Complex Analysis, Algebra & Topology

Zeros and Mappings

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1. Argument principle, Rouché's theorem, open mapping

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Lecture 8.1 - Winding number

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

The winding number is what the argument principle actually counts.

Lecture - 10.3 Argument Principle

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Argument principle, Rouché and zero counting free

In plain words

Walk round a loop and count how often the image circles the origin. That counts the zeros inside minus the poles, and Rouché turns the count into a comparison with something simpler.

Argument principle

For f meromorphic inside and on a simple closed $\gamma$ , with no zeros or poles on $\gamma$ :

$(1/2\pi{}i) \oint_\gamma f'/f$ dz = N − P = (number of zeros) − (number of poles), counted with multiplicity.

Equivalently, N − P is the winding number of the image curve $f(\gamma)$ about the origin.

Watch the integrand. For (zf)′/(zf) the function zf has one extra zero at the origin, so the integral is (N + 1) − P.

Rouché's theorem

If |f(z)| > |g(z)| on $\gamma$ then f and f + g have the same number of zeros inside.

Recipe. Given a polynomial, choose the term that dominates on the given circle:

Polynomial, circle	Dominant term	Zeros inside
$z^{100} - 50z^{30} + 40z^{10} + 6z + 1$ , \|z\| = 1	$-50z^{30} (50 > 48)$	30
$z^{7} - 5z^{3} + 12$ , \|z\| = 1	12 (12 > 6)	0
$z^{7} - 5z^{3} + 12$ , \|z\| = 2	$z^{7} (128 > 52)$	7
eᶻ − 4z − 1, \|z\| = 1	−4z (4 > e + 1)	1

Always check the inequality is strict on the whole circle.

Open mapping and inverse function

Non-constant holomorphic maps are open.
f injective holomorphic ⇒ f′ ≠ 0 and f⁻ $^{1}$ is holomorphic.
Conversely $f'(z_{0}) \ne 0$ gives local injectivity — but not global: eᶻ has non-vanishing derivative and is not injective.

Hurwitz

Locally uniform limits: if each $f_{n}$ is zero-free (or injective) and $f_{n} \to f$ ≢ constant, then f is zero-free (or injective).

Key takeaways

Argument principle counts N − P; check what the integrand's numerator really is.
Rouché: dominate on the circle, count the dominant term's zeros.
f′ ≠ 0 gives local, never global, injectivity.

Rouché in action: counting the zeros of z⁵ + 3z + 1interactivefree

Step 1 / 5The question

How many zeros does $p(z) = z^5 + 3z + 1$ have inside $|z| < 1$ ? And inside $|z| < 2$ ?

No formula finds these roots — but Rouché's theorem counts them without finding them.

Exam focus: Rouché is a counting tool: split the polynomial into a dominant term and the rest, verify the strict inequality on the circle, and read off the zero count. Watch for the extra factor when the integrand is (zf)′/(zf).

Practice

Part C2024 Decembershow ▾

Let P(z) be a non-constant polynomial over

\mathbb{C}

. Given R > 0, let S_R = {

z \in \mathbb{C}

: |P(z)| < R}. Which of the following statements are true?

A.S_R is an open subset of $\mathbb{C}$ .✓ Correct
B.S_R is a bounded subset of $\mathbb{C}$ .✓ Correct
C.|P(z)| = R for every z on the boundary of S_R.✓ Correct
D.Every connected component of S_R contains a zero of P(z).✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

Let f be a meromorphic function on an open set containing the unit circle C and its interior. Suppose f has no zeros and no poles on C, and let n_p and

n_{0}

denote the number of poles and zeros of f inside C. Which one of the following is true?

