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Complex analysis

Complex analysis is the calculus of functions of a complex variable $z = x + j y$ : which functions are differentiable throughout a region (analytic, tested with the Cauchy-Riemann equations), and what that analyticity gives for integration and series. It turns GATE questions on contour integrals, real definite integrals, harmonic conjugates and series expansions into short mechanical steps: find the singularities, decide which lie inside the contour or the region, and add up residues.

24 previous-year questions · 24 solved · 6 practice questions

Complex numbers, conventions and elementary functions

Conventions used in these notes. $j$ is the imaginary unit, $j^{2} = - 1$ . The principal argument $Arg z$ lies in $(- π, π]$ , and $Log z$ is the principal logarithm built on it. A closed contour $\oint_{C}$ is traversed counter-clockwise (the positive sense, with the enclosed region on the left) unless a question says otherwise.

A complex number $z = x + j y$ has real part $x = Re z$ , imaginary part $y = Im z$ (a real number), modulus $∣ z ∣ = r$ and argument $θ$ . Geometrically $z$ is the point $(x, y)$ of the complex plane, and $∣ z_{1} - z_{2} ∣$ is the distance between two points, which is how every circular contour in this topic is read.

CARTESIAN, POLAR AND EXPONENTIAL FORMS

z = x + j y = r (cos θ + j sin θ) = r e^{j θ}, r = ∣ z ∣ = x^{2} + y^{2}, e^{j θ} = cos θ + j sin θ

Conjugate $\overset{z}{ˉ} = x - j y = r e^{- j θ}$ , with $z \overset{z}{ˉ} = ∣ z ∣^{2}$ , $Re z = \frac{z + z ˉ}{2}$ and $Im z = \frac{z - z ˉ}{2 j}$ . The argument is fixed only up to adding $2 k π$ .

PRINCIPAL ARGUMENT

Arg z = ⎩ ⎨ ⎧ tan^{- 1} (y / x), tan^{- 1} (y / x) + π, tan^{- 1} (y / x) - π, π /2, - π /2, x > 0 x < 0, y \geq 0 x < 0, y < 0 x = 0, y > 0 x = 0, y < 0

Here $tan^{- 1}$ returns a value in $(- π /2, π /2)$ , and $Arg 0$ is undefined. Some texts take the principal range as $[0, 2 π)$ ; these notes use $(- π, π]$ throughout, so $Arg (- 1) = π$ .

PRODUCTS, QUOTIENTS AND THE TRIANGLE INEQUALITY

∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣ ∣ z_{2} ∣, \frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}, ar g (z_{1} z_{2}) = ar g z_{1} + ar g z_{2}, ar g \frac{z _{1}}{z _{2}} = ar g z_{1} - ar g z_{2}

The argument rules hold only up to multiples of $2 π$ : $Arg (z_{1} z_{2})$ can differ from $Arg z_{1} + Arg z_{2}$ by $\pm 2 π$ . Triangle inequality: $∣ z_{1} ∣ - ∣ z_{2} ∣ \leq ∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ .

DE MOIVRE'S THEOREM

(cos θ + j sin θ)^{n} = cos n θ + j sin n θ, (r e^{j θ})^{n} = r^{n} e^{j n θ}

For every integer $n$ . For a fractional power it gives only one of the values; use the roots formula below.

N-TH ROOTS

z^{1/ n} = r^{1/ n} e^{j (θ + 2 k π) / n}, k = 0, 1, \dots, n - 1

Every $z \neq = 0$ has exactly $n$ distinct $n$ -th roots. They lie on the circle of radius $r^{1/ n}$ (the positive real root), at angles spaced $2 π / n$ apart; for $n \geq 3$ they are the vertices of a regular $n$ -gon.

ROOTS OF UNITY

ω_{k} = e^{j 2 π k / n}, k = 0, 1, \dots, n - 1; k = 0 \sum n - 1 ω_{k} = 0 (n \geq 2), k = 0 \prod n - 1 ω_{k} = (- 1)^{n - 1}

With $ω = e^{j 2 π / n}$ the roots are $1, ω, ω^{2}, \dots, ω^{n - 1}$ and $ω^{n} = 1$ . Cube roots: $1 + ω + ω^{2} = 0$ with $ω = - \frac{1}{2} + j \frac{3}{2}$ .

EXPONENTIAL, TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS

e^{z} = e^{x} (cos y + j sin y), cos z = \frac{e ^{j z} + e ^{- j z}}{2}, sin z = \frac{e ^{j z} - e ^{- j z}}{2 j}, cos (j y) = cosh y, sin (j y) = j sinh y

$∣ e^{z} ∣ = e^{x}$ , and $y$ is one value of $ar g e^{z}$ ; $e^{z} \neq = 0$ for every $z$ , and $e^{z}$ has period $2 π j$ . $sin z$ and $cos z$ are entire but unbounded, so $∣ sin z ∣$ and $∣ cos z ∣$ can exceed 1, e.g. $cos (j) = cosh 1 \approx 1.543$ . The zeros of $sin z$ are $z = nπ$ and those of $cos z$ are $z = (n + \frac{1}{2}) π$ , with $n$ an integer; all of them are real.

COMPLEX LOGARITHM

ln z = ln ∣ z ∣ + j (Arg z + 2 k π), k \in Z; Log z = ln ∣ z ∣ + j Arg z

Defined for $z \neq = 0$ and multi-valued; $Log$ is the principal value ( $k = 0$ ). The principal branch is analytic everywhere except on its branch cut, the non-positive real axis (where $Arg$ jumps by $2 π$ ), and off the cut $(Log z)^{'} = 1/ z$ . Example: $Log (- 1) = j π$ . The rule $Log (z_{1} z_{2}) = Log z_{1} + Log z_{2}$ can fail by $\pm 2 π j$ .

COMPLEX POWERS

z^{c} = e^{c l n z}, principal value e^{c Log z}; j^{j} = e^{j \cdot j π /2} = e^{- π /2} \approx 0.208

Defined for $z \neq = 0$ . Multi-valued when $c$ is not an integer: for rational $c = p / q$ in lowest terms with $q \geq 2$ there are $q$ values; for irrational or non-real $c$ there are infinitely many. All values of $j^{j}$ are $e^{- π /2 - 2 k π}$ with $k \in Z$ , every one of them real.

Loci used to describe contours and regions

Equation or inequality	Set of points
$∣ z - a ∣ = ρ$	circle with centre $a$ and radius $ρ$
$∣ z - a ∣ < ρ$ ; $ρ_{1} < ∣ z - a ∣ < ρ_{2}$	open disc ; open annulus (ring) centred at $a$
$∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$	perpendicular bisector of the segment joining $z_{1}$ and $z_{2}$
$∣ z - z_{1} ∣ + ∣ z - z_{2} ∣ = 2 c$ with $2 c > ∣ z_{1} - z_{2} ∣$	ellipse with foci $z_{1}$ and $z_{2}$
$∣ z - z_{1} ∣ = k ∣ z - z_{2} ∣$ with $k > 0$ , $k \neq = 1$	a circle
$Re z > 0$ ; $Im z > 0$	right half-plane ; upper half-plane
point $z_{0}$ and circle $∣ z - a ∣ = ρ$	$z_{0}$ is inside when $∣ z_{0} - a ∣ < ρ$ , on the circle when $∣ z_{0} - a ∣ = ρ$ , outside when $∣ z_{0} - a ∣ > ρ$

WORKED EXAMPLE · ROOTS AND THE PRINCIPAL LOGARITHM

For $z = - 1 + j 3$ , find $Log z$ and all the cube roots of $z$ .

