How many zeros does have inside ? And inside ?
No formula finds these roots — but Rouché's theorem counts them without finding them.
Zeros and Mappings
1. Argument principle, Rouché's theorem, open mapping
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).
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)
Part B2023 Decembershow ▾How many roots does the polynomial have in the open disc { : |z| < 1}?
- A.100
- B.50
- C.30✓
- D.0
Solution
On |z| = 1, || = 50 > 1 + 40 + 6 + 1 = 48 ≥ ||, so by Rouché the polynomial has as many zeros inside as : thirty.
Part B2023 Decembershow ▾Previous-year question — unlock to view
Part Cshow ▾Let . Which of the following are true?
- A.All five zeros of p lie in |z| < 2.✓
- B.p has exactly one zero in |z| < 1/2.✓
- 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.✓
Solution
(1) On |z| = 2: || = 32 > 7 ≥ |3z + 1|, so Rouché against gives all five zeros inside. (2) On |z| = 1/2 the dominant term is 3z + 1: its minimum modulus there is 3·(1/2) − 1 = 1/2, while || = 1/32 < 1/2; Rouché against 3z + 1 gives exactly one zero (near −1/3) in |z| < 1/2. (3) fails: p(−1) = −3 < 0 and p(0) = 1 > 0, so the intermediate value theorem places a real zero in (−1, 0). (4) is the standard annulus argument: (zeros inside big circle) − (zeros inside small circle).
2. Conformal maps, Möbius transformations, Schwarz lemma
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.
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.
Part Bshow ▾The image of the upper half-plane under the map w = (z − i)/(z + i) is
- A.the unit disc |w| < 1✓
- 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.
Part Bshow ▾Let f : 𝔻 → 𝔻 be holomorphic with f(0) = 0 and f′(0) = 1. Then
- A.f(z) = z for all z✓
- B.f may be any rotation
- C. is possible
- D.no such f exists
Solution
Schwarz lemma: |f′(0)| = 1 forces f(z) = , and f′(0) = 1 gives .
Part Cshow ▾Which of the following are true about Möbius transformations?
- A.They map circles and lines to circles and lines✓
- B.They preserve the cross-ratio of four points✓
- C.Every Möbius transformation has at least one fixed point on the Riemann sphere✓
- D.They are conformal at every point of
Solution
(d) fails at the pole z = −d/c, where the map is not defined on it is conformal as a map of the sphere).
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.✓
- B.A Möbius transformation other than the identity has at most two fixed points in {}.✓
- C.Given two triples of distinct points, exactly one Möbius transformation maps the first triple to the second in order.✓
- D.T(z) = 1/z maps the punctured disc {0 < |z| < 1} onto the open unit disc.
Solution
(1) and (2) are standard — the fixed-point equation cz is at most quadratic. (3) is three-point transitivity via cross-ratios. (4) fails: 1/z sends 0 < |z| < 1 onto |w| > 1, the exterior of the closed disc.