By Daniel Alpay

ISBN-10: 3034800770

ISBN-13: 9783034800778

This is a set of routines within the concept of analytic services, with accomplished and distinctive strategies. we want to introduce the scholar to functions and facets of the idea of analytic services no longer consistently touched upon in a primary path. utilizing acceptable workouts express the scholars a few elements of what lies past a primary direction in complicated variables. We additionally talk about subject matters of curiosity for electric engineering scholars (for example, the conclusion of rational services and its connections to the idea of linear structures and nation area representations of such systems). Examples of significant Hilbert areas of analytic features (in specific the Hardy area and the Fock house) are given. The publication additionally features a half the place proper evidence from topology, practical research and Lebesgue integration are reviewed.

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**Additional resources for A Complex Analysis Problem Book **

**Example text**

515]. 5. Problem 1, p. 214]. 1. Prove that cos π 3π 5π 7π 9π 1 + cos + cos + cos + cos = . 3. Computing some sums 27 Hint. Let C denote the sum to be computed, and let S = sin π 3π 5π 7π 9π + sin + sin + sin + sin . 11 11 11 11 11 Using de Moivre’s formula, compute C + iS. 2. Show that, for a, b ∈ R (b = 0 mod π) n−1 cos(a + (2k + 1)b) = k=0 cos(a + bn) sin(bn) . 2) What does this formula become when b = mπ for some m ∈ Z. What does this formula become when a and b are assumed to be in C? 2), prove that for every real u diﬀerent from 0, n−1 cosh(2k + 1)u = k=0 cosh(nu) sinh(nu) .

Hence we have cos x cosh y = 0 and sin x sinh y = 0. 6) The function cosh y has no real roots, and therefore we have cos x = 0, and thus sin x = ±1. 6) leads to sinh y = 0. Thus, x= 2k + 1 π, 2 k ∈ Z, and y = 0. 5 we have sin x cosh y = 5 and cos x sinh y = 0. In the second equation, y = 0 would lead to sin x = 5, which cannot hold for real x. Therefore, x = π/2 + 2kπ (k ∈ Z), and y = arcosh 5. (c) The best procedure is to solve the equation eiz − e−iz = 2iz0 , where we have set z0 = a + ib. This is equivalent to (eiz )2 − 2iz0 eiz − 1 = 0.

1 + cosh 2y This inequality never holds for real y. Finally the points ±i correspond to the limit of tan(±iy) as y → ±∞. This shows that the image of L under tan is included in the set C \ {z = it , t ∈ R and |t| ≥ 1}. To show that equality holds, we use the fact that the function sin z has range the whole complex plane (see the previous exercise). More precisely, let u ∈ C \ {z = it , t ∈ R and |t| ≥ 1}. Then tan z = ±u ⇐⇒ sin2 z = u2 (1 − sin2 z) ⇐⇒ sin2 z = u2 . 1 + u2 44 Chapter 1. Complex Numbers: Algebra 2 u The number z ∈ ± π2 + iR if and only if 1+u 2 is real and of modulus greater than 1, that is, if and only if u = it with t ∈ R and |t| > 1.