A vector space is a set whose elements (vectors) can be added and multiplied by scalars, following ten simple rules. Rⁿ, sets of matrices, polynomials and functions are all vector spaces. This topic covers the definition, subspaces, linear combinations and span, linear dependence and independence, basis and dimension.
महत्त्वाचे मुद्दे
A vector space V over a field F (usually R): addition is closed, commutative and associative; a zero vector exists; every v has −v; scalar multiplication is closed; and α(u + v) = αu + αv, (α + β)v = αv + βv, (αβ)v = α(βv), 1·v = v.
Examples: Rⁿ; the set Mₘₓₙ of m × n matrices; Pₙ, polynomials of degree ≤ n; the set of continuous functions on [a, b].
Subspace: a non-empty subset W of V that is itself a vector space under the same operations.
Subspace test: W is a subspace if for all u, v ∈ W and scalars α, β, αu + βv ∈ W.
Every subspace contains the zero vector. So a set not containing 0 (like the line x + y = 1) is not a subspace.
Intersection of subspaces is a subspace; union usually is not. W₁ + W₂ is a subspace.
Linear combination: α₁v₁ + α₂v₂ + … + αₙvₙ. Span L(S): set of all linear combinations of vectors of S; it is the smallest subspace containing S.
Linearly independent: α₁v₁ + … + αₙvₙ = 0 only when all αᵢ = 0. Otherwise dependent.
Any set containing the zero vector is dependent. In Rⁿ, n vectors are independent iff the determinant of their components is non-zero.
Basis: a linearly independent set that spans V. Dimension = number of vectors in any basis (all bases have the same size).
dim Rⁿ = n; dim Pₙ = n + 1; dim Mₘₓₙ = mn.
In an n-dimensional space, any n independent vectors form a basis; any n + 1 vectors are dependent.
dim(W₁ + W₂) = dim W₁ + dim W₂ − dim(W₁ ∩ W₂).
महत्त्वाच्या संज्ञा
Scalar – element of the field F (a number).
Standard basis of R³ – e₁ = (1, 0, 0), e₂ = (0, 1, 0), e₃ = (0, 0, 1).
Direct sum – V = W₁ ⊕ W₂ when V = W₁ + W₂ and W₁ ∩ W₂ = {0}.
Vector space (सदिश अवकाश) – set with vector addition and scalar multiplication obeying the axioms.
सूत्रे
Subspace test: αu + βv ∈ W
Independence test in Rⁿ: det[v₁ v₂ … vₙ] ≠ 0
dim(W₁ + W₂) = dim W₁ + dim W₂ − dim(W₁ ∩ W₂)
५ महत्त्वाचे प्रश्न
Q1. Define a subspace and give an example of a subset of R² that is not a subspace.
A non-empty subset W of V is a subspace if it is a vector space under the same operations (equivalently αu + βv ∈ W). The line {(x, y) : x + y = 1} is not a subspace because it does not contain (0, 0).
Q2. Show that any set containing the zero vector is linearly dependent.
If S = {0, v₂, …, vₙ}, then 1·0 + 0·v₂ + … + 0·vₙ = 0 with a non-zero coefficient (1). So S is linearly dependent.
Q3. Show that (1, 2, 1), (2, 1, 0), (1, −1, 2) form a basis of R³.
Three vectors in R³ (dimension 3): enough to check independence.
Determinant with these as rows: 1(1·2 − 0·(−1)) − 2(2·2 − 0·1) + 1(2·(−1) − 1·1).
= 1(2) − 2(4) + 1(−3) = 2 − 8 − 3 = −9.
Determinant ≠ 0, so the vectors are linearly independent.
Three independent vectors in R³ form a basis.
Q4. Show that W = {(x, y, z) : x + y + z = 0} is a subspace of R³ and find its dimension.
(0, 0, 0) ∈ W, so W is non-empty.
Let u = (x₁, y₁, z₁), v = (x₂, y₂, z₂) ∈ W. Then αu + βv has coordinate sum α(x₁ + y₁ + z₁) + β(x₂ + y₂ + z₂) = 0.
So αu + βv ∈ W: W is a subspace.
Any element: z = −x − y, so (x, y, −x − y) = x(1, 0, −1) + y(0, 1, −1).
(1, 0, −1) and (0, 1, −1) are independent and span W, so dim W = 2 (a plane through the origin).
Q5. If W₁ and W₂ are subspaces of V, prove that W₁ ∩ W₂ is a subspace.
0 ∈ W₁ and 0 ∈ W₂, so 0 ∈ W₁ ∩ W₂ (non-empty).
Let u, v ∈ W₁ ∩ W₂ and α, β scalars.
Since W₁ is a subspace, αu + βv ∈ W₁; since W₂ is a subspace, αu + βv ∈ W₂.
So αu + βv ∈ W₁ ∩ W₂.
Hence W₁ ∩ W₂ is a subspace.
परीक्षेसाठी टीप
For subspace questions, first check that the zero vector is in the set — if not, you are done in one line. For basis questions in Rⁿ, the determinant test is the fastest proof of independence.
