By Daniel Alpay

This can be a number of workouts within the conception of analytic capabilities, with accomplished and exact ideas. we want to introduce the coed to functions and points of the idea of analytic capabilities now not continually touched upon in a primary direction. utilizing acceptable workouts we want to exhibit to the scholars a few points of what lies past a primary direction in complicated variables. We additionally talk about issues of curiosity for electric engineering scholars (for example, the conclusion of rational capabilities and its connections to the speculation of linear platforms and kingdom house representations of such systems). Examples of vital Hilbert areas of analytic features (in specific the Hardy house and the Fock area) are given. The ebook additionally encompasses a half the place correct proof from topology, useful research and Lebesgue integration are reviewed.

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

**Example text**

A) Show that tan z = sin(2x) sinh(2y) +i . 18) (b) What is the image of the strip L = {(x, y) ; x ∈ (−π/2, π/2) and y ∈ R} L1 = {(x, y) ; x ∈ (−π/4, π/4) and y ∈ R} under the function tan? (c) What is the image of the strip under the function tan? 6 in connection with the previous exercise. 3 Computing some sums The following exercise is taken from [69, p. 515]. 5. Problem 1, p. 214]. 1. Prove that cos 3π 5π 7π 9π 1 π + cos + cos + cos + cos = . 3. Computing some sums 27 Hint. Let C denote the sum to be computed, and let S = sin 3π 5π 7π 9π π + sin + sin + sin + sin .

22. (a) It suﬃces to compute (1 + z + · · · + z N )(1 − z) = 1 + z + · · · + z N − (z + z 2 + · · · + z N +1 ) = 1 − z N +1 . 37) we have 1 1 ≤ |1 − z| 1 − |z| for z ∈ D. 54) follows. 53) to obtain N N z 2n z 2n ≤ 2 + z n + z 5n 2 + z n + z 5n n=0 n=0 N ≤ |z|2n 2(1 − |z|) n=0 = 1 − |z|2N +2 , 2(1 − |z|)(1 − |z|2 ) and hence the result since 1 1 , ≤ 2 1 − |z| 1 − |z| z ∈ D. 1. 3): |ez | = |ex (cos y + i sin y)| = |ex | · | cos y + i sin y| = ex = eRe ≤ e|z| . z 40 Chapter 1. 3. We write z = ρeiθ Thus and zj = ρ1/n ei n−1 n−1 zjm = ρm/n j=0 ei θ+2πj n mθ+2πmj n , j = 0, .

S = sin and set π 3π 5π 7π 9π + sin + sin + sin + sin , 11 11 11 11 11 iπ π π + i sin = e 11 . 11 11 = −1. 6. Solutions 45 = = 1 1− 1− (1 − )(1 − ) π π (1 − cos ) + i sin 1− 11 11 . = = π 2(1 − Re ) 2(1 − cos ) 11 Hence 1 C= 2 π 1 π 11 and S = π = 2 cot 22 . 2. We set Cn to be the sum to be computed and n−1 Sn = sin(a + (2k + 1)b). k=0 Since b = 0 (mod π) we have that e2ib = 1 and we can write n−1 ei(a+(2k+1)b) Cn + iSn = k=0 n−1 = ei(a+b) (e2ib )k k=0 1 − e2ibn = ei(a+b) 1 − e2ib eibn e−ibn − eibn = ei(a+b) ib e e−ib − eib eibn sin bn = ei(a+b) ib e sin b i(a+nb) sin bn .