Let z 0 z_0 z 0 f f f f f f z 0 z_0 z 0
Classification by Laurent series:
Removable singularity : a n = 0 a_n = 0 a n = 0 n < 0 n \lt 0 n < 0 f ( z 0 ) = a 0 f(z_0) = a_0 f ( z 0 ) = a 0 Pole of order m m m : a − m ≠ 0 a_{-m} \neq 0 a − m = 0 a n = 0 a_n = 0 a n = 0 n < − m n \lt -m n < − m Essential singularity : infinitely many non-zero a n a_n a n n < 0 n \lt 0 n < 0 Proposition 8.1 (Riemann”s Removable Singularity Theorem). If f f f z 0 z_0 z 0 z 0 z_0 z 0
Proposition 8.2. z 0 z_0 z 0 m m m lim z → z 0 ( z − z 0 ) m f ( z ) \lim_{z \to z_0} (z - z_0)^m f(z) lim z → z 0 ( z − z 0 ) m f ( z )
Theorem 8.3 (Casorati-Weierstrass). If z 0 z_0 z 0 f f f f f f z 0 z_0 z 0
Solution Problem. Classify the singularities of f ( z ) = sin z z f(z) = \frac{\sin z}{z} f ( z ) = z s i n z
z = 0 z = 0 z = 0 sin z = z − z 3 / 6 + ⋯ \sin z = z - z^3/6 + \cdots sin z = z − z 3 /6 + ⋯ f ( z ) = 1 − z 2 / 6 + ⋯ f(z) = 1 - z^2/6 + \cdots f ( z ) = 1 − z 2 /6 + ⋯ z = 0 z = 0 z = 0 f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1
Problem. Classify the singularities of f ( z ) = e z − 1 z 2 f(z) = \frac{e^z - 1}{z^2} f ( z ) = z 2 e z − 1
z = 0 z = 0 z = 0 e z − 1 = z + z 2 / 2 + ⋯ e^z - 1 = z + z^2/2 + \cdots e z − 1 = z + z 2 /2 + ⋯ f ( z ) = 1 z + 1 2 + ⋯ f(z) = \frac{1}{z} + \frac{1}{2} + \cdots f ( z ) = z 1 + 2 1 + ⋯ 1 / z 1/z 1/ z z = 0 z = 0 z = 0 1 1 1
Problem. Classify the singularity of f ( z ) = e 1 / z f(z) = e^{1/z} f ( z ) = e 1/ z z = 0 z = 0 z = 0
e 1 / z = ∑ n = 0 ∞ 1 n ! z n = 1 + 1 z + 1 2 z 2 + ⋯ e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\, z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \cdots e 1/ z = ∑ n = 0 ∞ n ! z n 1 = 1 + z 1 + 2 z 2 1 + ⋯
Infinitely many negative powers ⇒ \Rightarrow ⇒ z = 0 z = 0 z = 0
Problem. Classify the singularities of f ( z ) = z + 1 z 3 ( z 2 + 1 ) f(z) = \frac{z + 1}{z^3(z^2 + 1)} f ( z ) = z 3 ( z 2 + 1 ) z + 1
z = 0 z = 0 z = 0 3 3 3 z = i z = i z = i z = − i z = -i z = − i
Problem. Determine the type of singularity of f ( z ) = z sin z f(z) = \frac{z}{\sin z} f ( z ) = s i n z z z = 0 z = 0 z = 0
sin z = z − z 3 / 6 + ⋯ \sin z = z - z^3/6 + \cdots sin z = z − z 3 /6 + ⋯ f ( z ) = 1 1 − z 2 / 6 + ⋯ = 1 + z 2 6 + ⋯ f(z) = \frac{1}{1 - z^2/6 + \cdots} = 1 + \frac{z^2}{6} + \cdots f ( z ) = 1 − z 2 /6 + ⋯ 1 = 1 + 6 z 2 + ⋯
No negative powers, so z = 0 z = 0 z = 0 f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1
