A contour (or piecewise smooth path) in C \mathbb{C} C γ : [ a , b ] → C \gamma : [a, b] \to \mathbb{C} γ : [ a , b ] → C
A simple closed contour is a contour with γ ( a ) = γ ( b ) \gamma(a) = \gamma(b) γ ( a ) = γ ( b )
Definition. For a contour γ \gamma γ f f f γ \gamma γ
∫ γ f ( z ) d z = ∫ a b f ( γ ( t ) ) γ " ( t ) d t \int_{\gamma} f(z)\, dz = \int_a^b f(\gamma(t))\gamma"(t)\, dt ∫ γ f ( z ) d z = ∫ a b f ( γ ( t )) γ " ( t ) d t
Proposition 4.1. The complex integral is linear:
∫ γ ( a f + b g ) d z = a ∫ γ f d z + b ∫ γ g d z \int_\gamma (af + bg)\, dz = a\int_\gamma f\, dz + b\int_\gamma g\, dz ∫ γ ( a f + b g ) d z = a ∫ γ f d z + b ∫ γ g d z
Proposition 4.2. Reversing orientation changes the sign:
∫ − γ f d z = − ∫ γ f d z \int_{-\gamma} f\, dz = -\int_\gamma f\, dz ∫ − γ f d z = − ∫ γ f d z
Proposition 4.3. Additivity over contours:
∫ γ 1 + γ 2 f d z = ∫ γ 1 f d z + ∫ γ 2 f d z \int_{\gamma_1 + \gamma_2} f\, dz = \int_{\gamma_1} f\, dz + \int_{\gamma_2} f\, dz ∫ γ 1 + γ 2 f d z = ∫ γ 1 f d z + ∫ γ 2 f d z
Proposition 4.4 (ML Inequality). If ∣ f ( z ) ∣ ≤ M |f(z)| \leq M ∣ f ( z ) ∣ ≤ M z z z γ \gamma γ L L L
∣ ∫ γ f ( z ) d z ∣ ≤ M L \left|\int_\gamma f(z)\, dz\right| \leq ML ∫ γ f ( z ) d z ≤ M L
Proof. ∣ ∫ a b f ( γ ( t ) ) γ ′ ( t ) d t ∣ ≤ ∫ a b ∣ f ( γ ( t ) ) ∣ ∣ γ ′ ( t ) ∣ d t ≤ M ∫ a b ∣ γ ′ ( t ) ∣ d t = M L \left|\int_a^b f(\gamma(t))\gamma'(t)\, dt\right| \leq \int_a^b |f(\gamma(t))||\gamma'(t)|\, dt \leq M \int_a^b |\gamma'(t)|\, dt = ML ∫ a b f ( γ ( t )) γ ′ ( t ) d t ≤ ∫ a b ∣ f ( γ ( t )) ∣∣ γ ′ ( t ) ∣ d t ≤ M ∫ a b ∣ γ ′ ( t ) ∣ d t = M L ■ \blacksquare ■
Solution Problem. Evaluate ∫ γ z 2 d z \int_\gamma z^2\, dz ∫ γ z 2 d z γ \gamma γ 0 0 0 1 + i 1 + i 1 + i
Solution. Parameterize γ ( t ) = t ( 1 + i ) \gamma(t) = t(1 + i) γ ( t ) = t ( 1 + i ) 0 ≤ t ≤ 1 0 \leq t \leq 1 0 ≤ t ≤ 1 γ ′ ( t ) = 1 + i \gamma'(t) = 1 + i γ ′ ( t ) = 1 + i
∫ γ z 2 d z = ∫ 0 1 ( t ( 1 + i ) ) 2 ( 1 + i ) d t = ( 1 + i ) 3 ∫ 0 1 t 2 d t = ( 1 + i ) 3 ⋅ 1 3 \int_\gamma z^2\, dz = \int_0^1 (t(1+i))^2 (1+i)\, dt = (1+i)^3 \int_0^1 t^2\, dt = (1+i)^3 \cdot \frac{1}{3} ∫ γ z 2 d z = ∫ 0 1 ( t ( 1 + i ) ) 2 ( 1 + i ) d t = ( 1 + i ) 3 ∫ 0 1 t 2 d t = ( 1 + i ) 3 ⋅ 3 1
( 1 + i ) 3 = ( 1 + i ) ( 1 + i ) 2 = ( 1 + i ) ( 2 i ) = 2 i + 2 i 2 = 2 i − 2 = − 2 + 2 i (1+i)^3 = (1+i)(1+i)^2 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i ( 1 + i ) 3 = ( 1 + i ) ( 1 + i ) 2 = ( 1 + i ) ( 2 i ) = 2 i + 2 i 2 = 2 i − 2 = − 2 + 2 i
