Unit-5: Complex Variable – Integration (Complete Theory + Formulas + Solved Examples)
Unit-5: Complex Variable – Integration
(Complete Formulas + Theory + Solved Examples)
1. Complex Line Integral
\[ \int_C f(z) \, dz = \int_a^b f(z(t)) z'(t) \, dt \] where z(t) is parametric representation of curve C from t = a to b.2. Cauchy-Goursat Theorem (Cauchy Integral Theorem)
If f(z) is analytic inside and on a simple closed contour C (positively oriented) and f'(z) is continuous, then \[ \oint_C f(z) \, dz = 0 \]3. Cauchy Integral Formula
If f(z) is analytic inside and on C, and z₀ is inside C: \[ f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} \, dz \] Derivatives: \[ f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz \quad (n = 1,2,\dots) \]4. Taylor’s Series
If f(z) is analytic inside |z − z₀| < R, \[ f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n \] where \[ a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \oint_C \frac{f(w)}{(w - z_0)^{n+1}} \, dw \] Radius of convergence = distance to nearest singularity.
Example 1: Taylor series of f(z) = eᶻ about z=0
\[ e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} \quad (|z| < \infty) \]
\[ e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} \quad (|z| < \infty) \]
5. Laurent’s Series
If f(z) is analytic in annulus r < |z − z₀| < R, \[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n \] where \[ a_n = \frac{1}{2\pi i} \oint_C \frac{f(w)}{(w - z_0)^{n+1}} \, dw \] Principal part: negative powers (n < 0)The coefficient a₋₁ is called the Residue at z₀
6. Singularities & Classification
| Type | Definition | Laurent Series | Example |
|---|---|---|---|
| Removable | Principal part = 0 | All a₋ₙ = 0 | sin z / z at 0 |
| Pole of order m | Highest negative power = -m | a₋ₘ ≠ 0, a₋ₖ=0 for k>m | 1/(z-1)³ → order 3 |
| Essential | Infinitely many negative powers | Infinite principal part | e^{1/z} at 0 |
| Branch Point | Multi-valued function | — | Log z, √z |
7. Zeros of Analytic Functions
Zero of order m at z₀: f(z₀)=f'(z₀)=…=f^{(m-1)}(z₀)=0, f^{(m)}(z₀) ≠ 0Then f(z) = (z - z₀)^m g(z), where g(z₀) ≠ 0 and g analytic.
8. Residues
Residue at z = z₀ → coefficient of 1/(z − z₀) in Laurent seriesResidue Formulas
| Case | Residue at z₀ |
|---|---|
| Simple pole (order 1) | \( \operatorname{Res}(f,z_0) = \lim_{z \to z_0} (z - z_0) f(z) \) |
| Pole of order m | \( \operatorname{Res}(f,z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z - z_0)^m f(z)] \) |
| f(z) = p(z)/q(z), simple pole where q(z₀)=0, p(z₀)≠0 | \( \operatorname{Res} = \frac{p(z_0)}{q'(z_0)} \) |
9. Cauchy’s Residue Theorem
If f(z) is analytic inside and on C except for finite number of isolated singularities inside C, then \[ \oint_C f(z) \, dz = 2\pi i \sum \operatorname{Residue\ at\ singularities\ inside\ C} \]
Example 2: Evaluate \( \oint_{|z|=2} \frac{dz}{z(z-1)} \)