Solution

$∣ z ∣ = 1 + 3 = 2$ . The point is in the second quadrant, so $Arg z = tan^{- 1} (\frac{3}{- 1}) + π = - \frac{π}{3} + π = \frac{2 π}{3}$ .
$Log z = ln 2 + j \frac{2 π}{3} \approx 0.693 + j 2.094$ . All the logarithms are $ln 2 + j (\frac{2 π}{3} + 2 k π)$ .
Cube roots: $z^{1/3} = 2^{1/3} e^{j (2 π /3 + 2 k π) /3} = 2^{1/3} e^{j (2 π /9 + 2 k π /3)}$ for $k = 0, 1, 2$ .
The angles are $\frac{2 π}{9}$ , $\frac{8 π}{9}$ and $\frac{14 π}{9}$ (the last is $- \frac{4 π}{9}$ as a principal argument), all on the circle of radius $2^{1/3} \approx 1.26$ and $12 0^{\circ}$ apart.

Answer: $Log (- 1 + j 3) = ln 2 + j \frac{2 π}{3}$ ; the cube roots are $2^{1/3} e^{j 2 π /9}$ , $2^{1/3} e^{j 8 π /9}$ and $2^{1/3} e^{- j 4 π /9}$ .

EXAM TIP

Argument trap: $tan^{- 1} (y / x)$ alone is correct only for $x > 0$ . For $z = - 1 - j$ , $tan^{- 1} (1) = π /4$ , but the point is in the third quadrant and $Arg z = - 3 π /4$ .

EXAM TIP

$∣ e^{j θ} ∣ = 1$ only for real $θ$ . For complex $z$ , $∣ e^{z} ∣ = e^{Re z}$ , so $∣ e^{j z} ∣ = e^{- y}$ . This is what makes $e^{j z}$ small in the upper half-plane, which the Fourier-type integrals at the end of these notes rely on.

Analytic functions, the Cauchy-Riemann equations and conformal mapping

A function of a complex variable assigns $w = f (z) = u (x, y) + j v (x, y)$ to each $z = x + j y$ , with $u$ and $v$ real. $lim_{z \to z_{0}} f (z) = L$ means $f (z) \to L$ however $z$ approaches $z_{0}$ in the plane; if two paths give different values, the limit does not exist. For $f = \overset{z}{ˉ} / z$ at $z_{0} = 0$ , the real axis gives $1$ and the imaginary axis gives $- 1$ , so there is no limit. $f$ is continuous at $z_{0}$ if $lim_{z \to z_{0}} f (z) = f (z_{0})$ , which holds exactly when $u$ and $v$ are both continuous there.

DERIVATIVE

f^{'} (z_{0}) = Δ z \to 0 lim \frac{f ( z _{0} + Δ z ) - f ( z _{0} )}{Δ z}

The limit must be the same for every direction of approach of $Δ z$ , which is far more restrictive than real differentiability. The sum, product, quotient and chain rules work as in real calculus, and differentiability implies continuity.

Analytic (holomorphic) at $z_{0}$ : differentiable at every point of some neighbourhood of $z_{0}$ , not merely at $z_{0}$ itself. Analytic in a domain: analytic at every point of it. Entire: analytic in the whole finite plane, e.g. polynomials, $e^{z}$ , $sin z$ and $cos z$ . A point where $f$ is not analytic is a singular point. A rational function is analytic everywhere except at the zeros of its denominator.

CAUCHY-RIEMANN EQUATIONS (CARTESIAN)

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}; f^{'} (z) = u_{x} + j v_{x} = v_{y} - j u_{y}

Necessary at every point where $f$ is differentiable. Sufficient for differentiability at $z_{0}$ when, in addition, $u_{x}, u_{y}, v_{x}, v_{y}$ exist in a neighbourhood of $z_{0}$ and are continuous at $z_{0}$ . $f$ is analytic in a domain where these conditions hold at every point. Subscripts denote partial derivatives.

CAUCHY-RIEMANN EQUATIONS (POLAR)

\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial θ}, \frac{\partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial θ}; f^{'} (z) = e^{- j θ} (u_{r} + j v_{r})

For $z = r e^{j θ}$ with $r \neq = 0$ and $f = u (r, θ) + j v (r, θ)$ . Use it when $f$ is naturally written in $r$ and $θ$ : for $Log z = ln r + j θ$ with $- π < θ < π$ (off the branch cut), $u_{r} = 1/ r = v_{θ} / r$ and $v_{r} = 0 = - u_{θ} / r$ , and $f^{'} = e^{- j θ} / r = 1/ z$ .

Testing whether a given f(z) is analytic

Write $f = u + j v$ and compute $u_{x}$ , $u_{y}$ , $v_{x}$ , $v_{y}$ .
Solve $u_{x} = v_{y}$ and $u_{y} = - v_{x}$ together. The solution set is where $f$ can be differentiable.
If the equations hold at every point of an open region (and the partials are continuous, as they are for polynomials, exponentials and trigonometric functions of $x$ and $y$ ), $f$ is analytic there.
If they hold only on a curve or at isolated points, $f$ is differentiable at those points (given continuous partials) but analytic nowhere, because a curve or a point contains no neighbourhood.
Shortcut: substitute $x = \frac{z + z ˉ}{2}$ and $y = \frac{z - z ˉ}{2 j}$ . The CR equations say exactly that $\frac{\partial f}{\partial z ˉ} = 0$ (treating $z$ and $\overset{z}{ˉ}$ as independent), so $f$ is analytic in a region only if $\overset{z}{ˉ}$ drops out, and it is differentiable only at the points where $\frac{\partial f}{\partial z ˉ} = 0$ . For $∣ z ∣^{2} = z \overset{z}{ˉ}$ , $\frac{\partial f}{\partial z ˉ} = z$ , which vanishes only at $z = 0$ .

HARMONIC FUNCTIONS

\nabla^{2} u = \frac{\partial ^{2} u}{\partial x ^{2}} + \frac{\partial ^{2} u}{\partial y ^{2}} = 0, \nabla^{2} v = 0

If $f = u + j v$ is analytic in a domain, both $u$ and $v$ are harmonic there (the derivative of an analytic function is again analytic, so the second partials exist and are continuous). $v$ is called a harmonic conjugate of $u$ . Where $f^{'} (z) \neq = 0$ , the level curves $u = c_{1}$ and $v = c_{2}$ cut at right angles. In polar form Laplace's equation is $u_{r r} + \frac{1}{r} u_{r} + \frac{1}{r ^{2}} u_{θ θ} = 0$ .

Finding a harmonic conjugate v of a given u

Check $u_{xx} + u_{y y} = 0$ . If it fails, no harmonic conjugate exists and $u$ cannot be the real part of an analytic function. If it holds in a simply connected domain, a conjugate exists there; in a domain with a hole it may fail to be single-valued (on $z \neq = 0$ , $u = ln ∣ z ∣$ is harmonic but its conjugate would be $ar g z$ ).
From $v_{y} = u_{x}$ : $v = \int u_{x} d y + g (x)$ , integrating with $x$ held constant.
Differentiate with respect to $x$ and impose $v_{x} = - u_{y}$ . This gives $g^{'} (x)$ , which must come out free of $y$ , and hence $g (x)$ .
The conjugate is unique up to an added real constant. Note the asymmetry: a harmonic conjugate of $v$ is $- u$ , not $u$ .

MILNE-THOMSON CONSTRUCTION

f^{'} (z) = u_{x} (z, 0) - j u_{y} (z, 0) \Rightarrow f (z) = \int [u_{x} (z, 0) - j u_{y} (z, 0)] d z + C

Builds $f$ directly from $u$ without finding $v$ : compute $u_{x}$ and $u_{y}$ , put $x = z$ and $y = 0$ , and integrate in $z$ . If $v$ is given instead, $f^{'} (z) = v_{y} (z, 0) + j v_{x} (z, 0)$ . $C$ is a complex constant. When $u$ is given, $Re C$ is fixed by $u$ itself and $Im C$ stays arbitrary unless a value such as $f (0)$ is also given (and the other way round when $v$ is given).

WORKED EXAMPLE · HARMONIC CONJUGATE BY MILNE-THOMSON

Show that $u = e^{x} cos y + x y$ is harmonic, and find the analytic function $f = u + j v$ with $f (0) = 1$ . Hence write down $v$ .