∫ γ z 2 d z = − 2 + 2 i 3 \int_\gamma z^2\, dz = \frac{-2 + 2i}{3} ∫ γ z 2 d z = 3 − 2 + 2 i ■ \blacksquare ■
Problem. Evaluate ∫ γ z ˉ d z \int_\gamma \bar{z}\, dz ∫ γ z ˉ d z γ \gamma γ
γ ( t ) = e i t \gamma(t) = e^{it} γ ( t ) = e i t 0 ≤ t ≤ 2 π 0 \leq t \leq 2\pi 0 ≤ t ≤ 2 π γ ′ ( t ) = i e i t \gamma'(t) = ie^{it} γ ′ ( t ) = i e i t z ˉ = e − i t \bar{z} = e^{-it} z ˉ = e − i t γ \gamma γ
∫ γ z ˉ d z = ∫ 0 2 π e − i t ⋅ i e i t d t = ∫ 0 2 π i d t = 2 π i \int_\gamma \bar{z}\, dz = \int_0^{2\pi} e^{-it} \cdot ie^{it}\, dt = \int_0^{2\pi} i\, dt = 2\pi i ∫ γ z ˉ d z = ∫ 0 2 π e − i t ⋅ i e i t d t = ∫ 0 2 π i d t = 2 π i
Note: Since z ˉ \bar{z} z ˉ
Problem. Evaluate ∫ γ d z z \int_\gamma \frac{dz}{z} ∫ γ z d z γ \gamma γ z = e i θ z = e^{i\theta} z = e i θ 0 ≤ θ ≤ π 0 \leq \theta \leq \pi 0 ≤ θ ≤ π
∫ 0 π i e i θ e i θ d θ = ∫ 0 π i d θ = i π \int_0^\pi \frac{ie^{i\theta}}{e^{i\theta}}\, d\theta = \int_0^\pi i\, d\theta = i\pi ∫ 0 π e i θ i e i θ d θ = ∫ 0 π i d θ = iπ
Problem. Evaluate ∫ γ z d z \int_\gamma z\, dz ∫ γ z d z γ \gamma γ 0 0 0 1 1 1 1 1 1 1 + i 1 + i 1 + i
γ 1 ( t ) = t \gamma_1(t) = t γ 1 ( t ) = t 0 ≤ t ≤ 1 0 \leq t \leq 1 0 ≤ t ≤ 1 ∫ 0 1 t ⋅ 1 d t = 1 2 \int_0^1 t \cdot 1\, dt = \frac{1}{2} ∫ 0 1 t ⋅ 1 d t = 2 1
γ 2 ( t ) = 1 + i t \gamma_2(t) = 1 + it γ 2 ( t ) = 1 + i t 0 ≤ t ≤ 1 0 \leq t \leq 1 0 ≤ t ≤ 1 ∫ 0 1 ( 1 + i t ) ⋅ i d t = ∫ 0 1 ( i − t ) d t = i − 1 2 \int_0^1 (1 + it) \cdot i\, dt = \int_0^1 (i - t)\, dt = i - \frac{1}{2} ∫ 0 1 ( 1 + i t ) ⋅ i d t = ∫ 0 1 ( i − t ) d t = i − 2 1
Total: 1 2 + i − 1 2 = i \frac{1}{2} + i - \frac{1}{2} = i 2 1 + i − 2 1 = i
Check: Since z z z 0 0 0 1 + i 1 + i 1 + i 1 2 ( 1 + i ) 2 = i \frac{1}{2}(1+i)^2 = i 2 1 ( 1 + i ) 2 = i ■ \blacksquare ■
Solution Problem. Use the ML inequality to show that lim R → ∞ ∫ C R e i z z d z = 0 \lim_{R \to \infty} \int_{C_R} \frac{e^{iz}}{z}\, dz = 0 lim R → ∞ ∫ C R z e i z d z = 0 C R C_R C R ∣ z ∣ = R |z| = R ∣ z ∣ = R I m ( z ) ≥ 0 \mathrm{Im}(z) \geq 0 Im ( z ) ≥ 0
On C R C_R C R z = R e i θ z = Re^{i\theta} z = R e i θ 0 ≤ θ ≤ π 0 \leq \theta \leq \pi 0 ≤ θ ≤ π ∣ e i z ∣ = ∣ e i R ( cos θ + i sin θ ) ∣ = e − R sin θ |e^{iz}| = |e^{iR(\cos\theta + i\sin\theta)}| = e^{-R\sin\theta} ∣ e i z ∣ = ∣ e i R ( c o s θ + i s i n θ ) ∣ = e − R s i n θ