A linear transformation is a function between vector spaces that preserves addition and scalar multiplication. Every linear transformation between finite-dimensional spaces can be represented by a matrix. This topic covers the definition and examples, kernel and range, the rank–nullity theorem, one-one and onto maps, isomorphism and the matrix of a transformation.
महत्त्वाचे मुद्दे
T: U → V is linear if T(u + v) = T(u) + T(v) and T(αu) = αT(u); equivalently T(αu + βv) = αT(u) + βT(v).
Every linear T maps 0 to 0. So T(x, y) = (x + 1, y) is not linear.
Examples: rotation and reflection in R², differentiation on polynomials, multiplication by a fixed matrix.
Kernel (null space): ker T = {u ∈ U : T(u) = 0}, a subspace of U. Its dimension is the nullity.
Range (image): R(T) = {T(u) : u ∈ U}, a subspace of V. Its dimension is the rank.
Rank–nullity theorem: rank T + nullity T = dim U.
T is one-one (injective) iff ker T = {0}.
T is onto (surjective) iff R(T) = V.
An isomorphism is a linear map that is one-one and onto. Two finite-dimensional spaces over the same field are isomorphic iff they have the same dimension.
A linear map is completely fixed by its values on a basis.
Matrix of T: write T of each basis vector of U as a combination of the basis of V; the coefficients form the columns of the matrix.
Composition of linear maps corresponds to multiplication of their matrices.
Eigenvalue of an operator: T(v) = λv for some non-zero v.
महत्त्वाच्या संज्ञा
Linear operator – linear map from a space to itself.
Linear transformation (रेखीय रूपांतरण) – map preserving vector operations.
सूत्रे
T(αu + βv) = αT(u) + βT(v)
dim U = rank(T) + nullity(T)
[T∘S] = [T][S] (matrices of compositions)
T one-one ⇔ ker T = {0}
५ महत्त्वाचे प्रश्न
Q1. Show that T(x, y) = (x + 1, y) is not a linear transformation.
T(0, 0) = (1, 0) ≠ (0, 0). A linear transformation must map the zero vector to the zero vector, so T is not linear.
Q2. State the rank–nullity theorem.
If T: U → V is a linear transformation and U is finite-dimensional, then rank(T) + nullity(T) = dim U, where rank is the dimension of the range and nullity the dimension of the kernel.
Q3. Show that T: R³ → R², T(x, y, z) = (x + y, y + z) is linear.
Let u = (x₁, y₁, z₁), v = (x₂, y₂, z₂), scalars α, β.
To prove linearity, use the single condition T(αu + βv) = αT(u) + βT(v). For the matrix, remember that images of basis vectors go into columns, not rows.
An inner product space is a vector space with an extra operation, the inner product, that generalises the dot product. It lets us measure length (norm), distance and angle, and talk about orthogonal vectors. The Cauchy–Schwarz inequality and the Gram–Schmidt process are the two most examined results.
महत्त्वाचे मुद्दे
An inner product on a real vector space V assigns a number ⟨u, v⟩ to each pair, such that: (i) ⟨u, u⟩ ≥ 0, and = 0 only for u = 0; (ii) ⟨u, v⟩ = ⟨v, u⟩; (iii) ⟨au + bv, w⟩ = a⟨u, w⟩ + b⟨v, w⟩. (Over C, (ii) becomes ⟨u, v⟩ = conjugate of ⟨v, u⟩.)
Standard examples: the dot product on Rⁿ; ⟨f, g⟩ = ∫ₐᵇ f(x)g(x) dx on continuous functions.
Norm: ‖u‖ = √⟨u, u⟩. Distance: d(u, v) = ‖u − v‖.
Cauchy–Schwarz inequality: |⟨u, v⟩| ≤ ‖u‖ ‖v‖, with equality iff u, v are linearly dependent.
Triangle inequality: ‖u + v‖ ≤ ‖u‖ + ‖v‖.
Angle: cos θ = ⟨u, v⟩/(‖u‖ ‖v‖). u and v are orthogonal if ⟨u, v⟩ = 0.
Pythagoras: if u ⊥ v then ‖u + v‖² = ‖u‖² + ‖v‖². Parallelogram law: ‖u + v‖² + ‖u − v‖² = 2‖u‖² + 2‖v‖².
An orthonormal set has unit vectors, each pair orthogonal. Orthogonal non-zero vectors are linearly independent.
Gram–Schmidt process turns any basis into an orthonormal basis.
Orthogonal complement W⊥ = {v : ⟨v, w⟩ = 0 for all w ∈ W}; for finite-dimensional V, V = W ⊕ W⊥.
महत्त्वाच्या संज्ञा
Norm (मानांक) – length of a vector.
Orthonormal basis – a basis of mutually perpendicular unit vectors.
For any vectors u, v in an inner product space, |⟨u, v⟩| ≤ ‖u‖ ‖v‖. Equality holds if and only if u and v are linearly dependent.
