Complex integration refers to the integration of complex-valued functions of a complex variable.
- It's a central topic in complex analysis.
- It plays a crucial role in understanding analytic functions, contour integrals, and major results such as Cauchy’s Theorem.
Instead of integrating along the real number line, you integrate along a path (contour) in the complex plane. If f(z) is a complex function and C is a curve in the complex plane, the integral is written:
\int_C f(z) \ dz
Complex Integration of a Real Variable
Complex integration starts with functions of a real variable that have complex values.
If
\int_{a}^{b} f(t)\,dt = \int_{a}^{b} u(t)\,dt + i \int_{a}^{b} v(t)\,dt
If u(t) and v(t) are continuous on [a,b], then f(t) is integrable.
Example: Integrate f(t) = 2t + it2. Here, u(t) = 2t and v(t) = t2. So, for 0 ≤ t ≤ 1.
\int_{0}^{1} (2t + i t^2)\, dt= \int_{0}^{1} 2t\, dt + i \int_{0}^{1} t^2\, dt\\ \left[ t^2 \right]_{0}^{1} + i \left[ \frac{t^3}{3} \right]_{0}^{1}= (1 - 0) + i \left( \frac{1}{3} - 0 \right)\\ 1 + \frac{i}{3}
Contour Integrals
In complex analysis, a contour integral is a type of integral where we integrate a complex function along a curve (called a contour) in the complex plane. It is similar to real integration, but instead of moving along a straight line on the real axis, we move along a path in two dimensions: the real and imaginary parts.
Suppose f(z) is a complex function, and C is a smooth curve parameterized by z(t) = x(t) + iy(t), for t ⋿ [a,b].
Then, the contour integral of f(z) along C is defined as:
\int_{C} f(z)\,dz = \int_{a}^{b} f(z(t))\,z'(t)\,dt
Where,
- z(t) traces the contour C,
- z′(t) = dz/dt
- f(z(t)) gives the value of the function at each point on the contour.
Properties of Contour Integrals
Let f(z) and g(z) be complex-valued functions defined on a contour C, and let α,β∈C.
1. Linearity: Contour integration is linear in nature.
\int_C \left( \alpha f(z) + \beta g(z) \right)\,dz = \alpha \int_C f(z)\,dz + \beta \int_C g(z)\,dz
2. Additivity over Contours: If a contour C is composed of two successive contours C1 and C2, then:
\int_C f(z)\,dz = \int_{C_1} f(z)\,dz + \int_{C_2} f(z)\,dz
3. Sign reversal property: Reversing the direction of the contour changes the sign of the integral:
\int_{-C} f(z)\,dz = - \int_C f(z)\,dz
4. Integral of Constant Function: If f(z)=k (constant), then:
\int_C k\,dz = k \int_C dz
5. Estimation (ML Inequality): If ∣f(z)∣≤M for all z ∈ C and the length of the contour is L, then:
\left| \int_C f(z)\,dz \right| \leq M \cdot L
Example : Consider the function f(z) = |z|2 =x2+y2 integrated along the curve parameterized by z(t) = t + it for 0 ≤ t ≤ 1.
This is the straight line segment joining the origin and the point 1+ i. We have here ̇dz(t)/dt = z'(t) = 1 + i.
So,
\int_{C}^{}f(z)z'(t)dt = \int_{0}^{1}(t^{2}+t^{2})(1+i)dt\\[3pts]2(1+i)\int_{0}^{1}t^{2}dt=\frac{2}{3}(1+i)
Example : Consider now the function f(z) = 1/z integrated along the smooth curve Γ parameterized by z(θ) = Re(i θ) for 0 ≤ θ ≤ 2π, where R ≠ 0.
The curve is a circle of radius R centered about the origin. Here f(z(θ)) = (1/R) e(−i θ) and dz(θ)/dθ = z'(θ) = i Re(i θ) = iz.
So, we have:
\int_{path}^{}f(z)z'(θ)dθ=\int_{0}^{2\pi }\frac{1}{z}izdθ\\[3pts]i\int_{0}^{2\pi }dθ = 2\pi i Note that the result is independent of R.