A. $(1/2\pi{}i) \int_C ($ zf)′/(zf) dz $= n_{0} - n_p + 1$ .✓ Correct
B. $(1/2\pi{}i) \int_C ($ zf)′/(zf) dz $= n_{0} - n_p - 1$ .
C. $(1/2\pi{}i) \int_C ($ zf)′/(zf) dz $= n_{0} - n_p$ .
D. $(1/2\pi{}i) \int_C ($ zf)′/(zf) dz $= n_p - n_{0}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

How many roots does the polynomial

z^{100} - 50z^{30} + 40z^{10} + 6z + 1

have in the open disc {

z \in \mathbb{C}

: |z| < 1}?

A.100
B.50
C.30✓ Correct
D.0

Solution

On |z| = 1, | $-50z^{30}$ | = 50 > 1 + 40 + 6 + 1 = 48 ≥ | $z^{100} + 40z^{10} + 6z + 1$ |, so by Rouché the polynomial has as many zeros inside as $-50z^{30}$ : thirty.

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Part Cshow ▾

Let f be holomorphic on a domain containing the closed disc |z| ≤ 2, with no zeros on |z| = 2, and let N be the number of zeros of f (with multiplicity) inside |z| < 2. Which of the following are true?

A. $N = (1/2\pi{}i) \oint_{|z|=2}$ f′(z)/f(z) dz.✓ Correct
B.If g is holomorphic on the same domain and |f(z) − g(z)| < |f(z)| on |z| = 2, then f and g have the same number of zeros inside |z| < 2.✓ Correct
C.A non-constant holomorphic function maps open sets to open sets.✓ Correct
D.If f is holomorphic and injective near a point $z_{0}$ , then $f'(z_{0}) = 0$ .

Solution

(1) is the argument principle (f has no poles, being holomorphic).

(2) is Rouché's theorem.

(3) is the Open Mapping theorem.

(4) has the implication backwards: local injectivity of a holomorphic map forces $f'(z_{0}) \ne 0$ — if $f'(z_{0}) = 0, f$ looks locally like z ↦ zᵏ (k ≥ 2) near $z_{0}$ , which is k-to-one on any punctured neighbourhood, not injective.

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Part Cshow ▾

Let

p(z) = z^{5} + 3z + 1

. Which of the following are true?

A.p(z) has exactly 5 zeros (counted with multiplicity) inside |z| < 2.✓ Correct
B.p(z) has exactly 1 zero inside |z| < 1.✓ Correct
C.All zeros of p(z) lie inside |z| < 2.✓ Correct
D.p(z) has a zero on |z| = 1.

Solution

On |z| = 2: | $z^{5}$ | = 32 and |3z + 1| ≤ 7 < 32, so by Rouché p has the same zero count as $z^{5}$ there — 5, giving (1).

On |z| = 1: |3z| = 3 and | $z^{5} + 1$ | ≤ 2 < 3, so p has the same zero count as 3z there — 1, giving (2).

Since p has degree 5, those 5 zeros inside |z| < 2 are all of them, giving (3).

(4) is false: the same strict inequality | $z^{5} + 1$ | ≤ 2 < 3 = |3z| on |z| = 1 rules out p(z) = 0 there too — equality would force | $z^{5}+1$ | = |3z| = 3, contradicting the bound.

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Part Cshow ▾

Let

p(z) = z^{5} + 3z + 1

. Which of the following are true?

A.All five zeros of p lie in |z| < 2.✓ Correct
B.p has exactly one zero in |z| < 1/2.✓ Correct
C.p has no real zero in the interval (−1, 0).
D.Zeros of p in an annulus can be counted by applying Rouché separately on the two boundary circles.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

2. Conformal maps, Möbius transformations, Schwarz lemma

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Lecture - 4.2 - Mobius Transformations

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Circles-to-circlines and cross-ratio.

Lecture 11.2 Automorphisms of the Unit disk

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

The disc automorphism group — memorise the formula.

Conformal maps and Möbius transformations free

In plain words

Conformal means angles survive. Möbius maps are best seen on the sphere, where a straight line is just a circle through the point at infinity, so circles always go to circles.