Solution

$u_{x} = e^{x} cos y + y$ and $u_{xx} = e^{x} cos y$ ; $u_{y} = - e^{x} sin y + x$ and $u_{y y} = - e^{x} cos y$ . So $u_{xx} + u_{y y} = 0$ and $u$ is harmonic.
Put $x = z$ , $y = 0$ : $u_{x} (z, 0) = e^{z}$ and $u_{y} (z, 0) = z$ .
$f^{'} (z) = u_{x} (z, 0) - j u_{y} (z, 0) = e^{z} - j z$ , so $f (z) = e^{z} - \frac{j}{2} z^{2} + C$ .
$f (0) = 1 + C = 1$ gives $C = 0$ .
Check: $- \frac{j}{2} z^{2} = - \frac{j}{2} (x^{2} - y^{2} + j 2 x y) = x y - \frac{j}{2} (x^{2} - y^{2})$ , whose real part is $x y$ , as required.
$v = Im f = e^{x} sin y - \frac{x ^{2} - y ^{2}}{2}$ . Verify: $v_{y} = e^{x} cos y + y = u_{x}$ and $v_{x} = e^{x} sin y - x = - u_{y}$ .

Answer: $f (z) = e^{z} - \frac{j}{2} z^{2}$ and $v = e^{x} sin y - \frac{x ^{2} - y ^{2}}{2}$ .

Conformal mapping (in brief). An analytic $w = f (z)$ maps curves in the $z$ -plane to curves in the $w$ -plane. At a point where $f^{'} (z_{0}) \neq = 0$ the map is conformal: the angle between any two curves through $z_{0}$ is preserved in size and in sense, and a small neighbourhood is magnified by $∣ f^{'} (z_{0}) ∣$ and rotated by $ar g f^{'} (z_{0})$ . At a critical point, where $f^{'} (z_{0}) = 0$ , angles are not preserved: if $f^{'} (z_{0}) = \dots = f^{(m - 1)} (z_{0}) = 0 \neq = f^{(m)} (z_{0})$ , angles at $z_{0}$ are multiplied by $m$ .

BILINEAR (MOBIUS) TRANSFORMATION

w = \frac{a z + b}{cz + d}, a d - b c \neq = 0

Conformal wherever $cz + d \neq = 0$ , since $w^{'} (z) = \frac{a d - b c}{( cz + d ) ^{2}}$ . For $c \neq = 0$ , $z = - d / c$ maps to $w = \infty$ and $z = \infty$ maps to $w = a / c$ . It maps circles and straight lines (a line counts as a circle through $\infty$ ) onto circles and straight lines, and it is fixed completely by the images of three distinct points. Its inverse $z = \frac{d w - b}{- c w + a}$ is also bilinear.

FIXED POINTS AND CROSS-RATIO

fixed points: c z^{2} + (d - a) z - b = 0; \frac{( w - w _{1} ) ( w _{2} - w _{3} )}{( w - w _{3} ) ( w _{2} - w _{1} )} = \frac{( z - z _{1} ) ( z _{2} - z _{3} )}{( z - z _{3} ) ( z _{2} - z _{1} )}

Fixed points solve $z = \frac{a z + b}{cz + d}$ ; there are at most two unless the map is the identity (for $c = 0$ and $a \neq = d$ there is one finite fixed point, and $\infty$ is also fixed). The cross-ratio equation gives the unique map sending $z_{1}, z_{2}, z_{3}$ to $w_{1}, w_{2}, w_{3}$ . If one of the given points is $\infty$ , delete the two factors that contain it.

Standard mappings

Map	Effect
$w = z + b$	translation by $b$
$w = a z$ with $a = ∣ a ∣ e^{j α}$	rotation by $α$ and magnification by $∣ a ∣$ about the origin
$w = 1/ z$	inversion: $∣ w ∣ = 1/∣ z ∣$ and $ar g w = - ar g z$ ; maps $∣ z ∣ = 1$ onto $∣ w ∣ = 1$ and $0 < ∣ z ∣ < 1$ onto $∣ w ∣ > 1$ , with $z = 0 \mapsto w = \infty$
$w = z^{2}$	doubles arguments: the first quadrant maps onto the upper half-plane; not conformal at $z = 0$
$w = e^{z}$	the strip $0 < Im z < π$ onto the upper half-plane; vertical lines to circles $∣ w ∣ = e^{x}$ , horizontal lines to rays from the origin

WORKED EXAMPLE · IMAGE OF THE UNIT CIRCLE UNDER A BILINEAR MAP

Find the images of the circle $∣ z ∣ = 1$ and of the disc $∣ z ∣ < 1$ under $w = \frac{z + 1}{z - 1}$ .

Solution

$z = 1$ lies on the circle and maps to $w = \infty$ , so the image of the circle is a straight line.
Two more points: $z = - 1 \mapsto w = 0$ , and $z = j \mapsto w = \frac{1 + j}{- 1 + j} = \frac{( 1 + j ) ( - 1 - j )}{( - 1 + j ) ( - 1 - j )} = \frac{- 2 j}{2} = - j$ .
The line through $0$ and $- j$ is the imaginary axis $Re w = 0$ .
The centre $z = 0$ maps to $w = - 1$ , which has $Re w < 0$ , so the disc maps onto the left half-plane.
Algebraic check: $w = \frac{( z + 1 ) ( z ˉ - 1 )}{∣ z - 1 ∣ ^{2}}$ gives $Re w = \frac{∣ z ∣ ^{2} - 1}{∣ z - 1 ∣ ^{2}}$ , which is $0$ on $∣ z ∣ = 1$ and negative for $∣ z ∣ < 1$ .

Answer: The circle $∣ z ∣ = 1$ (without the point $z = 1$ ) maps onto the imaginary axis, and the disc $∣ z ∣ < 1$ maps onto the half-plane $Re w < 0$ .

EXAM TIP

The CR equations at a single point do not make a function analytic. $f = ∣ z ∣^{2} = x^{2} + y^{2}$ satisfies them only at $z = 0$ , so it is differentiable there and analytic nowhere. Likewise $\overset{z}{ˉ}$ , $Re z$ and $Im z$ are nowhere analytic.

EXAM TIP

If $f$ is analytic in a domain and any one of $Re f$ , $Im f$ or $∣ f ∣$ is constant there, or $\overset{ˉ}{f}$ is also analytic, then $f$ is constant.

EXAM TIP

If $v$ is a harmonic conjugate of $u$ , then $uv$ and $u^{2} - v^{2}$ are harmonic (they are $\frac{1}{2} Im f^{2}$ and $Re f^{2}$ ), but $u^{2}$ alone is harmonic only when $u$ is constant, because $\nabla^{2} (u^{2}) = 2∣\nabla u ∣^{2}$ .

Complex integration: Cauchy's integral theorem and integral formula

A contour is a piecewise smooth curve; a simple closed contour does not cross itself. Integrals along a contour are computed by parametrising it. Orientation convention: $\oint_{C}$ means counter-clockwise (positive) unless stated, and reversing the direction changes the sign of the integral. In general the value depends on the path, unless the integrand is analytic in a simply connected domain containing the path.

CONTOUR INTEGRAL BY PARAMETRISATION

\int_{C} f (z) d z = \int_{a}^{b} f (z (t)) z^{'} (t) d t = \int_{C} (u d x - v d y) + j \int_{C} (v d x + u d y)

$z = z (t)$ , $a \leq t \leq b$ , traces $C$ in the given direction. Circle $∣ z - z_{0} ∣ = ρ$ counter-clockwise: $z = z_{0} + ρ e^{j θ}$ , $d z = j ρ e^{j θ} d θ$ , $0 \leq θ \leq 2 π$ . Segment from $z_{1}$ to $z_{2}$ : $z = z_{1} + t (z_{2} - z_{1})$ , $0 \leq t \leq 1$ .

THE BASIC CIRCLE INTEGRAL

\oint_{∣ z - z_{0} ∣ = ρ} (z - z_{0})^{n} d z = {2 π j, 0, n = - 1 n \neq = - 1 (n an integer)

Counter-clockwise, any radius $ρ > 0$ . Every term of a Laurent series integrates to zero except the $1/ (z - z_{0})$ term, which is why residues work.