Q2. Prove that orthogonal non-zero vectors are linearly independent.
Let a₁v₁ + … + aₙvₙ = 0 with ⟨vᵢ, vⱼ⟩ = 0 for i ≠ j. Take the inner product with vₖ: aₖ⟨vₖ, vₖ⟩ = 0. Since vₖ ≠ 0, ⟨vₖ, vₖ⟩ > 0, so aₖ = 0 for every k.
Q3. Verify the Cauchy–Schwarz inequality for u = (1, 2, 2) and v = (3, 0, 4) in R³ and find the angle.
Q5. With ⟨f, g⟩ = ∫₀¹ f(x)g(x) dx, find ⟨x, x²⟩, ‖x‖ and check Cauchy–Schwarz.
⟨x, x²⟩ = ∫₀¹ x³ dx = ¼.
‖x‖² = ∫₀¹ x² dx = ⅓, so ‖x‖ = 1/√3.
‖x²‖² = ∫₀¹ x⁴ dx = ⅕, so ‖x²‖ = 1/√5.
‖x‖ ‖x²‖ = 1/√15 ≈ 0.258 ≥ 0.25 ✓.
परीक्षेसाठी टीप
In Gram–Schmidt problems, check orthogonality after each step before normalising — one arithmetic slip spoils every later vector. Keep fractions, not decimals.
A complex function f(z) = u(x, y) + iv(x, y) is analytic in a region if it has a derivative at every point there. The Cauchy–Riemann equations give the test for differentiability. This topic covers limits and derivatives of complex functions, the Cauchy–Riemann equations in Cartesian and polar forms, harmonic functions, harmonic conjugates and the Milne-Thomson method.
महत्त्वाचे मुद्दे
z = x + iy; a complex function w = f(z) = u(x, y) + iv(x, y), where u and v are real.
Derivative: f'(z) = lim (Δz → 0) [f(z + Δz) − f(z)]/Δz; the limit must be the same along every direction of approach.
Analytic (holomorphic) function: differentiable at every point of a neighbourhood. A function analytic in the whole plane is entire (polynomials, eᶻ, sin z).
Cauchy–Riemann (C–R) equations: uₓ = vᵧ and uᵧ = −vₓ. They are necessary for differentiability.
Sufficient condition: if uₓ, uᵧ, vₓ, vᵧ are continuous and satisfy the C–R equations, f is analytic.
f'(z) = uₓ + ivₓ = vᵧ − iuᵧ.
Polar form: uᵣ = (1/r)v_θ and vᵣ = −(1/r)u_θ.
f(z) = z̄ = x − iy is continuous everywhere but differentiable nowhere (uₓ = 1, vᵧ = −1).
If f is analytic, u and v are harmonic: ∇²u = uₓₓ + uᵧᵧ = 0 and ∇²v = 0.
v is a harmonic conjugate of u if f = u + iv is analytic.
The curves u = constant and v = constant cut orthogonally.
Milne-Thomson method: find f(z) from u alone: f'(z) = uₓ(z, 0) − iuᵧ(z, 0), then integrate.
महत्त्वाच्या संज्ञा
Singular point – point where f fails to be analytic.
Harmonic function (संनादी फलन) – real function satisfying Laplace's equation.
Entire function – analytic at every point of the complex plane.
Orthogonal trajectories – families of curves cutting at right angles.
सूत्रे
C–R (Cartesian): uₓ = vᵧ, uᵧ = −vₓ
C–R (polar): uᵣ = (1/r)v_θ, vᵣ = −(1/r)u_θ
f'(z) = uₓ + ivₓ
Laplace: uₓₓ + uᵧᵧ = 0
Milne-Thomson: f'(z) = uₓ(z, 0) − iuᵧ(z, 0)
Conjugate: dv = vₓ dx + vᵧ dy = −uᵧ dx + uₓ dy
५ महत्त्वाचे प्रश्न
Q1. State the Cauchy–Riemann equations.
If f(z) = u + iv is differentiable at z = x + iy, then uₓ = vᵧ and uᵧ = −vₓ at that point. In polar form: uᵣ = (1/r)v_θ, vᵣ = −(1/r)u_θ.
Q2. Show that f(z) = z̄ is not analytic anywhere.
u = x, v = −y. uₓ = 1, vᵧ = −1. Since uₓ ≠ vᵧ at every point, the C–R equations fail everywhere, so z̄ is nowhere differentiable and hence not analytic.
Q3. Verify the C–R equations for f(z) = z² and find f'(z).
z² = (x + iy)² = x² − y² + 2ixy; u = x² − y², v = 2xy.
uₓ = 2x, vᵧ = 2x → uₓ = vᵧ ✓.
uᵧ = −2y, vₓ = 2y → uᵧ = −vₓ ✓.
Partial derivatives are continuous, so f is analytic everywhere.
f'(z) = uₓ + ivₓ = 2x + 2iy = 2z.
Q4. Show that u = x³ − 3xy² is harmonic, find its harmonic conjugate and f(z).