Conformality

A holomorphic f is conformal (angle- and orientation-preserving) exactly where f′ ≠ 0. At a zero of f′ of order k, angles are multiplied by k + 1.

Möbius transformations

T(z) = (az + b)/(cz + d), ad − bc ≠ 0. Facts to quote:

They form a group isomorphic to PGL $_{2}(\mathbb{C})$ , acting on the Riemann sphere.
They map circles and lines to circles and lines (a "circline" is a circle on the sphere).
Uniquely determined by the images of three points; preserve the cross-ratio.
Automorphisms of the unit disc: $e^{i\theta}(z$ − a)/(1 − āz) with |a| < 1.
Automorphisms of the upper half-plane: (az + b)/(cz + d) with a, b, c, d real, ad − bc > 0.
Automorphisms of $\mathbb{C}$ : az + b. Of the Riemann sphere: all Möbius maps.

The dictionary

From	To	Map
upper half-plane ℍ	unit disc 𝔻	(z − i)/(z + i)
𝔻	ℍ	i(1 − z)/(1 + z)
𝔻	right half-plane	(1 + z)/(1 − z)
sector $0 < \arg z < \pi/n$	ℍ	$z^{n}$
strip $0 < Im z < \pi$	ℍ	eᶻ
ℍ	strip	log z

Riemann mapping theorem

Every simply connected domain that is not all of $\mathbb{C}$ is biholomorphic to 𝔻. Hence $\mathbb{C}$ and 𝔻 are not biholomorphic (Liouville), although there are holomorphic surjections 𝔻 $\to \mathbb{C}$ and injections 𝔻 $\to \mathbb{C}$ .

Schwarz lemma

f : 𝔻 → 𝔻, f(0) = 0 ⇒ |f(z)| ≤ |z|, |f′(0)| ≤ 1; equality anywhere forces f(z) = $e^{i\theta}z$ .

Key takeaways

Conformal ⇔ f′ ≠ 0; a zero of f′ of order k multiplies angles by k + 1.
Möbius maps: circlines to circlines, three points determine them, cross-ratio is the invariant. $- \mathbb{C}$ and 𝔻 are simply connected but not biholomorphic — the exception in the Riemann mapping theorem.

A grid, deformed by a complex functioninteractive

This visual is part of Trap Analysis

z², 1/z, eᶻ, sin z — watch which ones preserve angles everywhere, and where each one doesn't.

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Möbius maps bending the gridinteractive

This visual is part of Trap Analysis

Slide the map on and watch straight lines bow into circles — every one still a circle or a line, angles preserved throughout.

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Exam focus: Know the dictionary of standard maps between disc, half-plane, strip and sector, and that Möbius maps send circles-and-lines to circles-and-lines and preserve cross-ratio.

Practice

Part B2025 Juneshow ▾

Let X be the image of the interval [0, 1] under the Möbius transformation f(z) = (z − i)/(z + i). Which of the following statements is true?

A.X is the line segment joining −1 and −i.
B.X = { $e^{i\theta}$ | $\theta \in [\pi, 3\pi/2]$ }.✓ Correct
C.X is the line segment joining −1 to 1.
D.X = { $e^{i\theta}$ | $\theta \in [-\pi/2, \pi]$ }.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Let 𝔻^× = {

z \in \mathbb{C} : 0 <

|z| < 1} be the punctured unit disk and f be a bijective holomorphic map of 𝔻^× onto itself. Which of the following statements are true?

A. $lim_{z\to0}$ f(z) does not exist.
B. $lim_{z\to0}$ f(z) exists and has absolute value ≤ 1.✓ Correct
C. $lim_{z\to0}$ f(z) = 0.✓ Correct
D.There exists $\theta \in \mathbb{R}$ such that f(z) = $e^{i\theta}z$ for all z ∈ 𝔻^×.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Let 𝔻={

z\in\mathbb{C}

:|z|<1} and f:𝔻→𝔻 be a holomorphic function which satisfies f(−1/2)=0. Which of the following statements are necessarily true?

