Winding number

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Let f(z)f(z) be analytic in the annulus Ω={1<∣z∣<2}\Omega = \left\{ 1 < \left|z\right| < 2\right\}. Assume that ff has no zeros in Ω\Omega. Show that there exists an integer nn and an analytic function gg in Ω\Omega such that, for all z∈Ωz \in \Omega, f(z)=zneg(z)f(z) = z^{n} e^{g(z)}.

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Let SS be the unilateral shift and let ff be an analytic function in a neighborhood of cl D\mathrm{cl}\ \mathbb {D} such that f(z)≠0f(z) \ne 0 if ∣z∣=1\left|z\right| = 1. Let γ(t)=f(exp⁡(2πit))\gamma (t) = f(\exp (2\pi i t)), 0≤t≤10 \le t \le 1. Show that σe(f(S))=f(∂D)={γ(t):0≤t≤1}\sigma_e(f(S)) = f(\partial \mathbb {D}) = \left\{ \gamma (t): 0 \le t \le 1\right\} and that if λ∉f(∂D)\lambda \notin f(\partial \mathbb {D}), ind⁡(f(S)−λ)=−n(γ;λ)\operatorname {ind}(f(S) - \lambda ) = -n(\gamma ; \lambda ), where n(γ;λ)=n(\gamma ; \lambda ) = the winding number of γ\gamma about λ\lambda. Moreover, show that if ind⁡(f(S)−λ)=0\operatorname {ind}(f(S) - \lambda ) = 0, then λ∉σ(f(S))\lambda \notin \sigma (f(S)).

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What are the components of the set of invertible elements in C(∂D)C(\partial \mathbb {D})?

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Compute I=∮Cx dy−y dxx2+y2I=\displaystyle \oint_{C} \dfrac {x\, dy-y\, dx}{x^{2}+y^{2}}, where CC is a simple closed contour traversed in the positive direction and not passing through the origin.

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Compute I=12π∮CX dY−Y dXX2+Y2I=\dfrac {1}{2\pi }\displaystyle \oint_{C} \dfrac {X\, dY-Y\, dX}{X^{2}+Y^{2}}, where X=ax+byX=ax+by, Y=cx+dyY=cx+dy, and CC is a simple closed contour enclosing the origin (ad−bc≠0ad-bc\neq 0).

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If the simple contour CC encloses the origin, X=φ(x,y)X=\varphi (x,y), Y=ψ(x,y)Y=\psi (x,y), and the curves φ(x,y)=0\varphi (x,y)=0 and ψ(x,y)=0\psi (x,y)=0 have several simple intersection points inside the contour CC, compute the integral II (see problem 4321).

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Find the circulation Γ\Gamma of the vector a→=grad⁡(arctan⁡yx)\overrightarrow {a}=\operatorname {grad}\left(\arctan \dfrac {y}{x}\right) along a closed contour CC: (1) CC does not encircle the OzOz axis; (2) CC encircles the OzOz axis.

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(a)
(b)

Let

Ω=D(0,1)∖{12,−12}. \Omega = D(0,1) \setminus \left\{ \frac12, -\frac12\right\} .

Find all analytic functions f:Ω→Ωf : \Omega \to \Omega with the following property: if γ\gamma is any cycle in Ω\Omega which is not homologous to zero (mod Ω\Omega), then f∗γf * \gamma is not homologous to zero (mod Ω\Omega).

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Find the value of the line integral ∫Cx dy−y dxx2+y2\displaystyle \int_{C} \dfrac {x \, dy-y \, dx}{x^{2}+y^{2}}, taken around a closed contour CC in the positive direction.

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The curves X(x,y)=0X(x,y)=0 and Y(x,y)=0Y(x,y)=0 have several simple points of intersection. Compute the value of the integral 12π∫CX dY−Y dXX2+Y2\dfrac {1}{2\pi }\displaystyle \int_{C} \dfrac {X\, dY-Y\, dX}{X^{2}+Y^{2}}, taken in the positive direction around a closed contour CC.

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We consider a closed, continuous, oriented curve in the zz-plane that avoids the origin. If, starting from an arbitrary point, zz describes the entire curve in the given direction (returning to its starting point) the argument of zz changes continuously and its total variation is a multiple, 2πn2\pi n, of 2π2\pi. The integer nn is called the winding number of the curve.