Q5. Using the Milne-Thomson method, find the analytic function whose real part is u = x² − y² + 2x.
uₓ = 2x + 2, uᵧ = −2y.
Put x = z, y = 0: uₓ(z, 0) = 2z + 2, uᵧ(z, 0) = 0.
f'(z) = uₓ(z, 0) − iuᵧ(z, 0) = 2z + 2.
Integrate: f(z) = z² + 2z + ic.
Check: real part of z² + 2z is x² − y² + 2x ✓.
परीक्षेसाठी टीप
Always check both C–R equations and mention continuity of the partial derivatives to claim analyticity. In harmonic conjugate problems, verify your answer by checking that f(z) comes out as a function of z alone.
A complex function w = f(z) maps points of the z-plane to the w-plane. When f is analytic with f′(z) ≠ 0, the mapping is conformal: it preserves angles in size and sense. Bilinear (Möbius) transformations w = (az + b)/(cz + d) are the most important conformal maps; they send circles and lines to circles and lines.
महत्त्वाचे मुद्दे
A mapping is conformal at z₀ if the angle between any two curves through z₀ equals the angle between their images, both in size and direction.
If f is analytic and f′(z₀) ≠ 0, then w = f(z) is conformal at z₀. Points where f′(z) = 0 are critical points (angles are multiplied there).
Basic maps: translation w = z + c; rotation w = e^(iα)z; magnification w = kz (k > 0); inversion w = 1/z.
Bilinear transformation: w = (az + b)/(cz + d) with ad − bc ≠ 0. It is a combination of translation, rotation, magnification and inversion.
Its inverse is also bilinear: z = (−dw + b)/(cw − a).
A bilinear map sends circles and straight lines (a line = circle through ∞) to circles or straight lines.
Cross-ratio (z₁, z₂, z₃, z₄) = (z₁ − z₂)(z₃ − z₄)/((z₁ − z₄)(z₃ − z₂)) is unchanged by a bilinear map.
There is exactly one bilinear map sending three given points z₁, z₂, z₃ to three given points w₁, w₂, w₃.
Fixed (invariant) points: solve z = (az + b)/(cz + d), i.e. cz² + (d − a)z − b = 0. One repeated fixed point → parabolic transformation.
महत्त्वाच्या संज्ञा
Conformal mapping (संरूपी प्रतिचित्रण) – an angle-preserving map.
Möbius transformation – another name for a bilinear transformation.
Critical point – a point where f′(z) = 0.
सूत्रे
w = (az + b)/(cz + d), ad − bc ≠ 0
(w − w₁)(w₂ − w₃)/((w − w₃)(w₂ − w₁)) = (z − z₁)(z₂ − z₃)/((z − z₃)(z₂ − z₁))
Fixed points: cz² + (d − a)z − b = 0
५ महत्त्वाचे प्रश्न
Q1. When is w = f(z) conformal?
At every point where f is analytic and f′(z) ≠ 0. For example, w = z² is conformal everywhere except at z = 0, where angles are doubled.
Q2. Find the image of the circle |z| = 2 under w = 1/z.
|w| = 1/|z| = ½. So the image is the circle |w| = ½.
Q3. Find the bilinear map that sends z = 0, 1, ∞ to w = −1, 0, 1.
Let w = (az + b)/(cz + d).
z = ∞ → w = a/c = 1, so a = c.
z = 0 → w = b/d = −1, so b = −d.
z = 1 → w = (a + b)/(c + d) = 0, so b = −a, hence d = a.
With a = c = d = 1, b = −1: w = (z − 1)/(z + 1).
Check: 0 → −1 ✓, 1 → 0 ✓, ∞ → 1 ✓.
Q4. Find the fixed points of w = (3z − 4)/(z − 1). What type is it?
Put w = z: z(z − 1) = 3z − 4, so z² − 4z + 4 = 0.
(z − 2)² = 0, so z = 2 is the only (repeated) fixed point.
One repeated fixed point, so the transformation is parabolic.
Q5. Under w = 1/z, find the image of the line x = 1.
z = 1/w. With w = u + iv: z = (u − iv)/(u² + v²).
So x = u/(u² + v²).
x = 1 gives u² + v² = u, i.e. (u − ½)² + v² = ¼.
The image is the circle with centre (½, 0) and radius ½ (it passes through w = 0, the image of z = ∞).
परीक्षेसाठी टीप
For "find the bilinear map" questions, either use the cross-ratio formula or solve for a, b, c, d as above — and always check all three given points at the end.
Complex integration is integration of a complex function along a curve (contour) in the complex plane. For analytic functions it gives powerful results: Cauchy's theorem, Cauchy's integral formula, Taylor and Laurent series, and the residue theorem, which makes many difficult integrals easy.
महत्त्वाचे मुद्दे
Contour integral: ∫_C f(z) dz along a curve C; if z = z(t), a ≤ t ≤ b, then ∫_C f(z) dz = ∫ₐᵇ f(z(t)) z'(t) dt.