A.|f(−1/5)| ≤ 1/5
B.|f(−1/5)| ≤ 1/3✓ Correct
C.|f′(−1/2)| ≤ 1/2
D.|f′(−1/2)| ≤ 4/3✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

For a complex number a such that 0 < |a| < 1, which of the following statements is true?

A.If |z| < 1, then |1 − āz| < |z − a|
B.If |z − a| = |1 − āz|, then |z| = 1✓ Correct
C.If |z| = 1, then |z − a| < |1 − āz|
D.If |1 − āz| < |z − a|, then |z| < 1

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the disc 𝔻 = {

z \in \mathbb{C}

: |z| < 1} and a non-constant holomorphic function f : 𝔻 → 𝔻. Suppose that f(0) = 0. For each n ≥ 1 and z ∈ 𝔻, define

f_{n}(z) = f(z^{n})

. Which of the following statements are true?

A.The series $\sum(n\ge1) f_{n}$ converges only at z = 0.
B.The series $\sum(n\ge1) f_{n}$ converges pointwise only on a countable set E ⊆ 𝔻 but not on 𝔻 \ E.
C.The series $\sum(n\ge1) f_{n}$ converges pointwise at all points of 𝔻 but not uniformly on some compact subsets of 𝔻.
D.The series $\sum(n\ge1) f_{n}$ converges uniformly on all compact subsets of 𝔻 to a holomorphic function on 𝔻.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Consider the half planes R₊ = {z = x + iy

\in \mathbb{C} : x > 0

} and R₋ = {z = x + iy

\in \mathbb{C} : x < 0

}, and the fractional linear transformations

T_{1}(z) = (z - 1)/(z + 1)

and

T_{2}(z) = (z + 1)/(z - 1)

. Let disc 𝔻 = {

z \in \mathbb{C}

: |z| < 1}. Which of the following statements is true?

A. $T_{1}$ and $T_{2}$ conformally map R₊ and R₋ respectively, onto the disc 𝔻✓ Correct
B. $T_{1}$ and $T_{2}$ conformally map R₋ and R₊ respectively, onto the disc 𝔻
C. $T_{1}$ and $T_{2}$ conformally map the disc 𝔻 onto, respectively R₊ and R₋
D. $T_{1}$ and $T_{2}$ conformally map the disc 𝔻 onto, respectively R₋ and R₊

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following are true about Möbius transformations?

A.They map circles and lines to circles and lines✓ Correct
B.They preserve the cross-ratio of four points✓ Correct
C.Every Möbius transformation has at least one fixed point on the Riemann sphere✓ Correct
D.They are conformal at every point of $\mathbb{C}$

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider Möbius transformations T(z) = (az + b)/(cz + d) with ad − bc ≠ 0. Which of the following are true?

A.A Möbius transformation maps the family of circles and lines to itself.✓ Correct
B.A Möbius transformation other than the identity has at most two fixed points in $\mathbb{C} \cup$ { $\infty$ }.✓ Correct
C.Given two triples of distinct points, exactly one Möbius transformation maps the first triple to the second in order.✓ Correct
D.T(z) = 1/z maps the punctured disc {0 < |z| < 1} onto the open unit disc.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The image of the upper half-plane under the map w = (z − i)/(z + i) is

A.the unit disc |w| < 1✓ Correct
B.the exterior |w| > 1
C.the right half-plane
D.the punctured plane

Solution

For Im z > 0, z is closer to i than to −i, so |z − i| < |z + i| and |w| < 1.

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Part Bshow ▾

Let f : 𝔻 → 𝔻 be holomorphic with f(0) = 0 and f′(0) = 1. Then

A.f(z) = z for all z✓ Correct
B.f may be any rotation
C. $f(z) = z^{2}$ is possible
D.no such f exists

The worked solution and the per-option analysis are in Trap Analysis.