Every ray from the origin intersects the curve in question at least ∣n∣\left|n\right| times.

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Here LL denotes a closed continuous curve without double points and D\mathfrak {D} the closed interior of LL. The winding number of a closed, continuous, oriented curve avoiding the origin is the integer nn such that the argument of ww changes by 2πn2\pi n as ww describes the entire curve once in the given direction.

The functions φ(z)\varphi (z) and ψ(z)\psi (z) are regular in D\mathfrak {D}, except possibly at finitely many poles, finite and different from 00 on the boundary LL of D\mathfrak {D}. We define f(z)=φ(z)ψ(z)f(z)=\varphi (z)\psi (z). The winding number of the image under w=f(z)w=f(z) of LL is equal to the sum of the winding numbers of the images of LL under w=φ(z)w=\varphi (z) and w=ψ(z)w=\psi (z) respectively.

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The winding number of a closed, continuous, oriented curve avoiding the origin is the integer nn such that the argument of ww changes by 2πn2\pi n as ww describes the entire curve once in the given direction.

The function f(z)f(z) is single-valued, regular and does not assume a certain value aa in the annulus r<∣z∣<Rr<\left|z\right|<R. All the closed continuous curves without double points that enclose the circle ∣z∣=r\left|z\right|=r and that lie inside the annulus are mapped by the function w=f(z)−aw=f(z)-a onto curves in the ww-plane with the same winding number.

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Let γ\gamma be a continuously differentiable closed curve in the complex plane, with parameter interval [a,b][a, b], and assume that γ(t)≠0\gamma (t) \neq 0 for every t∈[a,b]t \in [a, b]. Define the index of γ\gamma to be

Ind⁡(γ)=12πi∫abγ′(t)γ(t)dt \operatorname {Ind}(\gamma ) = \frac{1}{2 \pi i} \int _{a}^{b} \frac{\gamma ^{\prime }(t)}{\gamma (t)} d t

Prove that Ind (γ)(\gamma ) is always an integer. Hint: There exists φ\varphi on [a,b][a, b] with φ′=γ′/γ,φ(a)=0\varphi^{\prime } = \gamma^{\prime } / \gamma , \varphi (a) = 0. Hence γexp⁡(−φ)\gamma \exp (-\varphi ) is constant. Since γ(a)=γ(b)\gamma (a) = \gamma (b) it follows that exp⁡φ(b)=exp⁡φ(a)=1\exp \varphi (b) = \exp \varphi (a) = 1. Note that φ(b)=2πi\varphi (b) = 2 \pi i Ind (γ)(\gamma ).

Compute Ind (γ)(\gamma ) when γ(t)=eint,a=0,b=2π\gamma (t) = e^{i n t}, a = 0, b = 2 \pi. Explain why Ind (γ)(\gamma ) is often called the winding number of γ\gamma around 0.

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Let γ\gamma be as in Exercise 23, and assume in addition that the range of γ\gamma does not intersect the negative real axis. Prove that Ind (γ)=0(\gamma ) = 0. Hint: For 0≤c<∞0 \leq c < \infty, Ind (γ+c)(\gamma +c) is a continuous integer-valued function of cc. Also, Ind (γ+c)→0(\gamma +c) \rightarrow 0 as c→∞c \rightarrow \infty.

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Suppose γ1\gamma_{1} and γ2\gamma_{2} are curves as in Exercise 23, and

∣γ1(t)−γ2(t)∣<∣γ1(t)∣(a≤t≤b) \left|\gamma _{1}(t)-\gamma _{2}(t)\right| < \left|\gamma _{1}(t)\right| \quad (a \leq t \leq b)

Prove that Ind (γ1)=\left(\gamma_{1}\right) = Ind (γ2)\left(\gamma_{2}\right). Hint: Put γ=γ2/γ1\gamma = \gamma_{2} / \gamma_{1}. Then ∣1−γ∣<1\left|1-\gamma \right| < 1, hence Ind (γ)=0(\gamma ) = 0, by Exercise 24. Also,

γ′γ=γ2′γ2−γ1′γ1 \frac{\gamma ^{\prime }}{\gamma } = \frac{\gamma _{2}^{\prime }}{\gamma _{2}}-\frac{\gamma _{1}^{\prime }}{\gamma _{1}}
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Let γ\gamma be a closed curve in the complex plane (not necessarily diffierentiable) with parameter interval [0,2π][0,2 \pi ], such that γ(t)≠0\gamma (t) \neq 0 for every t∈[0,2π]t \in [0,2 \pi ].