Cauchy's (Cauchy–Goursat) theorem: if f is analytic inside and on a simple closed curve C, then ∮_C f(z) dz = 0.
Consequence: for analytic f in a simply connected region, the integral between two points does not depend on the path.
Cauchy's integral formula: if f is analytic inside and on C and a is inside C, f(a) = (1/2πi)∮_C f(z)/(z − a) dz.
Derivatives: f⁽ⁿ⁾(a) = (n!/2πi)∮_C f(z)/(z − a)ⁿ⁺¹ dz. So an analytic function has derivatives of all orders.
Basic result: ∮_C dz/(z − a) = 2πi if a is inside C, and 0 if outside.
Taylor series: an analytic function can be expanded about a point inside its disc of analyticity.
Laurent series: about a singular point, f(z) = Σ aₙ(z − a)ⁿ + Σ bₙ/(z − a)ⁿ in an annulus.
Singularities: removable (no negative powers), pole of order m (finitely many negative powers, highest (z − a)⁻ᵐ), essential (infinitely many, e.g. e^(1/z) at 0).
Residue at a: the coefficient b₁ of 1/(z − a) in the Laurent series.
Residue theorem: ∮_C f(z) dz = 2πi × (sum of residues at poles inside C).
Uses: evaluating real integrals such as ∫₀^2π dθ/(a + b cos θ) and ∫₋∞^∞ dx/(1 + x²).
महत्त्वाच्या संज्ञा
Simple closed curve – a closed curve that does not cross itself.
Pole (ध्रुव) – isolated singularity where |f| → ∞ in a definite way.
Residue – coefficient of 1/(z − a) in the Laurent expansion.
Positive direction – anticlockwise around the contour.
सूत्रे
f(a) = (1/2πi)∮ f(z)/(z − a) dz
f⁽ⁿ⁾(a) = (n!/2πi)∮ f(z)/(z − a)ⁿ⁺¹ dz
Residue at a simple pole: Res = lim (z → a) (z − a)f(z); for f = φ/ψ with ψ(a) = 0: Res = φ(a)/ψ'(a)
Residue at a pole of order m: Res = (1/(m − 1)!) lim d^(m−1)/dz^(m−1) [(z − a)ᵐ f(z)]
Residue theorem: ∮_C f(z) dz = 2πi Σ Res
५ महत्त्वाचे प्रश्न
Q1. State Cauchy's integral theorem.
If f(z) is analytic inside and on a simple closed contour C (with f' continuous, in the classical form), then ∮_C f(z) dz = 0.
Q2. Classify the singularity of f(z) = sin z/z at z = 0.
sin z/z = (z − z³/3! + …)/z = 1 − z²/3! + z⁴/5! − … There are no negative powers of z, so z = 0 is a removable singularity.
Q3. Evaluate ∮_C eᶻ/(z − 1) dz, where C is the circle |z| = 2.
f(z) = eᶻ is analytic everywhere; a = 1 lies inside |z| = 2.
Cauchy's integral formula: ∮ f(z)/(z − a) dz = 2πi f(a).
= 2πi e¹ = 2πie.
Q4. Evaluate ∮_C z/((z − 1)(z − 2)) dz, where C is |z| = 3, using the residue theorem.
Simple poles at z = 1 and z = 2, both inside |z| = 3.
Res at 1: lim (z → 1) z/(z − 2) = 1/(1 − 2) = −1.
Res at 2: lim (z → 2) z/(z − 1) = 2/1 = 2.
Sum of residues = −1 + 2 = 1.
∮ = 2πi × 1 = 2πi.
Q5. Evaluate ∮_C sin z/(z − π/2)³ dz, where C is |z| = 2.
a = π/2 ≈ 1.57 lies inside |z| = 2; f(z) = sin z is analytic.
Use the derivative formula with n = 2: ∮ f(z)/(z − a)³ dz = (2πi/2!) f''(a).
f''(z) = −sin z, so f''(π/2) = −1.
∮ = πi × (−1) = −πi.
परीक्षेसाठी टीप
Before using any formula, check which singular points lie inside the contour — mark them on a rough sketch. Points outside contribute nothing.
Cauchy's residue theorem gives a powerful way to evaluate real definite integrals that are hard by ordinary calculus. We change the real integral into a contour integral, find the poles inside the contour, add their residues and multiply by 2πi. Three standard types are asked in exams.
महत्त्वाचे मुद्दे
Residue theorem: if f is analytic inside and on a closed curve C except at poles a₁, …, aₙ inside C, then ∮_C f(z) dz = 2πi × (sum of residues at a₁, …, aₙ).
Residue at a simple pole a: lim (z − a)f(z); for f = P/Q with Q(a) = 0, Q′(a) ≠ 0, the residue is P(a)/Q′(a).
Residue at a pole of order m: (1/(m − 1)!) × lim d^(m−1)/dz^(m−1) [(z − a)ᵐ f(z)].
Type 1: ∫₀^(2π) F(cos θ, sin θ) dθ. Put z = e^(iθ): cos θ = (z + 1/z)/2, sin θ = (z − 1/z)/(2i), dθ = dz/(iz). The contour is |z| = 1.
