By Hardy G.H., Rogosinski W.W.

Vintage, graduate-level textual content discusses the Fourier sequence in Hilbert area, examines extra houses of trigonometrical Fourier sequence, and concludes with an in depth examine the purposes of formerly defined theorems

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M−1)! × 1 a b 2i +∞ (−1)m (m+n−1)! (m−1)! n=0 = 1 2i z−i m+n a b +∞ 2m+p−1)! (m−1)! p=−m = +∞ ap z − i = p=−m a b a b z−i n n 1 a b 2i n−m a b 2m+p z−i a b p p , which is the Laurent series expansion of f (z) in the set 0< z−i of centrum i a <2 b a , b a and radius R = 2 b a . e. a−1 = 1 (2m − 2)! (m − 1)! (−1)m−1 2i a b 2m−1 = 1 bm 2m − 2 m−1 1 · 2 a b 2m−1 1 · . i (c) We get from 1 1 ≤ m 2 2 |a + bz | (bR − a) for |z| = R > a , b the estimate lim R→+∞ CR dz 1 m ≤ lim m · πR = 0, R→+∞ (bR2 − a) (a + bz 2 ) proving that lim R→+∞ CR dz m = 0.

Com 37 Complex Funktions Examples c-6 Line integrals computed by means of residues 2 1 –2 –1 1 2 –1 –2 Figure 3: The curve |z| = 2 with the two poles insider. Determination of res f ; − A(z) = 1 z− π π . The pole z = − is simple. g. 2 2 and π 2 B(z) = cos z. Then π res f ; − 2 π 2 π − 2 A − = B = 1 π π π · − sin − − − 2 2 2 1 =− . π Alternatively we apply Rule I. Then res f ; − π 2 = 1 z− limπ z→− 2 = − 1 · π π 2 · z + π2 1 = π cos z −2 − 1 limz→− π2 d cos z dz 1 cos z − cos − π2 limz→− π2 z − − π2 1 1 1 =− · =− .

Z n z z z z 2 z3 for z = 0. 2 Find the residues at ∞ of the following functions: (a) 1 , z (1 − z 2 ) (b) (a) The function f (z) = res z4 (z 2 + 1) (c) z 2n , (1 + z)n 1 ;∞ z (1 − z 2 ) = 0. ⎛ 1 ;∞ z (1−z 2 ) n ∈ N. 1 has a zero of order 3 at ∞, so z (1 − z 2 ) Alternatively, res 2, ⎞ ⎜1 = −res ⎜ ⎝ z2 · 1 z 1 1− 1 z2 ⎟ ; 0⎟ ⎠ = −res z z 2 −1 ;0 = 0. (b) The Laurent series expansion of the function f (z) = z4 (z 2 + 1) 2 only contains even powers of z, so a−1 = 0, and thus res z4 (z 2 + 1) 2 ;∞ = 0.

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