{4} = \frac{1}{2}$$

By the residue theorem,

$$\int_{\vert z\vert=1}\frac{e^{2z}}{z^{2}(z^2 + 2z + 2)} dz = 2\pi i Res[0] = \pi i$$

(b) The circle $\vert z - i\vert = 2$ contains the singularity $z=0, -1+i$. Here $Res[0]$ is found by A. Thus, we find $Res[-1+i]$.

$$Res[-1+i] = \lim_{z \to -1+i}(z - (-1+i))(\frac{e^{2z}}{z^2(z^2 + 2z + 2)}$$

$$= \lim_{z \to -1+i}(\frac{e^{2z}}{z-(-1-i)})$$

$$= \frac{e^{2(-1+i)}}{(-1+i)^2 (-1 + i+1+i)}$$

$$= \frac{e^{2(-1+i)}}{4}$$

Thus by the residue theorem,

$$\int_{\vert z-i\vert=2}\frac{e^{2z}}{z^{2}(z^2 + 2z + 2)} dz = 2\pi i (Res[0] + Res[-1+i])$$

$$= 2\pi i(\frac{1}{2} + \frac{e^{2(-1+i)}}{4})$$

(c) The circle $\vert z\vert = 3$ contains all singularity $z=0, -1+i, -1 -i$. Note that $Res[0], Res[-1+i]$ are already found by A. So, we find $Res[-1-i]$.

$$Res[-1-i] = \lim_{z \to -1-i}(z - (-1-i))(\frac{e^{2z}}{z^2(z^2 + 2z + 2)}$$

$$= \lim_{z \to -1-i}(\frac{e^{2z}}{z-(-1+i)})$$

$$= \frac{e^{2(-1-i)}}{(-1-i)^2 (-1 - i+1-i)}$$

$$= \frac{e^{-2(1+i)}}{4}$$

Then by the residue theorem,

$$\int_{\vert z\vert=3}\frac{e^{2z}}{z^{2}(z^2 + 2z + 2)} dz = 2\pi i (Res[0] + Res[-1+i] + Res[-1-i])$$

$$= 2\pi i(\frac{1}{2} + \frac{e^{2(-1+i)}}{4} + \frac{e^{-2(1+i)}}{4})$$

$$= 2\pi i(\frac{1}{2} + \frac{e^{-2}e^{-2i} + e^{-2}e^{2i}}{4})$$

$$= 2\pi i(\frac{1}{2} + \frac{e^{-2}}{2}\frac{e^{2i}+e^{-2i}}{2})$$

$$= \pi i (1 + \frac{e^{-2}}\cos{2})$$