Type 2: ∫₋∞^∞ P(x)/Q(x) dx where deg Q ≥ deg P + 2 and Q has no real zeros. Integrate over a large semicircle in the upper half plane; the arc integral tends to 0. Answer = 2πi × (sum of residues in the upper half plane).
Type 3: ∫₋∞^∞ (P/Q) cos mx dx or sin mx dx (m > 0). Use f(z) = (P/Q)e^(imz); by Jordan's lemma the arc part → 0. Take the real part for cos and the imaginary part for sin.
Only poles inside the contour (inside |z| = 1, or in the upper half plane) are counted.
महत्त्वाच्या संज्ञा
Residue (अवशेष) – the coefficient of 1/(z − a) in the Laurent series about a.
Contour – a closed piecewise-smooth path.
Jordan's lemma – shows the arc integral of e^(imz)g(z) vanishes when g → 0.
सूत्रे
∫₀^(2π) dθ/(a + b cos θ) = 2π/√(a² − b²), a > |b|
∫₋∞^∞ dx/(x² + a²) = π/a
Residue of P/Q at a simple pole a = P(a)/Q′(a)
५ महत्त्वाचे प्रश्न
Q1. State Cauchy's residue theorem.
If f(z) is analytic inside and on a simple closed curve C except at a finite number of poles inside C, then ∮_C f(z) dz = 2πi times the sum of the residues of f at those poles.
Q2. Evaluate ∫₋∞^∞ dx/(1 + x²) using residues.
f(z) = 1/(1 + z²) has poles ±i; only z = i is in the upper half plane. Residue = 1/(2i). Integral = 2πi × 1/(2i) = π.
Q3. Evaluate ∫₀^(2π) dθ/(2 + cos θ).
Put z = e^(iθ): integral = ∮ [1/(2 + (z + 1/z)/2)] dz/(iz) = (2/i)∮ dz/(z² + 4z + 1).
Poles: z = −2 ± √3. Only α = −2 + √3 ≈ −0.27 lies inside |z| = 1.
Residue at α = 1/(2α + 4) = 1/(2√3).
Integral = (2/i) × 2πi × 1/(2√3) = 2π/√3.
Check with the formula 2π/√(a² − b²) = 2π/√(4 − 1) = 2π/√3 ✓.
Q4. Evaluate ∫₋∞^∞ dx/((x² + 1)(x² + 4)).
Poles in the upper half plane: z = i and z = 2i (both simple).
Residue at i: 1/(2i × (−1 + 4)) = 1/(6i).
Residue at 2i: 1/((−4 + 1) × 4i) = −1/(12i).
Sum = 2/(12i) − 1/(12i) = 1/(12i).
Integral = 2πi × 1/(12i) = π/6.
Q5. Evaluate ∫₋∞^∞ cos x/(x² + 1) dx.
Take f(z) = e^(iz)/(z² + 1). Upper-half-plane pole: z = i.
Real part: ∫ cos x/(x² + 1) dx = π/e. (Imaginary part gives ∫ sin x/(x² + 1) dx = 0, as expected for an odd integrand.)
परीक्षेसाठी टीप
Always list all poles first, then clearly mark which lie inside the contour — including an outside pole is the most common mistake. Check a Type 1 answer with the a + b cos θ formula when it applies.
This topic compares the "sizes" of infinite sets and studies the basic topology of the real line. A set is countable if its elements can be listed as a sequence. N, Z and Q are countable, but R is not. We also study neighbourhoods, interior and limit points, open and closed sets, and the Bolzano–Weierstrass and Heine–Borel theorems on R.
महत्त्वाचे मुद्दे
A set A is countable if it is finite or there is a one-one onto map f : N → A (then A is countably infinite, or denumerable). Otherwise A is uncountable.
Every subset of a countable set is countable. A countable union of countable sets is countable. N × N is countable.
Z and Q are countable; R, the interval (0, 1) and the set of irrational numbers are uncountable (Cantor's diagonal argument).
The set of algebraic numbers is countable, so "most" real numbers are transcendental.
Interior point: x is interior to A if some interval (x − ε, x + ε) ⊆ A. A is open if every point of A is interior.
Limit (accumulation) point: every neighbourhood of x contains a point of A other than x. The set of limit points is the derived set A′.
A is closed if it contains all its limit points, i.e. A′ ⊆ A. Equivalently, its complement is open. Closure Ā = A ∪ A′.
Any union of open sets is open; a finite intersection of open sets is open. ∩(−1/n, 1/n) = {0} shows an infinite intersection may not be open.
Bolzano–Weierstrass theorem: every bounded infinite subset of R has at least one limit point.
Heine–Borel theorem: a subset of R is compact (every open cover has a finite subcover) iff it is closed and bounded.
महत्त्वाच्या संज्ञा
Countable set (गणनीय संच) – a set that can be listed as a₁, a₂, a₃, …
Neighbourhood – a set containing an open interval around the point.