Poles: z=0 (simple, inside), z=1 (simple, inside)
Res at z=0: lim (z)(1/(z(z-1))) = 1/(0-1) = -1
Res at z=1: lim (z-1)/(z(z-1)) = 1/(1) = 1
Sum = -1 + 1 = 0
Integral = 2πi × 0 = 0
Poles: z=0 (simple, inside), z=1 (simple, inside)
Res at z=0: lim (z)(1/(z(z-1))) = 1/(0-1) = -1
Res at z=1: lim (z-1)/(z(z-1)) = 1/(1) = 1
Sum = -1 + 1 = 0
Integral = 2πi × 0 = 0
Example 3: \( \oint_{|z|=1} \frac{e^z}{z^3} dz \)
Pole of order 3 at z=0
Res = \frac{1}{2!} lim \frac{d^2}{dz^2} (e^z) = \frac{1}{2} e^z \big|_{z=0} = \frac{1}{2}
Integral = 2πi × (1/2) = πi
Pole of order 3 at z=0
Res = \frac{1}{2!} lim \frac{d^2}{dz^2} (e^z) = \frac{1}{2} e^z \big|_{z=0} = \frac{1}{2}
Integral = 2πi × (1/2) = πi
Example 4 (Real Integral): \( \int_{-\infty}^{\infty} \frac{x^2}{x^4 + 1} dx \)
Consider \( \oint \frac{z^2}{z^4 + 1} dz \) over semicircle in upper half-plane
Poles inside: z = e^{iπ/4}, e^{i3π/4}
Res at e^{iπ/4} = \frac{(e^{iπ/4})^2}{4 (e^{iπ/4})^3} = \frac{e^{iπ/2}}{4 e^{i3π/4}} = \frac{i}{4 e^{i3π/4}} = \frac{i}{4 \cdot (\frac{-1+i}{\sqrt{2}})} \)
After calculation, sum of residues = 1/(2√2) (1 - i)
Integral = 2πi × sum = π/√2
As R→∞, upper semicircle → 0 → real integral = π/√2
Answer: \frac{\pi}{\sqrt{2}}
Consider \( \oint \frac{z^2}{z^4 + 1} dz \) over semicircle in upper half-plane
Poles inside: z = e^{iπ/4}, e^{i3π/4}
Res at e^{iπ/4} = \frac{(e^{iπ/4})^2}{4 (e^{iπ/4})^3} = \frac{e^{iπ/2}}{4 e^{i3π/4}} = \frac{i}{4 e^{i3π/4}} = \frac{i}{4 \cdot (\frac{-1+i}{\sqrt{2}})} \)
After calculation, sum of residues = 1/(2√2) (1 - i)
Integral = 2πi × sum = π/√2
As R→∞, upper semicircle → 0 → real integral = π/√2
Answer: \frac{\pi}{\sqrt{2}}
Example 5: \( \int_0^{2\pi} \frac{d\theta}{a + b \cos \theta} \quad (a > b > 0) \)
z = e^{iθ}, dθ = dz/(i z), cos θ = (z + 1/z)/2
→ \( \oint \frac{dz}{i z (a + b(z + 1/z)/2)} = \oint \frac{2 dz}{i (2 a z + b z^2 + b)} \)
Pole inside unit circle: z = (−a + √(a² − b²))/b
Residue calculation gives:
Integral = \frac{2\pi}{\sqrt{a^2 - b^2}}
z = e^{iθ}, dθ = dz/(i z), cos θ = (z + 1/z)/2
→ \( \oint \frac{dz}{i z (a + b(z + 1/z)/2)} = \oint \frac{2 dz}{i (2 a z + b z^2 + b)} \)
Pole inside unit circle: z = (−a + √(a² − b²))/b
Residue calculation gives:
Integral = \frac{2\pi}{\sqrt{a^2 - b^2}}
Important Standard Residues
| f(z) | Singularity | Residue |
|---|---|---|
| 1/(z(z-a)) | z=0 | -1/a |
| cot πz | integer n | 1/π |
| e^{iz}/(z^2 + a^2) | z = ia | −π e^{-a}/a (for a>0) |
Summary of Power Series
| Series | Name | Valid Region |
|---|---|---|
| Only non-negative powers | Taylor | Inside circle of convergence |
| Both positive & negative powers | Laurent | Annulus |
| Only negative powers | Principal part | Determines singularity type |
Key Theorems Recap
- Cauchy Theorem → integral over closed path = 0 if analytic inside
- Cauchy Formula → values inside from boundary
- Residue Theorem → most powerful tool for contour integrals