Which statements about the point z = 0 are true?

e^{1/z} has an essential singularity at 0

z/(eᶻ − 1) has a removable singularity at 0

e^{1/z} has an essential singularity at 0

(1 − cos z)/z² has a pole at 0

1/sin(1/z) has an isolated singularity at 0

z/(eᶻ − 1) has a removable singularity at 0

Let f be analytic on the open unit disc 𝔻 with f(𝔻) ⊆ 𝔻 and f(0) = 0. Which statements must hold?

if f(1/2) = 1/2 then f(z) = z for all z in 𝔻

Which statements are true?

f(z) = z² is conformal at every z ≠ 0

every Möbius transformation maps circles and lines onto circles and lines

T(z) = (z − i)/(z + i) maps the upper half-plane onto the open unit disc

f(z) = z² is conformal at every z ≠ 0

f(z) = z² is conformal at z = 0

every Möbius transformation maps circles and lines onto circles and lines

T(z) = (z − i)/(z + i) maps the upper half-plane onto the open unit disc

Singularities, Residues, Rouché, the Argument Principle and Conformal Maps

The second half of Section 4 is about what happens at the points where a function stops being analytic, and about analytic functions as maps. Isolated singularities come in three kinds — removable, poles and essential — and the Laurent series tells them apart. The residue theorem turns a contour integral into a finite sum and, with the right contour, evaluates real integrals that calculus cannot touch. The argument principle counts zeros minus poles, and Rouché’s theorem turns that count into a comparison of two functions on a circle, which is how GATE asks “how many roots lie in this annulus”. Schwarz’s lemma is the sharp bound for self-maps of the disc. Finally, analytic functions with non-zero derivative are conformal, and the Möbius transformations are the conformal self-maps of the Riemann sphere, determined by the images of three points.

1. Isolated singularities

If f is analytic on 0 < |z − a| < r, the Laurent series Σcₙ(z − a)ⁿ classifies a: removable if no negative powers occur (equivalently f is bounded near a — Riemann’s theorem); a pole of order m if the most negative power is (z − a)^−m (equivalently |f| → ∞); essential if infinitely many negative powers occur. If g has a zero of order m at a, 1/g has a pole of order m there.

The three kinds, with the standard examples
Kind	Example at 0	Behaviour near the point
removable	sin z/z, (1 − cos z)/z², z/(eᶻ − 1)	bounded; extends analytically
pole of order m	1/zᵐ; z/(1 − cos z) is a simple pole	\|f\| → ∞
essential	e^1/z, sin(1/z), z² sin(1/z)	Casorati–Weierstrass: the values near a are dense in ℂ; Picard: every value but at most one is taken infinitely often

⚠️ Not every singularity is isolated

1/sin(1/z) has poles at z = 1/(kπ) for every non-zero integer k, and they accumulate at 0, so 0 is a non-isolated singularity: it is neither removable, a pole nor essential, and it has no residue. log z has a branch point at 0, which is not an isolated singularity of any single-valued branch either.

2. Residues and the residue theorem

The residue Res(f, a) is the Laurent coefficient c₋₁. Residue theorem: if f is analytic inside and on a positively oriented simple closed curve γ except at finitely many isolated singularities a₁, …, aₖ inside γ, then ∮_γ f dz = 2πi Σ Res(f, aⱼ).

Computing a residue
Situation	Formula	Example
simple pole	lim_z→a (z − a)f(z); for f = g/h with h(a) = 0 ≠ h′(a), g(a)/h′(a)	Res(cot z, 0) = cos 0/cos 0 = 1
pole of order m	(1/(m − 1)!) lim d^m−1/dz^m−1 [(z − a)ᵐf(z)]	Res(eᶻ/z³, 0) = 1/2
essential singularity	read c₋₁ from the Laurent series	z² sin(1/z) = z − 1/(6z) + …: residue −1/6
a quotient with a double zero below	expand: 1 − cos z = z²/2 − z⁴/24 + …	z/(1 − cos z) = 2/z + z/6 + …: residue 2

Example. ∮_|z|=3 (z² + 1)/(z(z − 2)) dz: both poles are inside, Res at 0 is 1/(−2) = −1/2 and Res at 2 is 5/2, so the integral is 2πi · 2 = 4πi. The residue at ∞ is −c₋₁ of the expansion in 1/z, and the sum of all residues of a rational function, including ∞, is 0.