Isolated point – a point of A that is not a limit point of A.
सूत्रे
Open: every x ∈ A has (x − ε, x + ε) ⊆ A
Closed: A′ ⊆ A; Ā = A ∪ A′
Q̄ = R, Q′ = R
५ महत्त्वाचे प्रश्न
Q1. Define a countable set and give one countable and one uncountable example.
A set is countable if it is finite or in one-one correspondence with N. Example: Z is countable; the interval (0, 1) is uncountable.
Q2. Find the limit points of A = {1/n : n ∈ N}. Is A open or closed?
The only limit point is 0. Since 0 ∉ A, A is not closed. A is not open either, because no interval around 1 lies inside A. So A is neither open nor closed; Ā = A ∪ {0}.
Q3. Show that Z is countable.
Define f : N → Z by f(n) = n/2 if n is even and f(n) = −(n − 1)/2 if n is odd.
This lists Z as 0, 1, −1, 2, −2, 3, … (f(1) = 0, f(2) = 1, f(3) = −1, …).
Every integer appears exactly once, so f is one-one and onto.
Hence Z is countable.
Q4. Outline Cantor's proof that (0, 1) is uncountable.
Suppose (0, 1) is countable: x₁, x₂, x₃, … with decimal expansions xₙ = 0.aₙ₁aₙ₂aₙ₃…
Build y = 0.b₁b₂b₃… where bₙ = 5 if aₙₙ ≠ 5, and bₙ = 4 if aₙₙ = 5.
Then y differs from each xₙ in the n-th digit, so y is not in the list.
But y ∈ (0, 1), a contradiction. Hence (0, 1) is uncountable, and so is R.
Q5. Classify as open, closed, both or neither: (a, b), [a, b], [a, b), R, Q.
(a, b): open (each point has a small interval inside); not closed (a is a limit point outside it).
[a, b]: closed (complement (−∞, a) ∪ (b, ∞) is open); not open (no interval around a lies inside).
[a, b): neither.
R: both open and closed (so is ∅).
Q: neither — every interval contains irrationals, and every real number is a limit point of Q.
परीक्षेसाठी टीप
For "open/closed" questions, test the end points: an end point inside the set spoils openness, an end point outside spoils closedness. Write the explicit listing map when proving a set countable.
A metric space is a set with a distance function (metric) that behaves like ordinary distance. It generalises ideas from real analysis — open and closed sets, convergence, completeness, continuity and compactness — to many settings. This topic covers the definition, standard examples, open balls and open/closed sets, sequences and completeness, and continuity.
महत्त्वाचे मुद्दे
A metric on X is a function d: X × X → R such that: (1) d(x, y) ≥ 0; (2) d(x, y) = 0 iff x = y; (3) d(x, y) = d(y, x); (4) d(x, z) ≤ d(x, y) + d(y, z) (triangle inequality).
Examples: usual metric on R, d(x, y) = |x − y|; Euclidean metric on R², √((x₁ − y₁)² + (x₂ − y₂)²); taxicab metric |x₁ − y₁| + |x₂ − y₂|; max metric max(|x₁ − y₁|, |x₂ − y₂|); discrete metric (0 if x = y, 1 otherwise).
A set G is open if every point of G has an open ball around it lying inside G. Every open ball is open.
A set F is closed if its complement is open; equivalently F contains all its limit points.
In the discrete metric every set is both open and closed.
Arbitrary unions and finite intersections of open sets are open.
Convergence: xₙ → x if d(xₙ, x) → 0. Limits are unique.
Cauchy sequence: d(xₘ, xₙ) → 0 as m, n → ∞. Every convergent sequence is Cauchy.
Complete metric space: every Cauchy sequence converges in it. R and Rⁿ are complete; Q is not (a sequence of rationals can converge to √2).
Continuity: f is continuous at a if for each ε > 0 there is δ > 0 with d(x, a) < δ ⇒ d(f(x), f(a)) < ε. Equivalently, the inverse image of every open set is open.
Compactness: every open cover has a finite subcover. In Rⁿ, a set is compact iff closed and bounded (Heine–Borel).
Banach's contraction principle: a contraction on a complete metric space has a unique fixed point.
महत्त्वाच्या संज्ञा
Limit point – every ball around it contains a point of the set other than itself.
Closure Ā – A together with its limit points.
Bounded set – contained in some ball.
Metric (मापक/अंतर फलन) – distance function obeying the four axioms.
Q2. Is Q (rationals) with the usual metric complete? Why?
No. The sequence 1, 1.4, 1.41, 1.414, … (decimal approximations of √2) is a Cauchy sequence of rationals, but its limit √2 is not rational. So it does not converge in Q.
Q3. Prove that the discrete metric is a metric.
d(x, y) = 0 if x = y, 1 if x ≠ y. Clearly d ≥ 0 and d = 0 iff x = y.
d(x, y) = d(y, x), since x = y iff y = x.
Triangle inequality: if x = z, d(x, z) = 0 ≤ anything.