Piecewise-smooth curve
If Γ1 is a smooth curve joining z0 to z1 and Γ2 is another smooth curve joining z1 to z2, then we can make a curve Γ, not necessarily smooth, by joining z0 to z2 by first going to the intermediate point z1 via Γ1 and then from there via Γ2 to our destination z2.
The resulting curve Γ is still continuous, but it may not be smooth, since the velocity need not be continuous at the intermediate point z1. Such curves can be called piecewise smooth or contours, as popularly called in the literature. We can, thus, construct curves that are not smooth but which can be made out of a finite number of smooth curves, called smooth components.
Let Γ be a contour with n smooth components {Γj} for j = 1, 2, . . . , n. If f(z) is a function continuous on Γ, then the contour integral of f along Γ is defined as
\int_{\Gamma }^{}f(z)dz\\[3pts]\sum_{j=1}^{n}\int_{\Gamma _{j}}^{}f(z)dz =\int_{\Gamma _{1}}^{}f(z)dz+\int_{\Gamma _{2}}^{}f(z)dz+....+\int_{\Gamma _{n}}^{}f(z)dz
Example : Integrate f(z) = |z|2 again, but over piecewise smooth contour.
Break the whole integration process into four parts.
For OA, dy = 0. so dz = dx. y = 0, so |z|2 = x2.
For AB, dx = 0. so dz = idy. x = 1 so |z|2 = 1+y2.
For BC, dy = 0, so dz = dx. y = 1 so |z|2 = x2+1.
For CO, dx = 0, so dz = idy. x = 0, so |z|2 = y2.
\oint_{C}^{}|z|^{2}dz\\[3pts]\int_{0}^{1}x^{2}dx+i\int_{0}^{1}(1+y^{2})dy+\int_{1}^{0}(1+x^{2})dx+i\int_{1}^{0}y^{2}dy\\[3pts] \frac{1}{3}+\frac{4i}{3}-\frac{4}{3}-\frac{i}{3}=-1+i
Path Independence Theorem
A complex function f(z) is said to be path independent in a domain D if its contour integral between two points depends only on the endpoints and not on the path taken.
This happens when f(z) has an antiderivative F(z) in the domain D, i.e., F'(z) = f(z)
In that case, for any contour Γ joining z0 to z1,
For a closed contour: If the contour is closed (i.e., z0=z1), then:
Related Articles
Solved Questions
Question 1: Integrate the function f(t) = 2t + it2 from 0 to 1.
Break the function into real and imaginary parts: u(t) = 2t and v(t) = t2.
Compute the integrals separately:
∫ (2t + it2) dt = ∫ 2t dt + i ∫ t2 dt
For the real part: ∫ 2t dt = 2 [t2 / 2] from 0 to 1 = [t2] from 0 to 1 = 1.
For the imaginary part: i ∫ t2 dt = i [t3 / 3] from 0 to 1 = i [1/3].
Combine the results: 1 + i/3.
Question 2: Integrate the function f(t) = iet from 0 to 2.
Compute the integral directly:
∫ iet dt = i ∫ et dt
∫ et dt = et
Apply the limits: i [et] from 0 to 2 = i (e2 - e0) = i (e2 - 1).
Result: i (e2 - 1).
Question 3: Compute the integral of f(z) = |z|2 = x2 + y2 along the contour parameterized by z(t) = t + it from 0 to 1.
Parameterize the contour and find z'(t):
z(t) = t + it so z'(t) = 1 + i.
Compute the integral:
∫ (x2 + y2) (1 + i) dt where x = t and y = t.
x2 + y2 = t2 + t2 = 2t2.
So, ∫ 2t2 (1 + i) dt = (1 + i) ∫ 2t2 dt.
∫ 2t2 dt = (2/3) t3 from 0 to 1 = 2/3.
Multiply by (1 + i): 2/3 (1 + i) = 2/3 + (2/3)i.
Result: 2/3 + (2/3)i.