3. Real integrals by residues

Trigonometric. ∫₀^2π R(cos θ, sin θ) dθ: put z = e^iθ, cos θ = (z + 1/z)/2, dθ = dz/(iz), and sum residues inside |z| = 1. Result to remember: ∫₀^2π dθ/(a + b cos θ) = 2π/√(a² − b²) for a > |b|; with a = 5, b = 4 it is 2π/3.
Rational on ℝ. If deg Q ≥ deg P + 2 and Q has no real zeros, ∫_−∞^∞ P/Q dx = 2πi Σ residues in the upper half-plane (close with a large semicircle, whose contribution → 0). ∫ dx/((x² + 1)(x² + 4)) = 2πi[1/(2i · 3) + 1/(−3 · 4i)] = π/6; ∫ dx/(x⁴ + 1) = π/√2.
Fourier type (Jordan’s lemma). ∫_−∞^∞ cos x/(x² + 1) dx = Re ∫ e^ix/(x² + 1) dx = Re[2πi · e⁻¹/(2i)] = π/e. Use e^ix, not cos z, because cos z grows like e^|y| in the upper half-plane.
Indented contours. A simple pole on the real axis contributes πi times its residue (half a circle): this gives ∫₀^∞ sin x/x dx = π/2.

4. The argument principle, Rouché’s theorem and Schwarz’s lemma

Argument principle. If f is meromorphic inside and on a simple closed curve γ with no zeros or poles on γ, then (1/2πi)∮_γ f′/f dz = Z − P, the number of zeros minus the number of poles inside, counted with multiplicity. For f = (z² − 1)/(z³(z − 3)) and |z| = 2 this is 2 − 3 = −1: the pole at 3 is outside.

Rouché’s theorem. If f and g are analytic inside and on γ and |f(z) − g(z)| < |f(z)| on γ, then f and g have the same number of zeros inside γ. In practice: split p = (dominant term) + (rest) on each circle. For z⁴ − 6z + 3: on |z| = 2, |z⁴| = 16 > |−6z + 3| ≤ 15, so 4 zeros in |z| < 2; on |z| = 1, |−6z| = 6 > |z⁴ + 3| ≤ 4, so 1 zero in |z| < 1; hence 3 zeros in 1 < |z| < 2 (none on |z| = 1, since the inequality is strict there).

🎯 Schwarz’s lemma and the automorphisms of the disc

If f: 𝔻 → 𝔻 is analytic and f(0) = 0, then g = f(z)/z has a removable singularity at 0 and |g| ≤ 1/r on |z| = r for every r < 1, so by the maximum modulus principle |g| ≤ 1: that is |f(z)| ≤ |z| and |f′(0)| ≤ 1, with equality anywhere (for some z ≠ 0, or at f′(0)) only for a rotation f = e^iθz. The analytic bijections of the disc are exactly e^iθ(z − a)/(1 − āz) with |a| < 1.

5. Conformal maps and Möbius transformations

An analytic f with f′(z₀) ≠ 0 is conformal at z₀: it preserves angles between curves, in size and orientation, because near z₀ it acts as multiplication by f′(z₀), a rotation and a scaling. At a critical point it is not: z² doubles angles at 0. Standard maps: z ↦ z² takes the first quadrant onto the upper half-plane; eᶻ takes the strip 0 < Im z < π onto the upper half-plane; the Cayley map (z − i)/(z + i) takes the upper half-plane onto the unit disc.

A Möbius transformation T(z) = (az + b)/(cz + d), ad − bc ≠ 0, is a bijection of the extended plane ℂ ∪ {∞}, a composition of translations, rotations–dilations and the inversion 1/z. It maps circles and lines to circles and lines.
There is exactly one Möbius map sending three distinct points to three distinct points; it preserves the cross-ratio. A Möbius map other than the identity has at most two fixed points.
Example. The map with 0 ↦ 0, 1 ↦ ∞, ∞ ↦ 1 is T(z) = z/(z − 1): the zero of the numerator gives 0, the zero of the denominator gives ∞, and the ratio of leading coefficients gives T(∞) = 1. Then T(−1) = 1/2.