Choose δ>0\delta > 0 so that ∣γ(t)∣>δ\left|\gamma (t)\right| > \delta for all t∈[0,2π]t \in [0,2 \pi ]. If P1P_{1} and P2P_{2} are trigonometric polynomials such that ∣Pȷ(t)−γ(t)∣<δ/4\left|P_{\jmath }(t)-\gamma (t)\right| < \delta / 4 for all t∈[0,2π]t \in [0,2 \pi ] (their existence is assured by Theorem 8.15), prove that

Ind⁡(P1)=Ind⁡(P2) \operatorname {Ind}\left(P_{1}\right) = \operatorname {Ind}\left(P_{2}\right)

by applying Exercise 25. Define this common value to be Ind (γ)(\gamma ). Prove that the statements of Exercises 24 and 25 hold without any differentiability assumption.

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Let ff be a continuous complex function defined in the complex plane. Suppose there is a positive integer nn and a complex number c≠0c \neq 0 such that

lim⁡∣z∣→∞z−nf(z)=c \lim _{\left|z\right| \rightarrow \infty } z^{-n} f(z) = c

Prove that f(z)=0f(z) = 0 for at least one complex number zz. Note that this is a generalization of Theorem 8.8. Hint: Assume f(z)≠0f(z) \neq 0 for all zz, define

γr(t)=f(rett) \gamma _{r}(t) = f\left(r e^{t t}\right)

for 0≤r<∞,0≤t≤2π0 \leq r < \infty , 0 \leq t \leq 2 \pi, and prove the following statements about the curves γr\gamma_{r} :

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(a)

Ind (γ0)=0\left(\gamma_{0}\right) = 0

(b)

Ind (γr)=n\left(\gamma_{r}\right) = n for all sufficiently large rr.

(c)

Ind (γr)\left(\gamma_{r}\right) is a continuous function of rr, on [0, ∞\infty).

Show that (a),(b)(a),(b), and (c)(c) are contradictory, since n>0n > 0.

Let DD be the closed unit disc in the complex plane. (Thus z∈Dz \in D if and only if ∣z∣≤1\left|z\right| \leq 1). Let gg be a continuous mapping of DD into the unit circle TT. (Thus, ∣g(z)∣=1\left|g(z)\right| = 1 for every z∈Dz \in D).

Prove that g(z)=−zg(z) = -z for at least one z∈Tz \in T. Hint: For 0≤r≤1,0≤t≤2π0 \leq r \leq 1,0 \leq t \leq 2 \pi, put

γr(t)=g(rett) \gamma _{r}(t) = g\left(r e^{t t}\right)

and put ψ(t)=e−ttγ1(t)\psi (t) = e^{-t t} \gamma_{1}(t). If g(z)≠−zg(z) \neq -z for every z∈Tz \in T, then ψ(t)≠−1\psi (t) \neq -1 for every t∈[0,2π]t \in [0,2 \pi ]. Hence Ind (ψ)=0(\psi ) = 0, by Exercises 24 and 26. It follows that Ind (γ1)=1\left(\gamma_{1}\right) = 1. But Ind (γ0)=0\left(\gamma_{0}\right) = 0. Derive a contradiction, as in Exercise 27.

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As in Example 10.36, consider the 1 -form

η=xdy−ydxx2+y2 \eta = \frac{x d y-y d x}{x^{2}+y^{2}}

in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}.

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(a)

Carry out the computation that leads to formula (113), and prove that dη=0d \eta = 0.

(b)

Let γ(t)=(rcos⁡t,rsin⁡t)\gamma (t) = (r \cos t, r \sin t), for some r>0r > 0, and let Γ\Gamma be a C′′\mathscr {C}^{\prime \prime }-curve in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}, with parameter interval [0,2π][0,2 \pi ], with Γ(0)=Γ(2π)\Gamma (0) = \Gamma (2 \pi ), such that the intervals [γ(t)[\gamma (t), Γ(t)]\Gamma (t)] do not contain 0 for any t∈[0,2π]t \in [0,2 \pi ]. Prove that

∫Γη=2π \int _{\Gamma } \eta = 2 \pi

Hint: For 0≤t≤2π,0≤u≤10 \leq t \leq 2 \pi , 0 \leq u \leq 1, define

Φ(t,u)=(1−u)Γ(t)+uγ(t). \Phi (t, u) = (1-u) \Gamma (t)+u \gamma (t).

