The residue of f f f α \alpha α a − 1 a_{-1} a − 1 ( z − α ) − 1 (z-\alpha)^{-1} ( z − α ) − 1 Laurent series of f f f α \alpha α
Res ( f , α ) = a − 1 \text{Res}(f, \alpha) = a_{-1} Res ( f , α ) = a − 1 At a removable singularity, all negative-power coefficients are 0 0 0 Res ( f , α ) = 0 \text{Res}(f, \alpha) = 0 Res ( f , α ) = 0
Cauchy’s Residue Theorem
Let C C C simple closed positively oriented contour . Let f f f C C C z 1 , z 2 , … , z n z_1, z_2, \ldots, z_n z 1 , z 2 , … , z n C C C
∮ C f ( z ) d z = 2 π i ∑ k = 1 n Res ( f , z k ) \oint_C f(z)\,\text{d}z = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k) ∮ C f ( z ) d z = 2 π i k = 1 ∑ n Res ( f , z k ) It generalizes Cauchy’s integral formula .
Residue at a Pole
Let α \alpha α n n n f f f
Res ( f , α ) = 1 ( n − 1 ) ! lim z → α d n − 1 d z n − 1 [ ( z − α ) n f ( z ) ] \text{Res}(f, \alpha) = \frac{1}{(n-1)!} \lim_{z \to \alpha} \frac{\text{d}^{n-1}}{\text{d}z^{n-1}} \left[(z-\alpha)^n f(z)\right] Res ( f , α ) = ( n − 1 )! 1 z → α lim d z n − 1 d n − 1 [ ( z − α ) n f ( z ) ] For a simple pole,
Res ( f , α ) = lim z → α ( z − α ) f ( z ) \text{Res}(f, \alpha) = \lim_{z \to \alpha} (z-\alpha) f(z) Res ( f , α ) = z → α lim ( z − α ) f ( z ) Examples:
Res ( exp ( 2 z ) , 0 ) = 2 \text{Res}\left(\exp\left(\frac{2}{z}\right), 0\right) = 2 Res ( exp ( z 2 ) , 0 ) = 2 Res ( 3 2 z + z 2 − z 3 , 0 ) = 3 2 \text{Res}\left(\frac{3}{2z + z^2 - z^3}, 0\right) = \frac{3}{2} Res ( 2 z + z 2 − z 3 3 , 0 ) = 2 3
Worked Examples
Residue from the Laurent Series
Find Res ( exp ( 2 z ) , 0 ) \text{Res}\left(\exp\left(\frac{2}{z}\right), 0\right) Res ( exp ( z 2 ) , 0 )
exp ( 2 z ) = ∑ n = 0 ∞ 2 n n ! z n = 1 + 2 z + 2 z 2 + ⋯ \exp\left(\frac{2}{z}\right) = \sum_{n=0}^{\infty} \frac{2^n}{n!\, z^n} = 1 + \frac{2}{z} + \frac{2}{z^2} + \cdots exp ( z 2 ) = n = 0 ∑ ∞ n ! z n 2 n = 1 + z 2 + z 2 2 + ⋯ The coefficient of z − 1 z^{-1} z − 1 2 2 2 2 2 2
Find Res ( π cot ( π z ) z 2 , 0 ) \text{Res}\left(\frac{\pi \cot(\pi z)}{z^2}, 0\right) Res ( z 2 π c o t ( π z ) , 0 )
π cot ( π z ) = 1 z − π 2 z 3 − ⋯ \pi \cot(\pi z) = \frac{1}{z} - \frac{\pi^2 z}{3} - \cdots π cot ( π z ) = z 1 − 3 π 2 z − ⋯ π cot ( π z ) z 2 = 1 z 3 − π 2 3 z − ⋯ \frac{\pi \cot(\pi z)}{z^2} = \frac{1}{z^3} - \frac{\pi^2}{3z} - \cdots z 2 π cot ( π z ) = z 3 1 − 3 z π 2 − ⋯ The residue is − π 2 3 -\frac{\pi^2}{3} − 3 π 2
Residue at a Simple Pole
Find Res ( 3 2 z + z 2 − z 3 , 0 ) \text{Res}\left(\frac{3}{2z + z^2 - z^3}, 0\right) Res ( 2 z + z 2 − z 3 3 , 0 )
The denominator is z ( 2 + z − z 2 ) z(2 + z - z^2) z ( 2 + z − z 2 ) z = 0 z = 0 z = 0
Res = lim z → 0 z ⋅ 3 z ( 2 + z − z 2 ) = 3 2 \text{Res} = \lim_{z \to 0} z \cdot \frac{3}{z(2 + z - z^2)} = \frac{3}{2} Res = z → 0 lim z ⋅ z ( 2 + z − z 2 ) 3 = 2 3 Residue at a Higher Order Pole
Find Res ( e z ( z − 2 ) 3 , 2 ) \text{Res}\left(\frac{e^z}{(z-2)^3}, 2\right) Res ( ( z − 2 ) 3 e z , 2 )
z = 2 z = 2 z = 2
Res = 1 2 ! lim z → 2 d 2 d z 2 e z = e 2 2 \text{Res} = \frac{1}{2!} \lim_{z \to 2} \frac{\text{d}^2}{\text{d}z^2} e^z = \frac{e^2}{2} Res = 2 ! 1 z → 2 lim d z 2 d 2 e z = 2 e 2 Removable Singularity