Key takeaways

Classify a singularity by its Laurent series: no negative powers is removable, finitely many is a pole, infinitely many is essential.
Residue at a simple pole is g(a)/h′(a); at a pole of order m use the (m − 1)-th derivative; at an essential singularity read c₋₁.
Real integrals: z = e^iθ for trigonometric integrands, the upper half-plane for rational ones, e^ix with Jordan’s lemma for Fourier types.
Rouché compares a dominant term with the rest on each circle; the annulus count is the difference of the two disc counts.
Möbius maps send circles and lines to circles and lines and are fixed by three points; Schwarz’s lemma bounds self-maps of the disc fixing 0.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The singularity of sin z/z at z = 0 is
1. removable
2. a simple pole
3. a pole of order 2
4. essential
Show answer
Answer: A — removable
sin z/z = 1 − z²/6 + z⁴/120 − … has no negative powers, so defining its value at 0 to be 1 makes it entire. The function is bounded near 0, which by Riemann’s theorem already forces removability.
Which statements about the point z = 0 are true?
1. e^1/z has an essential singularity at 0
2. (1 − cos z)/z² has a pole at 0
3. 1/sin(1/z) has an isolated singularity at 0
4. z/(eᶻ − 1) has a removable singularity at 0
Show answer
Answer: A — e^{1/z} has an essential singularity at 0; D — z/(eᶻ − 1) has a removable singularity at 0
(1) e^1/z = Σ z⁻ⁿ/n! has infinitely many negative powers. (2) (1 − cos z)/z² = 1/2 − z²/24 + …: removable. (3) The poles 1/(kπ) accumulate at 0, so 0 is not isolated. (4) z/(eᶻ − 1) = 1/(1 + z/2 + …) → 1: removable.
The residue of f(z) = z/(1 − cos z) at z = 0 is ____.
Numerical answer — type the value.
Show answer
Answer: 2
1 − cos z = (z²/2)(1 − z²/12 + …), so f = (2/z)(1 + z²/12 + …) = 2/z + z/6 + …: a simple pole with residue 2. Treating the denominator’s zero as simple and using g/h′ would divide by h′(0) = sin 0 = 0.
The residue of z² sin(1/z) at z = 0, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: -0.17
sin(1/z) = 1/z − 1/(6z³) + 1/(120z⁵) − …, so z² sin(1/z) = z − 1/(6z) + 1/(120z³) − …. The singularity is essential and the coefficient of 1/z is −1/6 = −0.1667, i.e. −0.17. There is no limit formula at an essential singularity; the series is the method. Typed, the answer is -0.17.
If ∮_|z|=3 (z² + 1)/(z(z − 2)) dz = kπi, the circle positively oriented, then k = ____.
Numerical answer — type the value.
Show answer
Answer: 4
Both simple poles, 0 and 2, lie inside |z| = 3. Res at 0 = (0 + 1)/(0 − 2) = −1/2; Res at 2 = (4 + 1)/2 = 5/2. The integral is 2πi(−1/2 + 5/2) = 4πi, so k = 4. Over |z| = 1 only the pole at 0 counts, giving −πi.
The value of ∫₀^2π dθ/(5 + 4 cos θ), correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.09
With z = e^iθ: ∮_|z|=1 dz/(iz(5 + 2(z + 1/z))) = ∮ dz/(i(2z² + 5z + 2)) = ∮ dz/(i(2z + 1)(z + 2)). Only z = −1/2 is inside, with residue 1/(i · 2 · (3/2)) = 1/(3i). The integral is 2πi/(3i) = 2π/3 = 2.094, i.e. 2.09, agreeing with 2π/√(a² − b²) = 2π/3.