ML-INEQUALITY

\int_{C} f (z) d z \leq M L, M = z \in C max ∣ f (z) ∣, L = length of C

Used to bound an integral, and to show that the integral over a large semicircular arc tends to zero when real integrals are evaluated by residues.

CAUCHY'S INTEGRAL THEOREM

\oint_{C} f (z) d z = 0

Conditions: $C$ is a simple closed contour and $f$ is analytic at every point on $C$ and inside it (more generally, in a simply connected domain containing $C$ ). Analyticity on $C$ alone is not enough: $1/ z$ is analytic on $∣ z ∣ = 1$ , yet $\oint_{∣ z ∣ = 1} d z / z = 2 π j$ .

Consequences. (1) Path independence: if $f$ is analytic in a simply connected domain $D$ , then $\int_{z_{1}}^{z_{2}} f (z) d z$ is the same for every path in $D$ and equals $F (z_{2}) - F (z_{1})$ for any antiderivative $F^{'} = f$ . (2) Deformation of contours: if $f$ is analytic in the ring-shaped region between two simple closed contours $C_{1}$ (outer) and $C_{2}$ (inner), and on both, then $\oint_{C_{1}} f d z = \oint_{C_{2}} f d z$ with both taken counter-clockwise; a contour may be shrunk or stretched freely as long as it never crosses a singularity. (3) For a region with holes, the integral over the outer boundary equals the sum of the integrals over the inner boundaries, all counter-clockwise.

CAUCHY'S INTEGRAL FORMULA

f (z_{0}) = \frac{1}{2 π j} \oint_{C} \frac{f ( z )}{z - z _{0}} d z ⟺ \oint_{C} \frac{f ( z )}{z - z _{0}} d z = 2 π j f (z_{0})

$f$ analytic on and inside the simple closed contour $C$ (counter-clockwise), and $z_{0}$ strictly inside $C$ . If $z_{0}$ is outside $C$ the integral is $0$ by Cauchy's theorem; if $z_{0}$ lies on $C$ the formula does not apply (in general the integral then does not exist as an ordinary integral).

CAUCHY'S INTEGRAL FORMULA FOR DERIVATIVES

f^{(n)} (z_{0}) = \frac{n !}{2 π j} \oint_{C} \frac{f ( z )}{( z - z _{0} ) ^{n + 1}} d z ⟺ \oint_{C} \frac{f ( z )}{( z - z _{0} ) ^{n + 1}} d z = \frac{2 π j}{n !} f^{(n)} (z_{0})

Same conditions, $n = 0, 1, 2, \dots$ It shows that an analytic function has derivatives of every order, all analytic. Mind the power $n + 1$ : for $(z - z_{0})^{3}$ in the denominator use $f^{''} (z_{0})$ and divide by $2!$ .

CAUCHY'S INEQUALITY AND ITS CONSEQUENCES

∣ f^{(n)} (z_{0}) ∣ \leq \frac{n ! M}{ρ ^{n}}, M = ∣ z - z_{0} ∣ = ρ max ∣ f (z) ∣

For $f$ analytic on and inside $∣ z - z_{0} ∣ = ρ$ . Consequences: a bounded entire function is constant (Liouville's theorem), which is why every non-constant polynomial has a zero; and a non-constant function analytic inside a bounded region and continuous up to its boundary attains its maximum modulus only on the boundary (maximum modulus principle).

Evaluating a closed contour integral of a quotient

Find every singularity: the zeros of the denominator, and any other point where the integrand is not analytic.
Test each one against the contour. For $∣ z - a ∣ = ρ$ , a point $z_{1}$ is inside when $∣ z_{1} - a ∣ < ρ$ . Discard those outside.
No singularity inside: the integral is $0$ by Cauchy's theorem.
One singularity $z_{0}$ inside, appearing as $(z - z_{0})^{n + 1}$ in the denominator: write the integrand as $\frac{φ ( z )}{( z - z _{0} ) ^{n + 1}}$ , where $φ$ collects everything else and is analytic inside $C$ , and apply the (derivative) formula to $φ$ .
Several singularities inside: split by partial fractions, or surround each with its own small circle (deformation) and add, or go straight to the residue theorem.
Clockwise contour: change the sign at the end.

WORKED EXAMPLE · INTEGRAL FORMULA WITH A DOUBLE POLE

Evaluate $I = \oint_{C} \frac{e ^{2 z}}{( z - 1 ) ^{2} ( z + 3 )} d z$ , where $C$ is the circle $∣ z ∣ = 2$ traversed counter-clockwise.

Solution

Singularities: $z = 1$ (double) and $z = - 3$ . Since $∣1∣ = 1 < 2$ and $∣ - 3∣ = 3 > 2$ , only $z = 1$ is inside $C$ .
Take $φ (z) = \frac{e ^{2 z}}{z + 3}$ , analytic on and inside $C$ . Then $I = \oint_{C} \frac{φ ( z )}{( z - 1 ) ^{2}} d z = \frac{2 π j}{1 !} φ^{'} (1)$ .
$φ^{'} (z) = \frac{2 e ^{2 z} ( z + 3 ) - e ^{2 z}}{( z + 3 ) ^{2}} = \frac{e ^{2 z} ( 2 z + 5 )}{( z + 3 ) ^{2}}$ , so $φ^{'} (1) = \frac{7 e ^{2}}{16}$ .
$I = 2 π j \cdot \frac{7 e ^{2}}{16} = j \frac{7 π e ^{2}}{8} \approx 20.31 j$ .

Answer: $I = j \frac{7 π e ^{2}}{8} \approx 20.31 j$

EXAM TIP

Non-analytic integrands cannot use Cauchy's theorem or formula; parametrise instead. On $∣ z ∣ = ρ$ you may replace $\overset{z}{ˉ}$ by $ρ^{2} / z$ : $\oint_{∣ z ∣ = ρ} \overset{z}{ˉ} d z = \oint_{∣ z ∣ = ρ} \frac{ρ ^{2}}{z} d z = 2 π j ρ^{2}$ , not $0$ .

EXAM TIP

For an analytic integrand, an open-path integral is just the antiderivative evaluated at the ends: $\int_{0}^{1 + j} z^{2} d z = \frac{( 1 + j ) ^{3}}{3} = \frac{- 2 + 2 j}{3}$ along every path. For $\int \overset{z}{ˉ} d z$ between the same two points the answer depends on the path, so the path must be given and used.

EXAM TIP

Read the contour carefully: $∣ z - 1∣ = 2$ encloses $z = 0$ and $z = 2.5$ but not $z = - 2$ (since $∣ - 2 - 1∣ = 3$ ). A factor such as $z^{2} + 1$ contributes the two points $\pm j$ ; here $∣ \pm j - 1∣ = 2 < 2$ , so both are inside.

Sequences, series and convergence tests

A sequence $z_{n} = x_{n} + j y_{n}$ converges to $L = a + j b$ exactly when $x_{n} \to a$ and $y_{n} \to b$ . A monotone real sequence converges if and only if it is bounded. A series $\sum z_{n}$ converges when its partial sums $S_{N} = \sum_{n = 1}^{N} z_{n}$ converge, and its sum is $lim S_{N}$ . A series converges absolutely if $\sum ∣ z_{n} ∣$ converges; absolute convergence implies convergence, and the ratio and root tests below prove absolute convergence. A convergent series that is not absolutely convergent is conditionally convergent, e.g. $\sum_{n = 1}^{\infty} (- 1)^{n + 1} / n = ln 2$ .

N-TH TERM TEST

\sum a_{n} converges \Rightarrow n \to \infty lim a_{n} = 0

Only a test for divergence: if $lim a_{n} \neq = 0$ , or the limit does not exist, the series diverges. The converse is false: $\sum 1/ n$ diverges although $1/ n \to 0$ .

