Questions
~2 questions per PSU paper
Difficulty
Medium-Hard
Importance
High yield for HPCL, NTPC, and BHEL
Overview
Complex variables involve functions defined over the complex plane, focusing on differentiability and integration in the complex domain. Mastering this topic is essential for PSU exams as it provides the mathematical foundation for signal processing and circuit analysis, specifically through residues and mappings.
Analytic Functions & Cauchy-Riemann
A function is analytic if it is differentiable at every point in a region. The Cauchy-Riemann equations are the primary conditions used to verify if a given complex function is analytic.
- f(z) = u + iv is analytic if du/dx = dv/dy and du/dy = -dv/dx
- Harmonic functions satisfy Laplace equation: d2u/dx2 + d2u/dy2 = 0
- An analytic function is necessarily continuous
- If f(z) is analytic, then u and v are harmonic conjugates
Cauchy's Integral Theorem & Formula
These tools allow for the evaluation of complex line integrals without direct parameterization. Cauchy’s integral formula is frequently used in exams to solve contour integrals by identifying singularities.
- Cauchy's Theorem: Integral over a closed contour for analytic f(z) is 0
- Cauchy's Integral Formula: f(a) = (1/2πi) * ∫ [f(z)/(z-a)] dz
- General Formula: f(n)(a) = (n!/2πi) * ∫ [f(z)/(z-a)^(n+1)] dz
- Contour must be closed and simple for theorem validity
Taylor & Laurent Series
Power series expansions are used to represent complex functions near specific points or singularities. Laurent series is particularly powerful because it accounts for poles, unlike the Taylor series.
- Taylor series expands about an ordinary point
- Laurent series: f(z) = ∑ a_n(z-z0)^n + ∑ b_n(z-z0)^(-n)
- Principal part of Laurent series determines the type of singularity
- Geometric series: 1/(1-z) = 1 + z + z^2 + ... for |z|<1
Residue Theorem
The Residue Theorem is the most important calculation technique for PSU exams, allowing for the evaluation of complex integrals by summing residues at isolated singularities.
- ∫ f(z) dz = 2πi * (Sum of Residues)
- Residue at simple pole: Res(f, a) = lim(z→a) (z-a)f(z)
- Residue for pole of order n: (1/(n-1)!) * lim(z→a) (d^(n-1)/dz^(n-1)) [(z-a)^n f(z)]
- Always check if the pole lies within the given contour
Formula Sheet
du/dx = dv/dy
du/dy = -dv/dx
∇^2u = 0
f(a) = (1/2πi) ∮ [f(z)/(z-a)] dz
∮ f(z) dz = 2πi Σ Res(f, a_i)
Res(f, a) = lim_{z->a} (z-a)f(z)
Res(f, a) = 1/(n-1)! * lim_{z->a} [d^{n-1}/dz^{n-1} ((z-a)^n f(z))]
Exam Tip
When solving contour integrals, always identify poles inside the contour first; 90% of exam questions are solved simply by calculating residues at these interior poles.
Common Mistakes
- Failing to check if a pole actually lies inside the given contour before calculating the residue.
- Forgetting to multiply the final residue sum by 2πi in Cauchy's residue theorem problems.
- Incorrectly identifying the order of a pole, leading to the use of the wrong residue formula.
More Revision Notes
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