The value of ∫_−∞^∞ dx/((x² + 1)(x² + 4)), correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.52
Close in the upper half-plane; the poles there are i and 2i. Res at i = 1/(2i · 3) = 1/(6i); Res at 2i = 1/((−3)(4i)) = −1/(12i). The sum is 1/(12i), and the integral is 2πi/(12i) = π/6 = 0.5236, i.e. 0.52.
The value of ∫_−∞^∞ cos x/(x² + 1) dx is
1. π/e
2. πe
3. π
4. π/(2e)
Show answer
Answer: A — π/e
Integrate e^iz/(z² + 1) over the upper half-plane; Jordan’s lemma kills the arc. The only pole inside is i, with residue e⁻¹/(2i), so ∫ e^ix/(x² + 1) dx = 2πi · e⁻¹/(2i) = π/e, which is real, and the cosine integral is its real part. Using cos z directly fails because |cos z| grows on the arc.
The number of zeros, counted with multiplicity, of p(z) = z⁴ − 6z + 3 in the annulus 1 < |z| < 2 is ____.
Numerical answer — type the value.
Show answer
Answer: 3
On |z| = 2: |z⁴| = 16 and |−6z + 3| ≤ 15 < 16, so p has 4 zeros in |z| < 2 (Rouché with z⁴). On |z| = 1: |−6z| = 6 and |z⁴ + 3| ≤ 4 < 6, so p has 1 zero in |z| < 1 (Rouché with −6z), and none on |z| = 1. Hence 4 − 1 = 3.
For f(z) = (z² − 1)/(z³(z − 3)), the value of (1/2πi)∮_|z|=2 f′(z)/f(z) dz, the circle positively oriented, is ____.
Numerical answer — type the value.
Show answer
Answer: -1
By the argument principle the integral counts Z − P inside |z| = 2. Zeros: ±1, two simple zeros. Poles: z = 0 of order 3; z = 3 lies outside. So Z − P = 2 − 3 = −1. Counting the pole at 3 would give −2. Typed, the answer is -1.
Let f be analytic on the open unit disc 𝔻 with f(𝔻) ⊆ 𝔻 and f(0) = 0. Which statements must hold?
1. |f(1/2)| ≤ 1/2
2. |f′(0)| ≤ 1
3. if f(1/2) = 1/2 then f(z) = z for all z in 𝔻
4. f′(0) can equal 2
Show answer
Answer: A — |f(1/2)| ≤ 1/2; B — |f′(0)| ≤ 1; C — if f(1/2) = 1/2 then f(z) = z for all z in 𝔻
Schwarz’s lemma gives |f(z)| ≤ |z| and |f′(0)| ≤ 1, so (1), (2) hold and (4) fails. If |f(z₀)| = |z₀| for some z₀ ≠ 0, f is a rotation e^iθz; f(1/2) = 1/2 forces e^iθ = 1, so (3) holds.
T is the Möbius transformation with T(0) = 0, T(1) = ∞ and T(∞) = 1. The value of T(−1) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
A zero at 0 and a pole at 1 give T(z) = kz/(z − 1); T(∞) = k = 1. So T(z) = z/(z − 1) and T(−1) = −1/(−2) = 1/2. The Möbius map through three prescribed points is unique, so this is the only answer.
Which statements are true?
1. f(z) = z² is conformal at every z ≠ 0
2. f(z) = z² is conformal at z = 0
3. every Möbius transformation maps circles and lines onto circles and lines
4. T(z) = (z − i)/(z + i) maps the upper half-plane onto the open unit disc
Show answer
Answer: A — f(z) = z² is conformal at every z ≠ 0; C — every Möbius transformation maps circles and lines onto circles and lines; D — T(z) = (z − i)/(z + i) maps the upper half-plane onto the open unit disc
(1) f′(z) = 2z ≠ 0 there. (2) f′(0) = 0 and angles at 0 are doubled (the positive real and imaginary axes, at 90°, go to rays at 180°). (3) holds because each generator (translation, dilation-rotation, 1/z) does. (4) For Im z > 0, z is closer to i than to −i, so |T(z)| < 1, and T is a bijection onto the disc.