GEOMETRIC AND P-SERIES BENCHMARKS

n = 0 \sum \infty r^{n} = \frac{1}{1 - r} (∣ r ∣ < 1), n = 1 \sum \infty \frac{1}{n ^{p}} converges ⟺ p > 1

The geometric series (real or complex $r$ ) diverges for $∣ r ∣ \geq 1$ . $\sum 1/ n$ ( $p = 1$ ) diverges, while $\sum 1/ n^{2} = π^{2} /6$ . These two families are the usual benchmarks for the comparison tests.

Convergence tests for a series of terms a_n

Test	Condition	Conclusion
Comparison	$0 \leq a_{n} \leq b_{n}$ for all large $n$	$\sum b_{n}$ converges $\Rightarrow$ $\sum a_{n}$ converges; $\sum a_{n}$ diverges $\Rightarrow$ $\sum b_{n}$ diverges
Limit comparison	$a_{n}, b_{n} > 0$ and $a_{n} / b_{n} \to c$ with $0 < c < \infty$	both converge or both diverge
Ratio	$L = lim ∣ a_{n + 1} / a_{n} ∣$	$L < 1$ : converges absolutely; $L > 1$ (or $\infty$ ): diverges; $L = 1$ : no conclusion
Root	$L = lim ∣ a_{n} ∣^{1/ n}$	the same three cases as the ratio test
Integral	$a_{n} = f (n)$ with $f$ positive, continuous and decreasing for $x \geq 1$	$\sum a_{n}$ and $\int_{1}^{\infty} f (x) d x$ converge or diverge together
Alternating series	$\sum (- 1)^{n + 1} b_{n}$ with $b_{n} \geq 0$ decreasing and $b_{n} \to 0$	converges; the error after $N$ terms is at most $b_{N + 1}$

LIMITS USED WITH THE RATIO AND ROOT TESTS

n \to \infty lim n^{1/ n} = 1, n \to \infty lim (1 + \frac{1}{n})^{n} = e, n \to \infty lim \frac{( n ! ) ^{1/ n}}{n} = \frac{1}{e}

Handy when $a_{n}$ contains $n^{k}$ , $n^{n}$ or $n!$ . For fixed $k$ and $a > 1$ , $n^{k}$ grows more slowly than $a^{n}$ , which grows more slowly than $n!$ , which grows more slowly than $n^{n}$ .

RADIUS OF CONVERGENCE OF A POWER SERIES

n = 0 \sum \infty a_{n} (z - z_{0})^{n} : \frac{1}{R} = n \to \infty lim \frac{a _{n + 1}}{a _{n}} or \frac{1}{R} = n \to \infty lim ∣ a_{n} ∣^{1/ n}

Use whichever limit exists (the ratio form needs $a_{n} \neq = 0$ for all large $n$ ); $R = \infty$ if the limit is $0$ and $R = 0$ if it is $\infty$ ; in general $1/ R = lim sup ∣ a_{n} ∣^{1/ n}$ . The series converges absolutely for $∣ z - z_{0} ∣ < R$ and diverges for $∣ z - z_{0} ∣ > R$ ; points on the circle $∣ z - z_{0} ∣ = R$ must be tested separately. Inside the disc it may be differentiated or integrated term by term without changing $R$ .

WORKED EXAMPLE · A REAL SERIES AND A POWER SERIES WITH GAPS

(a) Does $\sum_{n = 1}^{\infty} \frac{n ^{2}}{2 ^{n}}$ converge? (b) Find the radius of convergence of $\sum_{n = 0}^{\infty} \frac{n + 1}{4 ^{n}} (z - 2 j)^{2 n}$ .

Solution

(a) Ratio test: $\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 ) ^{2}}{2 ^{n + 1}} \cdot \frac{2 ^{n}}{n ^{2}} = \frac{1}{2} (1 + \frac{1}{n})^{2} \to \frac{1}{2} < 1$ , so the series converges (its sum is $6$ ).
(b) Only even powers of $(z - 2 j)$ appear, so the coefficient ratio $a_{n + 1} / a_{n}$ in powers $(z - 2 j)^{n}$ is undefined (every odd coefficient is zero). Apply the ratio test to the terms $t_{n}$ themselves.
$\frac{t _{n + 1}}{t _{n}} = \frac{n + 2}{n + 1} \cdot \frac{∣ z - 2 j ∣ ^{2}}{4} \to \frac{∣ z - 2 j ∣ ^{2}}{4}$ .
Convergence needs $\frac{∣ z - 2 j ∣ ^{2}}{4} < 1$ , i.e. $∣ z - 2 j ∣ < 2$ . Hence $R = 2$ , for a disc centred at $2 j$ .
Equivalently, with $w = (z - 2 j)^{2}$ the series $\sum (n + 1) (w /4)^{n}$ has radius $4$ in $w$ , and $∣ w ∣ < 4 ⟺ ∣ z - 2 j ∣ < 2$ .

Answer: (a) It converges (ratio limit $1/2$ ). (b) $R = 2$ .

EXAM TIP

A ratio limit of $1$ decides nothing: $\sum 1/ n$ (divergent) and $\sum 1/ n^{2}$ (convergent) both give $L = 1$ . Switch to comparison, the integral test or the alternating series test.

EXAM TIP

When the only singularities of $f$ are poles or essential singularities (as for rational functions, $tan z$ or $e^{1/ (z - 1)}$ ), the radius of convergence of the Taylor series of $f$ about $z_{0}$ is the distance from $z_{0}$ to the nearest of them, complex ones included. $\frac{1}{1 + x ^{2}} = 1 - x^{2} + x^{4} - \dots$ has $R = 1$ about $0$ although the function is smooth for every real $x$ , because of the poles at $\pm j$ .

Taylor and Laurent series

Near every point where it is analytic, a function equals the sum of its own power series. Around a point where $f$ is analytic you get a Taylor series, with non-negative powers only. Around an isolated singularity, or in any ring where $f$ is analytic, you get a Laurent series, which also has negative powers. The region of validity is part of the answer: the same function has different series in different regions.

TAYLOR SERIES

f (z) = n = 0 \sum \infty \frac{f ^{(n)} ( z _{0} )}{n !} (z - z_{0})^{n}, ∣ z - z_{0} ∣ < R

The series equals $f$ throughout the largest open disc about $z_{0}$ in which $f$ is analytic. When the singularity on the edge of that disc is a pole or an essential singularity, that disc is exactly the disc of convergence, so $R$ is the distance from $z_{0}$ to that singularity (a removable singularity does not limit $R$ ). $z_{0} = 0$ gives the Maclaurin series. The expansion is unique, so any method that produces a power series (substitution, multiplication, term-by-term differentiation or integration) gives the Taylor series.

STANDARD EXPANSIONS ABOUT 0

e^{z} \frac{1}{1 - z} = n = 0 \sum \infty \frac{z ^{n}}{n !}, = n = 0 \sum \infty z^{n}, sin z \frac{1}{1 + z} = n = 0 \sum \infty \frac{( - 1 ) ^{n} z ^{2 n + 1}}{( 2 n + 1 )!}, = n = 0 \sum \infty (- 1)^{n} z^{n}, cos z Log (1 + z) = n = 0 \sum \infty \frac{( - 1 ) ^{n} z ^{2 n}}{( 2 n )!} (all z) = n = 1 \sum \infty \frac{( - 1 ) ^{n + 1} z ^{n}}{n} (∣ z ∣ < 1)

$sinh z$ and $cosh z$ have the same series as $sin z$ and $cos z$ without the alternating signs. Binomial series: $(1 + z)^{α} = 1 + α z + \frac{α ( α - 1 )}{2 !} z^{2} + \dots$ for $∣ z ∣ < 1$ (principal value).

