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HOMEWORK 5.
Due Monday, April 6.
- Do Problem 1 on page 80 of the text.
Solution: Let M (r) = sup{|f (z)| : |z| = r}. If r > R then M (r) ≤ M rn
so |f (z)| ≤ M rn^ for z ∈ B(0, r). By Thm 2.14 |f (m)(0)| ≤ m!^ M r
n rm^ →^ 0, as r → ∞, if m > n.
- Do Problem 9 on page 80 of the text.
Solution: By Thm 2.30, there exists a function V so that F = U + iV is an entire function. Then e−F^ is entire and |e−F^ | = e−U^ ≤ 1, so by Liouville’s thm e−F^ (z)^ = C. It folows that e−U^ = |C| and U = − log |C|.
- Do Problem 10 on page 80 of the text.
Solution: Suppose ∃a ∈ G s.t. g(a) 6 = 0. By continuity, ∃r > 0 s.t. g 6 = 0 in B(a, r), whence 1/g is analytic in B(a, r). It follows that p(z) = |f (z)|^2 is analytic in B(a, r), since p = f · (f g) · (1/g). Now p ∈ R and p′(z) =
limh→ 0 p(z^ +^ h h)^ −^ p(z) so, if h ∈ R then Im{p′(z)} = 0 while if h ∈ iR then Re{p′(z)} = 0. This implies that |f (z)| = A for z ∈ B(a, r) and, consequently, f (B(a, r)) ⊂ {z : |z| = A}. By the Maximum Modulus, f is constant in B(a, r), and by Uniqueness thm in G.
- Do Problem 2 on page 83 of the text.
Solution: Let zn = 1/n^2 and define Cn = {z : |z − zn| = 1/n^2 }. Now let γ be the curve that is obtained by tracing, in succession, the circles Cn.
- Do Problem 8 on page 87 of the text.
Solution: Since f is continuous, Lemma 2.7 (p.71) implies that, for any closed rectifiable γ,
γ f^ = lim^
γ fn^ = 0. By Morera’s thm,^ f^ is analytic.
- Do Problem 4 on page 95 of the text.
Solution: For γ a closed, rectifiable curve in G, define Γ(s, t) = γ(s)
t + |^1 γ^ −(s^ )t|
- Do Problem 11 on page 96 of the text.
Solution: Apply Corollary 5.9 (p. 86) with f (z) = ez^ − e−z^ , G = C, a = 0,
and k = 3. Then
γ
ez^ − e−z z^4 dz^ =
2 πi 3 n(γ,^ 0). Answers: (a) 2πi/3; (b) and (c) 4πi/3.
- ∫ Let G be a region and f : G → C a continuous function. Suppose that
L f^ = 0 whenever^ L^ is the boundary of a rectangle whose sides are parallel to the coordinate axes. Prove or disprove: f is analytic in G.
Solution: Let z = x+iy ∈ G, select a square in G that contains z, and denote its lower left vertex by a+ib. Let Γ be the union of line segments [a+ib, x+ib] and [x + ib, x + iy], and let Γ′^ be the union of [a + ib, a + iy] and [a + iy, x + iy]. Let F (z) =
Γ f^ =^
Γ′^ f^. Now^ F^ (z) =^
Γ f^ =^
∫ (^) x a f^ (t^ +^ ib)^ dt^ +^ i^
∫ (^) y b f^ (x^ +^ it)^ dt so Fy = if ; also F (z) =
Γ′^ f^ =^ i^
∫ (^) y b f^ (a^ +^ it)^ dt^ +^
∫ (^) x a f^ (t^ +^ iy)^ dt^ so^ Fx^ =^ f^. Write F = U + iV and notice that Fx = −iFy is equivalent to U, V satisfying C-R conditions.