Definition. The residue of f f f z 0 z_0 z 0 a − 1 a_{-1} a − 1
R e s ( f , z 0 ) = a − 1 = 1 2 π i ∫ γ f ( z ) d z \mathrm{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\int_\gamma f(z)\, dz Res ( f , z 0 ) = a − 1 = 2 π i 1 ∫ γ f ( z ) d z
Where γ \gamma γ z 0 z_0 z 0
For a simple pole at z 0 z_0 z 0
R e s ( f , z 0 ) = lim z → z 0 ( z − z 0 ) f ( z ) \mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z) Res ( f , z 0 ) = lim z → z 0 ( z − z 0 ) f ( z )
If f = g / h f = g/h f = g / h g ( z 0 ) ≠ 0 g(z_0) \neq 0 g ( z 0 ) = 0 h ( z 0 ) = 0 h(z_0) = 0 h ( z 0 ) = 0 h ′ ( z 0 ) ≠ 0 h'(z_0) \neq 0 h ′ ( z 0 ) = 0
R e s ( f , z 0 ) = g ( z 0 ) h ′ ( z 0 ) \mathrm{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)} Res ( f , z 0 ) = h ′ ( z 0 ) g ( z 0 )
For a pole of order m m m z 0 z_0 z 0
R e s ( f , z 0 ) = 1 ( m − 1 ) ! lim z → z 0 d m − 1 d z m − 1 [ ( z − z 0 ) m f ( z ) ] \mathrm{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] Res ( f , z 0 ) = ( m − 1 )! 1 lim z → z 0 d z m − 1 d m − 1 [ ( z − z 0 ) m f ( z ) ]
Solution Problem. Find the residue of f ( z ) = z z 2 + 4 z + 3 f(z) = \frac{z}{z^2 + 4z + 3} f ( z ) = z 2 + 4 z + 3 z
z 2 + 4 z + 3 = ( z + 1 ) ( z + 3 ) z^2 + 4z + 3 = (z + 1)(z + 3) z 2 + 4 z + 3 = ( z + 1 ) ( z + 3 ) z = − 1 z = -1 z = − 1 z = − 3 z = -3 z = − 3
At z = − 1 z = -1 z = − 1 R e s = lim z → − 1 z z + 3 = − 1 2 \mathrm{Res} = \lim_{z \to -1} \frac{z}{z + 3} = \frac{-1}{2} Res = lim z → − 1 z + 3 z = 2 − 1 z = − 3 z = -3 z = − 3 R e s = lim z → − 3 z z + 1 = − 3 − 2 = 3 2 \mathrm{Res} = \lim_{z \to -3} \frac{z}{z + 1} = \frac{-3}{-2} = \frac{3}{2} Res = lim z → − 3 z + 1 z = − 2 − 3 = 2 3
Problem. Find the residue of f ( z ) = e z ( z − 1 ) 2 ( z − 2 ) f(z) = \frac{e^z}{(z - 1)^2(z - 2)} f ( z ) = ( z − 1 ) 2 ( z − 2 ) e z
At z = 1 z = 1 z = 1 2 2 2 R e s = d d z [ e z z − 2 ] z = 1 = e z ( z − 2 ) − e z ( z − 2 ) 2 ∣ z = 1 = − e − e 1 = − 2 e \mathrm{Res} = \frac{d}{dz}\left[\frac{e^z}{z - 2}\right]_{z=1} = \frac{e^z(z - 2) - e^z}{(z-2)^2}\Big|_{z=1} = \frac{-e - e}{1} = -2e Res = d z d [ z − 2 e z ] z = 1 = ( z − 2 ) 2 e z ( z − 2 ) − e z z = 1 = 1 − e − e = − 2 e
At z = 2 z = 2 z = 2 R e s = e 2 ( 2 − 1 ) 2 = e 2 \mathrm{Res} = \frac{e^2}{(2-1)^2} = e^2 Res = ( 2 − 1 ) 2 e 2 = e 2
Theorem 8.4 (Residue Theorem). If f f f γ \gamma γ z 1 , z 2 , … , z n z_1, z_2, \ldots, z_n z 1 , z 2 , … , z n γ \gamma γ