LAURENT SERIES

f (z) = n = - \infty \sum \infty a_{n} (z - z_{0})^{n} = n = 0 \sum \infty a_{n} (z - z_{0})^{n} + n = 1 \sum \infty \frac{a _{- n}}{( z - z _{0} ) ^{n}}, a_{n} = \frac{1}{2 π j} \oint_{C} \frac{f ( z )}{( z - z _{0} ) ^{n + 1}} d z

Valid in an annulus $r_{1} < ∣ z - z_{0} ∣ < r_{2}$ in which $f$ is analytic ( $r_{1}$ may be $0$ and $r_{2}$ may be $\infty$ ); $C$ is any closed counter-clockwise contour around $z_{0}$ inside the annulus. The negative-power part is the principal part. In a given annulus the expansion is unique, but each annulus bounded by the singularities has its own series. In practice the coefficients come from known series, not from the integral.

Expanding a rational function in a given region

Split into partial fractions. For an expansion about $z_{0} \neq = 0$ , first put $t = z - z_{0}$ and expand in powers of $t$ .
Sort the singularities by distance from the centre. The region asked for tells you, for each term $\frac{1}{z - b}$ , whether $∣ z ∣ < ∣ b ∣$ or $∣ z ∣ > ∣ b ∣$ there.
If $∣ z ∣ < ∣ b ∣$ : $\frac{1}{z - b} = - \frac{1}{b} \cdot \frac{1}{1 - z / b} = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{b ^{n + 1}}$ (non-negative powers).
If $∣ z ∣ > ∣ b ∣$ : $\frac{1}{z - b} = \frac{1}{z} \cdot \frac{1}{1 - b / z} = \sum_{n = 0}^{\infty} \frac{b ^{n}}{z ^{n + 1}}$ (negative powers).
A repeated factor $\frac{1}{( z - b ) ^{2}}$ : differentiate the series for $\frac{1}{z - b}$ term by term and change the sign, since $\frac{d}{d z} \frac{1}{z - b} = - \frac{1}{( z - b ) ^{2}}$ .

WORKED EXAMPLE · ONE FUNCTION, THREE LAURENT SERIES

Expand $f (z) = \frac{1}{( z - 1 ) ( z - 2 )}$ in powers of $z$ in (i) $∣ z ∣ < 1$ , (ii) $1 < ∣ z ∣ < 2$ , (iii) $∣ z ∣ > 2$ . Use (ii) to evaluate $\oint_{∣ z ∣ = 1.5} f (z) d z$ .

Solution

Partial fractions: $f (z) = \frac{1}{z - 2} - \frac{1}{z - 1}$ .
(i) $∣ z ∣ < 1$ , so also $∣ z ∣ < 2$ : $f = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 1}} + \sum_{n = 0}^{\infty} z^{n} = \sum_{n = 0}^{\infty} (1 - 2^{- (n + 1)}) z^{n} = \frac{1}{2} + \frac{3}{4} z + \frac{7}{8} z^{2} + \dots$ , a Taylor series.
(ii) $1 < ∣ z ∣ < 2$ : $\frac{1}{z - 2} = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 1}}$ as before, but now $\frac{1}{z - 1} = \sum_{n = 0}^{\infty} \frac{1}{z ^{n + 1}}$ . So $f = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 1}} - \sum_{n = 1}^{\infty} \frac{1}{z ^{n}}$ .
(iii) $∣ z ∣ > 2$ : both terms need negative powers, $f = \sum_{n = 0}^{\infty} \frac{2 ^{n} - 1}{z ^{n + 1}} = \frac{1}{z ^{2}} + \frac{3}{z ^{3}} + \frac{7}{z ^{4}} + \dots$ (the $1/ z$ coefficient is $0$ ).
The circle $∣ z ∣ = 1.5$ lies in the annulus of (ii), where the coefficient of $1/ z$ is $a_{- 1} = - 1$ . Integrating term by term, only that term survives: $\oint_{∣ z ∣ = 1.5} f d z = 2 π j \cdot (- 1) = - 2 π j$ .
Check with residues: only the pole $z = 1$ is inside, with residue $\frac{1}{1 - 2} = - 1$ , which gives the same $- 2 π j$ .

Answer: (i) $\sum_{n \geq 0} (1 - 2^{- (n + 1)}) z^{n}$ ; (ii) $- \sum_{n \geq 0} z^{n} / 2^{n + 1} - \sum_{n \geq 1} z^{- n}$ ; (iii) $\sum_{n \geq 0} (2^{n} - 1) / z^{n + 1}$ ; the integral is $- 2 π j$ .

EXAM TIP

The coefficient $a_{- 1}$ of a Laurent series is the residue at the centre $z_{0}$ only when the series is valid in a punctured disc $0 < ∣ z - z_{0} ∣ < r$ . In an annulus $r_{1} < ∣ z - z_{0} ∣ < r_{2}$ with $r_{1} > 0$ , $2 π j a_{- 1}$ is the integral around any circle in the annulus, so $a_{- 1}$ is the sum of the residues inside the inner circle $∣ z - z_{0} ∣ = r_{1}$ . In part (ii) above the centre $0$ is not a singularity at all; $a_{- 1} = - 1$ there is the residue of the pole at $z = 1$ , the only singularity inside $∣ z ∣ = 1$ . Always use the series valid in the region that contains the contour.

EXAM TIP

For terms built from $1/ z$ , substitute into a known series: $e^{1/ z} = \sum_{n = 0}^{\infty} \frac{1}{n ! z ^{n}}$ and $sin (1/ z) = \frac{1}{z} - \frac{1}{3 ! z ^{3}} + \dots$ , both valid for $∣ z ∣ > 0$ and both with infinitely many negative powers.

Zeros, singularities and residues

$z_{0}$ is a zero of order $m$ of an analytic $f$ if $f (z_{0}) = f^{'} (z_{0}) = \dots = f^{(m - 1)} (z_{0}) = 0$ and $f^{(m)} (z_{0}) \neq = 0$ ; equivalently $f (z) = (z - z_{0})^{m} g (z)$ with $g$ analytic and $g (z_{0}) \neq = 0$ . The zeros of a non-constant function analytic in a domain are isolated. $z_{0}$ is an isolated singularity if $f$ is analytic in some punctured disc $0 < ∣ z - z_{0} ∣ < r$ but not at $z_{0}$ . Not every singularity is isolated: $Log z$ is singular at every point of its branch cut, and $z = 0$ is a limit of the poles $z = 1/ (nπ)$ of $1/ sin (1/ z)$ . Residues are defined only at isolated singularities.

Classifying an isolated singularity from the Laurent series in a punctured disc

Type	Principal part	Behaviour as $z \to z_{0}$	Example
Removable	none (every $a_{- n} = 0$ )	$f (z)$ tends to a finite limit; defining $f (z_{0})$ as that limit makes $f$ analytic there	$\frac{sin z}{z}$ at $0$ (limit $1$ )
Pole of order $m$	finitely many terms, the lowest being $a_{- m} (z - z_{0})^{- m}$ with $a_{- m} \neq = 0$	$∣ f (z) ∣ \to \infty$ ; $(z - z_{0})^{m} f (z)$ tends to a finite non-zero limit	$\frac{e ^{z}}{( z - 1 ) ^{3}}$ at $1$ ( $m = 3$ ); a pole of order 1 is a simple pole
Essential	infinitely many non-zero terms	no limit, finite or infinite; near $z_{0}$ , $f$ comes arbitrarily close to every complex value	$e^{1/ z}$ and $sin (1/ z)$ at $0$

ORDER OF A POLE FROM ORDERS OF ZEROS

f = \frac{p}{q}, p has a zero of order k and q a zero of order m at z_{0} \Rightarrow {pole of order m - k, removable singularity, m > k m \leq k

Here $p$ and $q$ are analytic at $z_{0}$ ; take $k = 0$ when $p (z_{0}) \neq = 0$ . Find orders from series: $1 - cos z = \frac{z ^{2}}{2} - \dots$ has a zero of order 2 at $0$ , so $\frac{1 - cos z}{z ^{5}}$ has a pole of order 3 there, not 5.