∣ ∫ C R e i z z d z ∣ ≤ ∫ 0 π e − R sin θ R ⋅ R d θ = ∫ 0 π e − R sin θ d θ \left|\int_{C_R} \frac{e^{iz}}{z}\, dz\right| \leq \int_0^\pi \frac{e^{-R\sin\theta}}{R} \cdot R\, d\theta = \int_0^\pi e^{-R\sin\theta}\, d\theta ∫ C R z e i z d z ≤ ∫ 0 π R e − R s i n θ ⋅ R d θ = ∫ 0 π e − R s i n θ d θ
By Jordan’s inequality sin θ ≥ 2 θ π \sin\theta \geq \frac{2\theta}{\pi} sin θ ≥ π 2 θ θ ∈ [ 0 , π / 2 ] \theta \in [0, \pi/2] θ ∈ [ 0 , π /2 ]
≤ 2 ∫ 0 π / 2 e − 2 R θ / π d θ = π R ( 1 − e − R ) → 0 \leq 2\int_0^{\pi/2} e^{-2R\theta/\pi}\, d\theta = \frac{\pi}{R}(1 - e^{-R}) \to 0 ≤ 2 ∫ 0 π /2 e − 2 R θ / π d θ = R π ( 1 − e − R ) → 0 R → ∞ R \to \infty R → ∞ ■ \blacksquare ■
Problem. Bound ∣ ∫ γ d z z 2 + 4 ∣ \left|\int_\gamma \frac{dz}{z^2 + 4}\right| ∫ γ z 2 + 4 d z γ \gamma γ ∣ z ∣ = 3 |z| = 3 ∣ z ∣ = 3
On γ \gamma γ ∣ z 2 + 4 ∣ ≥ ∣ z ∣ 2 − 4 = 9 − 4 = 5 |z^2 + 4| \geq |z|^2 - 4 = 9 - 4 = 5 ∣ z 2 + 4∣ ≥ ∣ z ∣ 2 − 4 = 9 − 4 = 5 ∣ 1 z 2 + 4 ∣ ≤ 1 5 \left|\frac{1}{z^2 + 4}\right| \leq \frac{1}{5} z 2 + 4 1 ≤ 5 1
Length of γ \gamma γ L = 2 π ⋅ 3 = 6 π L = 2\pi \cdot 3 = 6\pi L = 2 π ⋅ 3 = 6 π
∣ ∫ γ d z z 2 + 4 ∣ ≤ 1 5 ⋅ 6 π = 6 π 5 \left|\int_\gamma \frac{dz}{z^2 + 4}\right| \leq \frac{1}{5} \cdot 6\pi = \frac{6\pi}{5} ∫ γ z 2 + 4 d z ≤ 5 1 ⋅ 6 π = 5 6 π
When f f f F F F F ′ = f F' = f F ′ = f
∫ γ f ( z ) d z = F ( γ ( b ) ) − F ( γ ( a ) ) \int_\gamma f(z)\, dz = F(\gamma(b)) - F(\gamma(a)) ∫ γ f ( z ) d z = F ( γ ( b )) − F ( γ ( a ))
This follows from the fundamental theorem of calculus applied to F ( γ ( t ) ) F(\gamma(t)) F ( γ ( t ))
Solution Problem. Evaluate ∫ γ cos z d z \int_\gamma \cos z\, dz ∫ γ cos z d z γ \gamma γ 0 0 0 π + i \pi + i π + i
Since cos z \cos z cos z sin z \sin z sin z
∫ γ cos z d z = sin ( π + i ) − sin ( 0 ) = sin ( π + i ) \int_\gamma \cos z\, dz = \sin(\pi + i) - \sin(0) = \sin(\pi + i) ∫ γ cos z d z = sin ( π + i ) − sin ( 0 ) = sin ( π + i )
sin ( π + i ) = sin π cosh 1 + i cos π sinh 1 = − i sinh 1 \sin(\pi + i) = \sin\pi\cosh 1 + i\cos\pi\sinh 1 = -i\sinh 1 sin ( π + i ) = sin π cosh 1 + i cos π sinh 1 = − i sinh 1
So the integral equals − i sinh 1 -i\sinh 1 − i sinh 1
Problem. Evaluate ∫ γ e 2 z d z \int_\gamma e^{2z}\, dz ∫ γ e 2 z d z γ \gamma γ 1 1 1 i i i
Antiderivative: 1 2 e 2 z \frac{1}{2}e^{2z} 2 1 e 2 z
∫ γ e 2 z d z = 1 2 ( e 2 i − e 2 ) \int_\gamma e^{2z}\, dz = \frac{1}{2}(e^{2i} - e^{2}) ∫ γ e 2 z d z = 2 1 ( e 2 i − e 2 )