# Download Complex Functions Examples c-7 - Applications of the by Leif Mejlbro PDF

By Leif Mejlbro

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Extra info for Complex Functions Examples c-7 - Applications of the Calculus of Residues

Example text

E. the simple poles are 3i, −3i, i and − i. com 52 Complex Funktions Examples c-7 Improper integral, where the integrand is a rational function None of these lies on the x-axis. Since f (z) is analytic outside the poles and since we have the estimate |f (z)| ≤ C |z|2 for |z| ≥ 4 og Im(z) ≥ 0, we conclude that +∞ −∞ x2 − x + 2 dx = 2πi {res(f ; i) + res(f ; 3i)} x4 + 10x2 + 9 z2 − z + 2 z2 − z + 2 + lim z→i (z + i) (z 2 + 9) z→3i (z 2 + 1) (z + 3i) π 5π {3 − 3i + 9 + 31 − 2} = . 24 12 = 2π = lim −1 − i + 2 −9 − 3i + 2 + 2i · 8 −8 · 6i = 2πi Alternatively, one may apply the traditional real method of integration, by using that we have proved that the integral exists.

1 3 3 1 has the simple poles +1 π π z1 = −1, z2 = exp −i , z3 = exp i . 3 3 π lies inside γR , so it follows by the residuum theorem that If R > 1, then only z3 = exp i 3 1) The function f (z) = f (z) dz γR z3 1 2πi z3 = · 3z32 3 z33 √ 2πi 1 3 π √ =− +i = 3−i . com 62 Complex Funktions Examples c-7 Improper integral, where the integrand is a rational function 2) We get along II the estimate II 1 2π dz ≤ 3 · R→0 R −1 3 z3 + 1 for R → +∞. 3) Along III we choose the parametric description (R − r) exp i 2π 3 r ∈ [0, R], , so R f (z) dz = 0 III 2π 2π 3 dr = − exp i 3 1 + (R − r)3 exp(2πi) − exp i R dx .

We employ more than 1500 people worldwide and offer global reach and local knowledge along with our all-encompassing list of services. 6 1) Find the domain of analyticity of the function f (z) = Log z . z2 − 1 Explain why f has a removable singularity at z = 1. 2) Let Cr,R denote the simple, closed curve on the ﬁgure, where 0 < r < R < +∞. 2 Compute the line integral f (z) dz. g. by letting r → 0+ and R → +∞ in (1). 1) Clearly, f is deﬁned and analytic, when z ∈ C \ (R− ∪ {0, 1}) , and the singularity at z = 1 is at most a simple pole, +∞ f (z) = res(f ; 1) Log z = + an (z − 1)n , z−1 z2 − 1 n=0 0 < |z − 1| < 1.