RESIDUE

Res_{z = z_{0}} f (z) = a_{- 1} = \frac{1}{2 π j} \oint_{C} f (z) d z

$a_{- 1}$ is the coefficient of $(z - z_{0})^{- 1}$ in the Laurent series valid in $0 < ∣ z - z_{0} ∣ < r$ ; $C$ is a small counter-clockwise circle around $z_{0}$ enclosing no other singularity. At a removable singularity the residue is $0$ . At an essential singularity it must be read from the Laurent series: $Res_{z = 0} e^{1/ z} = 1$ and $Res_{z = 0} z^{2} e^{1/ z} = \frac{1}{3 !} = \frac{1}{6}$ .

RESIDUE AT A SIMPLE POLE

Res_{z = z_{0}} f = z \to z_{0} lim (z - z_{0}) f (z); f = \frac{p ( z )}{q ( z )}, p (z_{0}) \neq = 0, q (z_{0}) = 0, q^{'} (z_{0}) \neq = 0 \Rightarrow Res_{z = z_{0}} f = \frac{p ( z _{0} )}{q ^{'} ( z _{0} )}

Here $p$ and $q$ are analytic at $z_{0}$ . The $p / q^{'}$ form avoids factorising: $Res_{z = nπ} \frac{1}{sin z} = \frac{1}{cos nπ} = (- 1)^{n}$ , $Res_{z = 0} cot z = \frac{cos 0}{cos 0} = 1$ , and for $\frac{1}{z ^{4} + 1}$ at a root $z_{k}$ the residue is $\frac{1}{4 z _{k}^{3}} = - \frac{z _{k}}{4}$ , since $z_{k}^{4} = - 1$ .

RESIDUE AT A POLE OF ORDER M

Res_{z = z_{0}} f = \frac{1}{( m - 1 )!} z \to z_{0} lim \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - z_{0})^{m} f (z)]

Using a larger $m$ than the true order still gives the right answer (with more work); using a smaller one does not. For high orders, or when $f$ contains $sin$ , $cos$ or $e^{z}$ , reading $a_{- 1}$ from a series is usually faster: $Res_{z = 0} \frac{sin z}{z ^{4}}$ is the coefficient of $z^{3}$ in $sin z$ , which is $- \frac{1}{6}$ .

RESIDUE AT INFINITY

Res_{z = \infty} f = - Res_{w = 0} [\frac{1}{w ^{2}} f (\frac{1}{w})]; finite z_{k} \sum Res_{z = z_{k}} f + Res_{z = \infty} f = 0

Equivalently $Res_{z = \infty} f = - b_{1}$ , where $b_{1}$ is the coefficient of $1/ z$ in the expansion of $f$ valid for large $∣ z ∣$ . The sum rule holds when $f$ has finitely many singularities. So if a contour $C$ encloses every finite singularity, $\oint_{C} f d z = - 2 π j Res_{z = \infty} f$ . For a rational $f$ whose denominator degree is at least the numerator degree plus 2, $Res_{z = \infty} f = 0$ and the finite residues add to zero.

WORKED EXAMPLE · RESIDUES AT A SIMPLE AND A DOUBLE POLE

Find the residues of $f (z) = \frac{z + 1}{z ^{2} ( z - 2 )}$ at each of its poles, and evaluate $\oint_{∣ z ∣ = 1} f (z) d z$ and $\oint_{∣ z ∣ = 3} f (z) d z$ .

Solution

Poles: $z = 2$ (simple) and $z = 0$ (order 2, since the numerator equals $1 \neq = 0$ there).
At $z = 2$ : $Res = lim_{z \to 2} \frac{z + 1}{z ^{2}} = \frac{3}{4}$ .
At $z = 0$ ( $m = 2$ ): $Res = lim_{z \to 0} \frac{d}{d z} [\frac{z + 1}{z - 2}] = lim_{z \to 0} \frac{( z - 2 ) - ( z + 1 )}{( z - 2 ) ^{2}} = - \frac{3}{4}$ .
$∣ z ∣ = 1$ encloses only $z = 0$ : $\oint = 2 π j (- \frac{3}{4}) = - \frac{3 π j}{2}$ .
$∣ z ∣ = 3$ encloses both poles: $\oint = 2 π j (\frac{3}{4} - \frac{3}{4}) = 0$ . This agrees with the sum rule: the denominator has degree 3 and the numerator degree 1, so $Res_{z = \infty} f = 0$ .

Answer: Residue $- \frac{3}{4}$ at $z = 0$ and $\frac{3}{4}$ at $z = 2$ ; the integrals are $- \frac{3 π j}{2}$ and $0$ .

EXAM TIP

Do not read the order of a pole off the denominator alone. $\frac{sin z}{z ^{2}}$ has a simple pole at $0$ (residue $1$ ), and $\frac{e ^{z} - 1}{z ^{3}}$ has a pole of order 2 (residue $\frac{1}{2}$ ), because the numerators vanish at $0$ .

EXAM TIP

The type of a singularity and the value of its residue are separate questions: $e^{1/ z}$ has an essential singularity at $0$ with residue $1$ , and a pole can have residue $0$ , e.g. $\frac{1}{z ^{2}}$ at $0$ .

EXAM TIP

The residue at infinity can be non-zero even when $f \to 0$ at infinity: $Res_{z = \infty} \frac{1}{z} = - 1$ , consistent with the residue $1$ at $z = 0$ .

The residue theorem and real integrals

RESIDUE THEOREM

\oint_{C} f (z) d z = 2 π j k \sum Res_{z = z_{k}} f (z)

$C$ is a simple closed contour traversed counter-clockwise; $f$ is analytic on $C$ and inside it except at finitely many isolated singularities $z_{k}$ inside $C$ . Only singularities strictly inside count. Cauchy's theorem and both integral formulas are special cases.

ARGUMENT PRINCIPLE

\frac{1}{2 π j} \oint_{C} \frac{f ^{'} ( z )}{f ( z )} d z = N - P

$C$ is a simple closed contour traversed counter-clockwise; $f$ is analytic on and inside $C$ except for poles inside it, and has no zeros or poles on $C$ . $N$ and $P$ are the numbers of zeros and poles of $f$ inside $C$ , each counted as many times as its order. Reason: at a zero of order $n$ , $f^{'} / f$ has a simple pole with residue $n$ , and at a pole of order $p$ one with residue $- p$ . The left side also equals the number of times the image curve $f (C)$ winds counter-clockwise around $w = 0$ . Related (Rouche's theorem): if $f$ and $g$ are analytic on and inside $C$ and $∣ g (z) ∣ < ∣ f (z) ∣$ on $C$ , then $f$ and $f + g$ have the same number of zeros inside $C$ ; e.g. on $∣ z ∣ = 2$ , $∣3 z^{2} + 1∣ \leq 13 < 32 = ∣ z^{5} ∣$ , so all five zeros of $z^{5} + 3 z^{2} + 1$ lie in $∣ z ∣ < 2$ .

Real integrals by residues (in brief). The standard types below turn a real integral into a contour integral. When the original integrand is real, the answer must be real; an imaginary answer means a lost factor of $j$ .

INTEGRALS OVER 0 TO 2 PI

\int_{0}^{2 π} F (cos θ, sin θ) d θ = \oint_{∣ z ∣ = 1} F (\frac{z + z ^{- 1}}{2}, \frac{z - z ^{- 1}}{2 j}) \frac{d z}{j z}

Substitute $z = e^{j θ}$ , so $d θ = \frac{d z}{j z}$ and $θ : 0 \to 2 π$ traces the unit circle once counter-clockwise. $F$ must be a rational function of $cos θ$ and $sin θ$ , finite for all $θ$ . Keep only the poles with $∣ z ∣ < 1$ . Also useful: $cos n θ = \frac{z ^{n} + z ^{- n}}{2}$ and $sin n θ = \frac{z ^{n} - z ^{- n}}{2 j}$ .

