Questions
0–1 questions per paper
Difficulty
Medium
Importance
Low yield, focus only if remaining syllabus is complete.
Overview
Complex Variables deal with functions of a complex variable z = x + iy, extending calculus into the complex plane. While appearing less frequently in PSU exams compared to Calculus or Linear Algebra, mastering the conditions for analyticity and residue integration is essential for scoring easy marks on occasional questions. Focus on identifying singularities and applying standard expansion theorems.
Analytic Functions
A function is analytic at a point if it is differentiable in some neighborhood of that point. For engineering exams, the Cauchy-Riemann equations serve as the primary tool to verify if a function given in terms of u(x,y) and v(x,y) is analytic.
- Cauchy-Riemann equations: du/dx = dv/dy and du/dy = -dv/dx
- Necessary condition for analyticity is satisfying C-R equations
- Sufficient condition is satisfying C-R equations with continuous partial derivatives
- Harmonic functions satisfy Laplace equation: del^2 u = 0
Cauchy's Integral Theorem & Formula
These theorems allow the evaluation of contour integrals of complex functions without direct parameterization. If a function is analytic within and on a simple closed curve, its line integral is zero.
- Cauchy's Integral Theorem: integral f(z) dz = 0 if f is analytic
- Cauchy's Integral Formula: f(a) = (1/2*pi*i) * integral [f(z)/(z-a)] dz
- Generalization for derivatives: f^n(a) = (n! / 2*pi*i) * integral [f(z)/(z-a)^(n+1)] dz
- Residue Theorem: integral f(z) dz = 2*pi*i * sum of residues
Taylor & Laurent Series
These series provide power series representations for functions, crucial for evaluating residues. Taylor series is used for functions analytic within a disk, while Laurent series expands functions around singular points (poles).
- Taylor series centered at a: sum [f^n(a)/n!] * (z-a)^n
- Laurent series includes both positive and negative powers of (z-a)
- Residue is the coefficient 'a_{-1}' of the (z-a)^(-1) term
- Singularities can be removable, poles, or essential
Formula Sheet
z = x + iy
Cauchy-Riemann: ux = vy, uy = -vx
f(a) = (1/2*pi*i) * integral [f(z)/(z-a)] dz
Residue at simple pole: lim(z->a) (z-a)f(z)
Residue at pole of order m: (1/(m-1)!) * lim(z->a) [d^(m-1)/dz^(m-1)] * ((z-a)^m * f(z))
integral f(z) dz = 2*pi*i * sum(Residues at poles inside C)
Exam Tip
Always identify the order of the pole first; the residue at a simple pole is simply limit as z tends to a of (z-a)f(z).
Common Mistakes
- Forgetting to include 2*pi*i when applying the Residue Theorem.
- Confusing the order of partial derivatives in Cauchy-Riemann equations (wrong sign).
- Failing to check if a singularity lies inside the contour of integration before applying the residue formula.
More Revision Notes
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