sin z z \frac{\sin z}{z} z s i n z 0 0 0 Res ( sin z z , 0 ) = 0 \text{Res}\left(\frac{\sin z}{z}, 0\right) = 0 Res ( z s i n z , 0 ) = 0
Contour Integral over 2 Poles
Evaluate ∫ C e z z 2 ( z − 1 ) d z \int_C \frac{e^z}{z^2 (z-1)}\,\text{d}z ∫ C z 2 ( z − 1 ) e z d z C C C ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
Both z = 0 z = 0 z = 0 z = 1 z = 1 z = 1 C C C
Res ( f , 1 ) = lim z → 1 e z z 2 = e \text{Res}(f, 1) = \lim_{z \to 1} \frac{e^z}{z^2} = e Res ( f , 1 ) = z → 1 lim z 2 e z = e Res ( f , 0 ) = lim z → 0 d d z e z z − 1 = lim z → 0 e z ( z − 1 ) − e z ( z − 1 ) 2 = − 2 \text{Res}(f, 0) = \lim_{z \to 0} \frac{\text{d}}{\text{d}z} \frac{e^z}{z-1} = \lim_{z \to 0} \frac{e^z (z-1) - e^z}{(z-1)^2} = -2 Res ( f , 0 ) = z → 0 lim d z d z − 1 e z = z → 0 lim ( z − 1 ) 2 e z ( z − 1 ) − e z = − 2 ∫ C f ( z ) d z = 2 π i ( e − 2 ) \int_C f(z)\,\text{d}z = 2\pi i (e - 2) ∫ C f ( z ) d z = 2 π i ( e − 2 ) Contour Away from the Origin
Evaluate ∫ C d z z 4 + 4 \int_C \frac{\text{d}z}{z^4 + 4} ∫ C z 4 + 4 d z C C C ∣ z − 1 ∣ = 2 |z-1| = 2 ∣ z − 1∣ = 2
The roots of z 4 = − 4 z^4 = -4 z 4 = − 4 ± 1 ± i \pm 1 \pm i ± 1 ± i 1 1 1 1 1 1 1 1 1 5 \sqrt{5} 5 5 \sqrt{5} 5 1 + i 1 + i 1 + i 1 − i 1 - i 1 − i C C C
For a simple pole z k z_k z k z k 4 = − 4 z_k^4 = -4 z k 4 = − 4
Res ( f , z k ) = 1 4 z k 3 = z k 4 z k 4 = − z k 16 \text{Res}(f, z_k) = \frac{1}{4z_k^3} = \frac{z_k}{4z_k^4} = -\frac{z_k}{16} Res ( f , z k ) = 4 z k 3 1 = 4 z k 4 z k = − 16 z k Res ( f , 1 + i ) + Res ( f , 1 − i ) = − ( 1 + i ) + ( 1 − i ) 16 = − 1 8 \text{Res}(f, 1+i) + \text{Res}(f, 1-i) = -\frac{(1+i) + (1-i)}{16} = -\frac{1}{8} Res ( f , 1 + i ) + Res ( f , 1 − i ) = − 16 ( 1 + i ) + ( 1 − i ) = − 8 1 ∫ C f ( z ) d z = 2 π i ( − 1 8 ) = − π i 4 \int_C f(z)\,\text{d}z = 2\pi i \left(-\frac{1}{8}\right) = -\frac{\pi i}{4} ∫ C f ( z ) d z = 2 π i ( − 8 1 ) = − 4 π i Contour Integral over 3 Poles
Evaluate ∫ C d z z 4 + z 3 − 2 z 2 \int_C \frac{\text{d}z}{z^4 + z^3 - 2z^2} ∫ C z 4 + z 3 − 2 z 2 d z C C C ∣ z ∣ = 3 |z| = 3 ∣ z ∣ = 3
Factor the denominator.
z 4 + z 3 − 2 z 2 = z 2 ( z − 1 ) ( z + 2 ) z^4 + z^3 - 2z^2 = z^2 (z-1)(z+2) z 4 + z 3 − 2 z 2 = z 2 ( z − 1 ) ( z + 2 ) All singularities lie inside C C C z = 0 z = 0 z = 0 z = 1 z = 1 z = 1 z = − 2 z = -2 z = − 2
Res ( f , 1 ) = lim z → 1 1 z 2 ( z + 2 ) = 1 3 \text{Res}(f, 1) = \lim_{z \to 1} \frac{1}{z^2(z+2)} = \frac{1}{3} Res ( f , 1 ) = z → 1 lim z 2 ( z + 2 ) 1 = 3 1 Res ( f , − 2 ) = lim z → − 2 1 z 2 ( z − 1 ) = − 1 12 \text{Res}(f, -2) = \lim_{z \to -2} \frac{1}{z^2(z-1)} = -\frac{1}{12} Res ( f , − 2 ) = z → − 2 lim z 2 ( z − 1 ) 1 = − 12 1 Res ( f , 0 ) = lim z → 0 d d z 1 z 2 + z − 2 = lim z → 0 − ( 2 z + 1 ) ( z 2 + z − 2 ) 2 = − 1 4 \text{Res}(f, 0) = \lim_{z \to 0} \frac{\text{d}}{\text{d}z} \frac{1}{z^2 + z - 2} = \lim_{z \to 0} \frac{-(2z+1)}{(z^2+z-2)^2} = -\frac{1}{4} Res ( f , 0 ) = z → 0 lim d z d z 2 + z − 2 1 = z → 0 lim ( z 2 + z − 2 ) 2 − ( 2 z + 1 ) = − 4 1 ∮ C f ( z ) d z = 2 π i ( 1 3 − 1 12 − 1 4 ) = 0 \oint_C f(z)\,\text{d}z = 2\pi i \left(\frac{1}{3} - \frac{1}{12} - \frac{1}{4}\right) = 0 ∮ C f ( z ) d z = 2 π i ( 3 1 − 12 1 − 4 1 ) = 0 