∫ γ f ( z ) d z = 2 π i ∑ k = 1 n R e s ( f , z k ) \int_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^{n} \mathrm{Res}(f, z_k) ∫ γ f ( z ) d z = 2 π i ∑ k = 1 n Res ( f , z k )
Proof. For each singularity z k z_k z k γ k \gamma_k γ k γ \gamma γ γ k \gamma_k γ k
∫ γ f d z = ∑ k = 1 n ∫ γ k f d z = ∑ k = 1 n 2 π i ⋅ R e s ( f , z k ) \int_\gamma f\, dz = \sum_{k=1}^n \int_{\gamma_k} f\, dz = \sum_{k=1}^n 2\pi i \cdot \mathrm{Res}(f, z_k) ∫ γ f d z = ∑ k = 1 n ∫ γ k f d z = ∑ k = 1 n 2 π i ⋅ Res ( f , z k ) ■ \blacksquare ■
Solution Problem 1. Evaluate ∫ γ e z z ( z − 1 ) 2 d z \int_\gamma \frac{e^z}{z(z-1)^2}\, dz ∫ γ z ( z − 1 ) 2 e z d z γ \gamma γ ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Solution. Singularities inside γ \gamma γ z = 0 z = 0 z = 0 z = 1 z = 1 z = 1 2 2 2
At z = 0 z = 0 z = 0 R e s = lim z → 0 e z ( z − 1 ) 2 = 1 ( − 1 ) 2 = 1 \mathrm{Res} = \lim_{z \to 0} \frac{e^z}{(z-1)^2} = \frac{1}{(-1)^2} = 1 Res = lim z → 0 ( z − 1 ) 2 e z = ( − 1 ) 2 1 = 1
At z = 1 z = 1 z = 1 R e s ( f , 1 ) = d d z [ ( z − 1 ) 2 ⋅ e z z ( z − 1 ) 2 ] z = 1 = d d z [ e z z ] z = 1 = e z ⋅ z − e z z 2 ∣ z = 1 = e − e 1 = 0 \mathrm{Res}(f, 1) = \frac{d}{dz}\left[(z-1)^2 \cdot \frac{e^z}{z(z-1)^2}\right]_{z=1} = \frac{d}{dz}\left[\frac{e^z}{z}\right]_{z=1} = \frac{e^z \cdot z - e^z}{z^2}\Big|_{z=1} = \frac{e - e}{1} = 0 Res ( f , 1 ) = d z d [ ( z − 1 ) 2 ⋅ z ( z − 1 ) 2 e z ] z = 1 = d z d [ z e z ] z = 1 = z 2 e z ⋅ z − e z z = 1 = 1 e − e = 0
∫ γ f d z = 2 π i ( 1 + 0 ) = 2 π i \int_\gamma f\, dz = 2\pi i(1 + 0) = 2\pi i ∫ γ f d z = 2 π i ( 1 + 0 ) = 2 π i ■ \blacksquare ■
Problem 2. Evaluate ∫ γ 1 z 4 + 1 d z \int_\gamma \frac{1}{z^4 + 1}\, dz ∫ γ z 4 + 1 1 d z γ \gamma γ ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Solution. The poles are the fourth roots of − 1 -1 − 1 z k = e i π / 4 + i k π / 2 z_k = e^{i\pi/4 + ik\pi/2} z k = e iπ /4 + ik π /2 k = 0 , 1 , 2 , 3 k = 0, 1, 2, 3 k = 0 , 1 , 2 , 3 ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Each is a simple pole with R e s ( f , z k ) = 1 4 z k 3 \mathrm{Res}(f, z_k) = \frac{1}{4z_k^3} Res ( f , z k ) = 4 z k 3 1 z k 4 = − 1 z_k^4 = -1 z k 4 = − 1 z k − 3 = − z k z_k^{-3} = -z_k z k − 3 = − z k − 1 4 ∑ z k = 0 -\frac{1}{4}\sum z_k = 0 − 4 1 ∑ z k = 0
∫ γ d z z 4 + 1 = 2 π i ⋅ 0 = 0 \int_\gamma \frac{dz}{z^4 + 1} = 2\pi i \cdot 0 = 0 ∫ γ z 4 + 1 d z = 2 π i ⋅ 0 = 0 ■ \blacksquare ■