If x ≠ z, then y differs from at least one of x, z, so d(x, y) + d(y, z) ≥ 1 = d(x, z).
All four axioms hold.
Q4. Prove that every convergent sequence in a metric space is a Cauchy sequence.
Let xₙ → x. Given ε > 0, there is N with d(xₙ, x) < ε/2 for all n ≥ N.
For m, n ≥ N, by the triangle inequality: d(xₘ, xₙ) ≤ d(xₘ, x) + d(x, xₙ).
< ε/2 + ε/2 = ε.
So (xₙ) is Cauchy. (The converse fails in incomplete spaces like Q.)
Q5. Show that d'(x, y) = |x − y|/(1 + |x − y|) is a metric on R.
d' ≥ 0, d' = 0 iff x = y, and d' is symmetric — clear from |x − y|.
For the triangle inequality, f(t) = t/(1 + t) is increasing for t ≥ 0 (f'(t) = 1/(1 + t)² > 0).
Let a = |x − y|, b = |y − z|; then |x − z| ≤ a + b, so d'(x, z) ≤ f(a + b).
f(a + b) = a/(1 + a + b) + b/(1 + a + b) ≤ a/(1 + a) + b/(1 + b).
So d'(x, z) ≤ d'(x, y) + d'(y, z). Hence d' is a metric (bounded by 1).
परीक्षेसाठी टीप
Proving something is a metric means checking all four axioms in order — the triangle inequality is where the marks are. Learn the standard examples (Q incomplete, discrete metric) as ready counterexamples.
Continuity, uniform continuity and connectedness extend familiar ideas from R to any metric space. Continuous functions preserve compactness and connectedness, which gives general versions of the extreme value and intermediate value theorems. This topic builds on open sets, compactness and completeness studied earlier.
महत्त्वाचे मुद्दे
f : (X, d) → (Y, ρ) is continuous at a if for every ε > 0 there is δ > 0 with d(x, a) < δ ⇒ ρ(f(x), f(a)) < ε.
Sequential criterion: f is continuous at a iff xₙ → a implies f(xₙ) → f(a).
Open-set criterion: f is continuous on X iff f⁻¹(G) is open in X for every open G in Y (equivalently, inverse images of closed sets are closed).
f is uniformly continuous if one δ works for all points: d(x, y) < δ ⇒ ρ(f(x), f(y)) < ε. Uniform continuity implies continuity, not conversely.
A continuous function on a compact metric space is uniformly continuous, is bounded, and attains its maximum and minimum.
The continuous image of a compact set is compact.
X is connected if it cannot be written as A ∪ B with A, B non-empty, disjoint and open. Otherwise it is disconnected.
A subset of R is connected iff it is an interval.
The continuous image of a connected set is connected. This gives the intermediate value theorem.
A discrete metric space with more than one point is not connected.
महत्त्वाच्या संज्ञा
Uniform continuity (एकसमान सातत्य) – continuity with δ independent of the point.
Separation – a split of X into two disjoint non-empty open sets.
Homeomorphism – a continuous bijection with continuous inverse.
f : X → Y is uniformly continuous if for every ε > 0 there is a δ > 0 (depending only on ε) such that d(x, y) < δ implies ρ(f(x), f(y)) < ε for all x, y ∈ X.
Q2. Show that Q (with the usual metric) is not connected.
Take A = Q ∩ (−∞, √2) and B = Q ∩ (√2, ∞). Both are open in Q, non-empty, disjoint, and A ∪ B = Q because √2 ∉ Q. So Q is disconnected.
Q3. Prove that f(x) = x² is uniformly continuous on [0, 1].
For x, y ∈ [0, 1]: |x² − y²| = |x + y| |x − y| ≤ 2|x − y|.
Given ε > 0, choose δ = ε/2.
If |x − y| < δ then |x² − y²| < 2δ = ε.
δ does not depend on x or y, so f is uniformly continuous on [0, 1].
Q4. Show that f(x) = 1/x is not uniformly continuous on (0, 1).
Take xₙ = 1/n and yₙ = 1/(n + 1), both in (0, 1) for n ≥ 2.
|xₙ − yₙ| = 1/(n(n + 1)) → 0.
But |f(xₙ) − f(yₙ)| = |n − (n + 1)| = 1 for every n.
So for ε = 1/2 no δ can work. Hence f is not uniformly continuous (though it is continuous on (0, 1)).
Q5. Prove that the continuous image of a connected space is connected.
Let f : X → Y be continuous and onto, X connected. Suppose Y = A ∪ B is a separation.
Then X = f⁻¹(A) ∪ f⁻¹(B). Both sets are open (continuity), disjoint, and non-empty (f is onto).
This is a separation of X, contradicting connectedness.
Hence Y = f(X) is connected. On R this gives the intermediate value theorem.
परीक्षेसाठी टीप
To disprove uniform continuity, use two sequences that get closer while their images stay a fixed distance apart. To prove it, try to get a bound |f(x) − f(y)| ≤ K|x − y|.