Then Φ\Phi is a 2 -surface in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\} whose parameter domain is the indicated rectangle. Because of cancellations (as in Example 10.32),

∂Φ=Γ−γ \partial \Phi = \Gamma -\gamma

Use Stokes' theorem to deduce that

∫Γη=∫γη \int _{\Gamma } \eta = \int _{\gamma } \eta

because dη=0d \eta = 0.

(c)

Take Γ(t)=(acos⁡t,bsin⁡t)\Gamma (t) = (a \cos t, b \sin t) where a>0,b>0a > 0, b > 0 are fixed. Use part (b) to show that

∫02πaba2cos⁡2t+b2sin⁡2tdt=2π \int _{0}^{2 \pi } \frac{a b}{a^{2} \cos ^{2} t+b^{2} \sin ^{2} t} d t = 2 \pi
(d)

Show that

η=d(arctan⁡yx) \eta = d\left(\arctan \frac{y}{x}\right)

in any convex open set in which x≠0x \neq 0, and that

η=d(−arctan⁡xy) \eta = d\left(-\arctan \frac{x}{y}\right)

in any convex open set in which y≠0y \neq 0. Explain why this justifies the notation η=dθ\eta = d \theta, in spite of the fact that η\eta is not exact in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}.

(e)

Show that (b) can be derived from (d).

(f)

If Γ\Gamma is any closed C′\mathscr {C}^{\prime }-curve in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}, prove that

12π∫Γη=Ind⁡(Γ) \frac{1}{2 \pi } \int _{\Gamma } \eta = \operatorname {Ind}(\Gamma )

(See Exercise 23 of Chap. 8 for the definition of the index of a curve).

Suppose Γ\Gamma is a closed curve in the plane, with parameter interval [0,2π][0,2 \pi ]. Take α∉Γ∗\alpha \notin \Gamma^{*}. Approximate Γ\Gamma uniformly by trigonometric polynomials Γn\Gamma_{n}. Show that Ind⁡Γn(α)=Ind⁡Γm(α)\operatorname {Ind}_{\Gamma_{n}}(\alpha ) = \operatorname {Ind}_{\Gamma_{m}}(\alpha ) if mm and nn are sufficiently large. Define this common value to be Ind⁡Γ(α)\operatorname {Ind}_{\Gamma }(\alpha ). Prove that the result does not depend on the choice of {Γn}\left\{ \Gamma_{n}\right\}; prove that Lemma 10.39 is now true for closed curves, and use this to give a different proof of Theorem 10.40.

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Let Ω\Omega be the plane minus two points, and show that some closed paths Γ\Gamma in Ω\Omega satisfy assumption (1) of Theorem 10.35 without being null-homotopic in Ω\Omega.

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Suppose f∈H(Ω)f \in H(\Omega ), Γ\Gamma is a cycle in Ω\Omega such that Ind⁡Γ(α)=0\operatorname {Ind}_{\Gamma }(\alpha ) = 0 for all α∉Ω,∣f(ζ)∣≤1\alpha \notin \Omega ,\left|f(\zeta )\right| \leq 1 for every ζ∈Γ∗\zeta \in \Gamma^{*}, and Ind⁡Γ(z)≠0\operatorname {Ind}_{\Gamma }(z) \neq 0. Prove that ∣f(z)∣≤1\left|f(z)\right| \leq 1.

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Suppose A=C(T)A = C(T), the algebra of all continuous complex functions on the unit circle TT, with the supremum norm. Show that two invertible members of C(T)C(T) are in the same coset of G1G_{1} if and only if they are homotopic mappings of TT into the set of all nonzero complex numbers. Deduce from this that G/G1G / G_{1} is isomorphic to the additive group of the integers. (The notation is as in Theorem 10.34).

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Prove the equality

Δγarg⁡z=Δγarg⁡(z−1) \Delta _{\gamma } \arg z = \Delta _{\gamma } \arg (z-1)

for any simple piecewise-smooth closed curve γ\gamma such that the domain bounded by the curve γ\gamma contains the segment [0,1][0,1].

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