By Mejlbro L.

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1 − |r| ei t Hence, I(−r) = I(r), and we see that I(r) is not deﬁned for r = ±1, and that ⎧ for |r| < 1, ⎨ 2π 2π it 1+re dt = r ∈ R \ {−1, 1}. 2 Find the Laurent series of ϕ(θ) = 1 − r cos θ , 1 + r2 − 2r cos θ ψ(θ) = 1 , when |z| > 1. Then ﬁnd the Fourier series of 1−z r sin θ , 1 + r2 − 2r cos θ r > 1. It follows from |z| > 1 that 1 1 =− · 1−z z +∞ 1 1− 1 z =− +∞ 1 1 1 =− , n zn z n=0 z n=1 |z| > 1. Then put z = r ei θ , r > 1, to get 1 1 1 − r cos θ + i r sin θ 1 − r cos θ + i · r sin θ = = ϕ(θ) + i ψ(θ), = = i θ 2 2 1−z 1−re (1 − r cos θ) + (r sin θ) 1 + r2 − 2r cos θ and +∞ +∞ +∞ +∞ 1 1 1 −i n θ 1 1 =− =− e =− cos nθ + i sin nθ.

1) It follows from 6z 2 − z − 1 = 6 z − 1 2 z+ 1 3 , 1 1 1 1 that the singular points of the function are − and . