Muestra

f = lim

n→∞

Z

T

fn = (4.4) lim

n→∞0 = 0

The first equality follows by Lemma 2.7 p. 71 since T is closed and rectifiable and since each{ fⁿ} ∈ A(G), we have fnis continuous for each fnand by the uniform convergence to a function f , we get f is continuous (Theorem 6.1 p. 29).

Finally, we can apply Morera’s Theorem 5.10 p. 86 to obtain that f is analytic in G, since f is continuous andR

γ f = 0 for every triangular path T in G.

Exercise 9. Show that if f : C→ C is a continuous function such that f is analytic off [−1, 1] then f is an entire function.

Solution. By Morera’s Theorem, it suffice to show that for every triangular path T in C, we haveR

T f = 0, since we already assume f is continuous. There are several cases to consider:

Case 1: A triangle does not intersect I. In this case, we obtain by Cauchy’s Theorem, thatR

T f = 0, since we can find an open neighborhood G containing T (is closed and rectifiable) such that f ∈ A(G) (by assumption) and n(T ; w) = 0∀w ∈ C\G (by Theorem 4.4 p. 82).

Case 2: A triangle touches I exactly at one point P. This single point of intersection P is a removable singularity, since f is continuous. Again apply Cauchy’s Theorem to obtainR

T f = 0. (Procedure: Translate the triangle by±i depending if it lies above or below the x-axis. By Cauchy’s TheoremR

T f = 0 over the translated triangle and let → 0 since f is continuous).

Case 3: One edge of the triangle touches I. Like in case 2 we can translate the triangle by± We can also argue this way: Let G be an open neighborhood containing the triangle T . Let{Tⁿ} be a sequence of triangles that are not intersecting I, but whose limit is the given T . Then by the continuity of f , we get

Z

T

f = lim

n→∞

Z

Tn

f = lim

n→∞0 = 0, where the first equality follows by Lemma 2.7 and the second one by case 1.

Case 4: A triangle T is cut into 2 parts by I. Then, we can always decompose the triangle T in three parts.

Two triangles are of the kind explained in case 3 and one is of the kind explained in case 2 (a sketch might help). Thus,R

T f = 0 again.

Case 5: A triangle T contains parts of I (also here a sketch might help). In this instance, we can decompose the triangle into 5 parts. Two triangles are of the kind explained in case 3 and the other three triangles are of the kind explained in case 2. Hence,R

T f = 0.

Summary: SinceR

T f = 0 for every triangular path in C and f is continuous, we get by Morera’s Theorem:

f ∈ A(C), that is f is an entire function.

Exercise 10. Use Cauchy’s Integral Formula to prove the Cayley–Hamilton Theorem: If A is an n× n matrix over C and f (z) = det(z− A) is the characteristic polynomial of A then f (A) = 0. (This exercise was taken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390–391).

Solution. Not available.

4.6 The homotopic version of Cauchy’s Theorem and simple connec-tivity

Exercise 1. Let G be a region and let σ₁, σ₂ : [0, 1] → G be the constant curves σ1(t) ≡ a, σ2(t) ≡ b.

Show that if γ is closed rectifiable curve in G and γ∼ σ¹, then γ∼ σ². (Hint: connect a and b by a curve.)

Solution. Assume γ is a closed and rectifiable curve in G and γ∼ σ¹. Since σ1(t)≡ a and σ²(t) ≡ b, we

Exercise 2. Show that if we remove the requirement “Γ(0, t) = Γ(1, t) for all t” from Definition6.1 then the curve γ₀(t) = e^2πit, 0≤ t ≤ 1, is homotopic to the constant curve γ1(t)≡ 1 in the region G = C − {0}.

Solution. Not available.

Exercise 3. 3. LetC = all rectifiable curves in G joining a to b and show that Definition 6.11 gives an equivalence relation onC.

Solution. Not available.

Exercise 4. Let G = C− {0} and show that every closed curve in G is homotopic to a closed curve whose trace is contained in{z : |z| = 1}.

Solution. Not available.

Exercise 5. Evaluate the integralR

γ dz

z²+1where γ(θ) = 2| cos 2θ|e^iθfor 0≤ θ ≤ 2π.

Solution. A sketch shows that the two zeros of z²+ 1 (they are±i) are inside of the closed and rectifiable curve γ (The region looks like a clover with four leaves ). Using partial fraction decompositions gives

Z

since n(γ; i) = n(γ;−i) since i and −i are contained in the region generated by γ. Hence Z

Solution. A sketch reveals that the zero−πi is inside the region and the zero πi is outside the region. Using partial fraction decomposition yields

Z

since iπ is not contained in the region generated by γ, so n(γ; iπ) = 0 and−iπ is contained in the region, so

n(γ;−iπ) = 1. Hence, Z

γ

dz

z²+π² =−1.

Exercise 7. Let f (z) = [(z−¹2−i)· (z−1−³2i)· (z−1−2ⁱ)· (z−³2−i)]⁻¹and let γ be the polygon [0, 2, 2+2i, 2i, 0].

FindR

γ f .

Solution. Not available.

Exercise 8. Let G = C− {a, b}, a , b, and let γ be the curve in the figure below.

(a) Show that n(γ; a) = n(γ; b) = 0.

(b) Convince yourself that γ is not homotopic to zero. (Notice that the word is “convince” and not “prove”.

Can you prove it?) Notice that this example shows that it is possible to have a closed curve γ in a region such that n(γ; z) = 0 for all z not in G without γ being homotopic to zero. That is, the converse to Corollary 6.10 is false.

Solution. Let γ be the path depicted on p. 96. We can write it as a sum of 6 paths. Two of them will be closed and have a and b in their unbounded component. Therefore, two integrals will be zero and we will have another 2 pair of non-closed paths. The first pair begins at the leftmost crossing pair and goes around a in opposite direction. The second pair begins at the middle crossing pair and goes around b in opposite direction. They also meet at the rightmost crossing point. If we integrate over the path around a is equivalent to evaluateR

γ1−γ2

1

z−a dz where γ1(t) = a + re^it, 0≤ t ≤ π and γ²(t) = a + re^it, π≤ t ≤ 2π for some r > 0. It is easily seen that we have

Z

γ₁

1 z− a dz =

Z

γ₂

1

z− a dz = πi and therefore n(γ; a) = 0 and similarly we obtain n(γ; b) = 0.

Exercise 9. Let G be a region and let γ0and γ1be two closed smooth curves in G. Suppose γ0∼ γ¹, and Γ satisfies (6.2). Also suppose that γt(s) = Γ(s, t) is smooth for each t. If w∈ C − G define h(t) = n(γ^t; w) and show that h : [0, 1]→ Z is continuous.

Solution. Not available.

Exercise 10. Find all possible values ofR

γ dz

1+z², where γ is any closed rectifiable curve in C not passing through±i.

Solution. Let γ be a closed rectifiable curve in C not passing through±i. Using partial fraction decompo-sition and the definition of the winding number we obtain

Z

γ

dz 1 + z²

= 1 2i

Z

γ

1 z− 1 dz−

Z

γ

1 z + i dz

!

= 1

2i(2πin(γ; i)− 2πin(γ; −i)) = π (n(γ; i) − n(γ; −i)) . Exercise 11. EvaluateR

γ e^z−e^−z

z⁴ dz where γ is one of the curves depicted below. (Justify your answer.) Solution. Using Corollary 5.9 p. 86 we have∀a ∈ G − {γ}

Z

γ

f (z)

(z− a)^k+1 dz = 2πi f^(k)(a)n(γ; a)1 k!,

where γ is a closed rectifiable curve in G (open) such that n(γ; w) = 0 for all w∈ C − G and f : G → C is analytic.

In our case a = 0 and γ satisfies the above assumptions for exercise a), b) and c). Let f (z) = e^z− e^−zwhich is clearly analytic on any region G. We have f⁽³⁾(0) = e⁰+ e⁻⁰= 2 and using the formula above for k = 3

we obtain Z

γ

e^z− e^−z

z⁴ dz = 2πi· 2n(γ; 0)1 6 =2πi

3 n(γ; 0).

Hence, n(γ; 0) = 1 for part a) and thus the result is ^2πi₃ . We have n(γ; 0) = 1 for part b) and thus the result is^4πi₃ . We obtain the same result for part c).