STANDARD RESULT OVER 0 TO 2 PI

\int_{0}^{2 π} \frac{d θ}{a + b cos θ} = \int_{0}^{2 π} \frac{d θ}{a + b sin θ} = \frac{2 π}{a ^{2} - b ^{2}}, a > ∣ b ∣

$a$ and $b$ real. If $a < - ∣ b ∣$ the value is $- \frac{2 π}{a ^{2} - b ^{2}}$ . If $∣ a ∣ \leq ∣ b ∣$ the integrand is infinite somewhere on the interval and the integral does not exist as a proper integral. Over $0$ to $π$ the cosine version gives half, $\frac{π}{a ^{2} - b ^{2}}$ .

IMPROPER INTEGRALS OF RATIONAL FUNCTIONS

\int_{- \infty}^{\infty} \frac{P ( x )}{Q ( x )} d x = 2 π j Im z_{k} > 0 \sum Res_{z = z_{k}} \frac{P ( z )}{Q ( z )}

Conditions: $P$ and $Q$ polynomials, $de g Q \geq de g P + 2$ , and $Q$ has no real zeros. Close the real axis with a large semicircle in the upper half-plane; its contribution vanishes by the ML-inequality. Count only the poles with $Im z > 0$ . For an even integrand, $\int_{0}^{\infty}$ is half of $\int_{- \infty}^{\infty}$ .

FOURIER-TYPE INTEGRALS

\int_{- \infty}^{\infty} \frac{P ( x )}{Q ( x )} e^{j m x} d x = 2 π j Im z_{k} > 0 \sum Res_{z = z_{k}} [\frac{P ( z )}{Q ( z )} e^{j m z}], m > 0

Conditions: $de g Q \geq de g P + 1$ and $Q$ has no real zeros. With $P$ , $Q$ real, take real and imaginary parts to get $\int \frac{P}{Q} cos m x d x$ and $\int \frac{P}{Q} sin m x d x$ . Use $e^{j m z}$ , never $cos m z$ , on the contour: $∣ e^{j m z} ∣ = e^{- m y} \leq 1$ in the upper half-plane, whereas $cos m z$ grows there. Standard result: $\int_{- \infty}^{\infty} \frac{cos m x}{x ^{2} + a ^{2}} d x = \frac{π}{a} e^{- ma}$ for $a > 0$ , $m \geq 0$ .

SIMPLE POLES ON THE REAL AXIS (PRINCIPAL VALUE)

PV \int_{- \infty}^{\infty} g (x) d x = 2 π j Im z_{k} > 0 \sum Res_{z = z_{k}} g + π j real x_{k} \sum Res_{z = x_{k}} g, g (z) = \frac{P ( z )}{Q ( z )} e^{j m z}, m \geq 0

Use this when $Q$ has real zeros $x_{k}$ and every one of them is simple. Degree conditions as above: $de g Q \geq de g P + 1$ when $m > 0$ , and $de g Q \geq de g P + 2$ when $m = 0$ . Each real pole is bypassed by a small semicircle above it, which contributes $- π j$ times its residue; that is where the half-weight $π j$ comes from. Example: $g = e^{j m z} / z$ with $m > 0$ has residue $1$ at $z = 0$ and no other pole, so $PV \int_{- \infty}^{\infty} \frac{e ^{j m x}}{x} d x = π j$ ; its imaginary part gives $\int_{- \infty}^{\infty} \frac{sin m x}{x} d x = π$ , and hence $\int_{0}^{\infty} \frac{sin x}{x} d x = \frac{π}{2}$ .

Procedure for a real integral

Identify the type and check its conditions: the degree condition, no pole on the path of integration (or only simple real poles, with the principal-value form), $a > ∣ b ∣$ .
Write the complex integrand and find its poles.
Keep only the relevant poles: $∣ z ∣ < 1$ for the $0$ to $2 π$ type, $Im z > 0$ for the infinite-range types.
Compute the residues (simple pole: $p / q^{'}$ ; repeated pole: the derivative formula), multiply their sum by $2 π j$ , and include any constant factor such as the $1/ j$ from $d θ = d z / (j z)$ .
Check that the result is real and has a sensible sign: a positive integrand must give a positive answer.

WORKED EXAMPLE · AN INTEGRAL OVER 0 TO 2 PI

Evaluate $\int_{0}^{2 π} \frac{d θ}{5 + 4 cos θ}$ by residues, and check it against the standard result.

Solution

Put $z = e^{j θ}$ : $5 + 4 cos θ = 5 + 2 (z + \frac{1}{z}) = \frac{2 z ^{2} + 5 z + 2}{z}$ , and $d θ = \frac{d z}{j z}$ .
$I = \oint_{∣ z ∣ = 1} \frac{z}{2 z ^{2} + 5 z + 2} \cdot \frac{d z}{j z} = \frac{1}{j} \oint_{∣ z ∣ = 1} \frac{d z}{( 2 z + 1 ) ( z + 2 )}$ .
Poles: $z = - \frac{1}{2}$ (inside the unit circle) and $z = - 2$ (outside).
$Res_{z = - 1/2} \frac{1}{( 2 z + 1 ) ( z + 2 )} = \frac{1}{2 ( z + 2 )}_{z = - 1/2} = \frac{1}{2 \cdot \frac{3}{2}} = \frac{1}{3}$ .
$I = \frac{1}{j} \cdot 2 π j \cdot \frac{1}{3} = \frac{2 π}{3}$ . Standard result: $\frac{2 π}{25 - 16} = \frac{2 π}{3}$ .

Answer: $\frac{2 π}{3} \approx 2.094$

WORKED EXAMPLE · AN IMPROPER RATIONAL INTEGRAL

Evaluate $\int_{0}^{\infty} \frac{d x}{( x ^{2} + 1 ) ^{2}}$ .

Solution

The integrand is even, $de g Q - de g P = 4 \geq 2$ and there are no real poles, so $\int_{0}^{\infty} = \frac{1}{2} \int_{- \infty}^{\infty}$ and the residue method applies.
$\frac{1}{( z ^{2} + 1 ) ^{2}} = \frac{1}{( z - j ) ^{2} ( z + j ) ^{2}}$ . The only pole in the upper half-plane is $z = j$ , of order 2.
$Res_{z = j} = \frac{d}{d z} [\frac{1}{( z + j ) ^{2}}]_{z = j} = \frac{- 2}{( 2 j ) ^{3}} = \frac{- 2}{- 8 j} = \frac{1}{4 j}$ .
$\int_{- \infty}^{\infty} = 2 π j \cdot \frac{1}{4 j} = \frac{π}{2}$ , so the required integral is $\frac{π}{4}$ .

Answer: $\frac{π}{4} \approx 0.785$

EXAM TIP

Factor carefully before choosing poles. $z^{4} + 1 = 0$ gives $z = e^{j π /4}, e^{j 3 π /4}, e^{j 5 π /4}, e^{j 7 π /4}$ , and only the first two are in the upper half-plane. $z^{2} + 2 z + 5 = 0$ gives $- 1 \pm 2 j$ , and only $- 1 + 2 j$ counts.

EXAM TIP

In the $0$ to $2 π$ type, the factor $\frac{1}{j}$ from $d θ = \frac{d z}{j z}$ is easy to lose; it is what makes $2 π j \times$ (sum of residues) real.

EXAM TIP

A clockwise contour, or one described as negatively oriented, flips the sign of $2 π j \sum Res$ .

PREVIOUS-YEAR QUESTIONS

24 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

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GATE 2026 EE Q.62

GATE 2026 · EEQ.62 · 2 marks · NAT

The magnitude of the contour integral

\oint_{C} (\frac{( z + 1 ) ^{2}}{( z - i ) ( z - 2 )}) d z

over the contour $C : ∣ z - 2 - i ∣ = 3/2$ is ______

(Round off to two decimal places)

Note: $z$ is a complex variable and $i = - 1$

Answer and solution

GATE 2026 EE Q.62: answer and solution

See the question in the paper

Complex analysis

Complex numbers, conventions and elementary functions

Analytic functions, the Cauchy-Riemann equations and conformal mapping

Complex integration: Cauchy's integral theorem and integral formula

Sequences, series and convergence tests

Taylor and Laurent series

Zeros, singularities and residues

The residue theorem and real integrals

24 questions from the EC, EE and IN papers

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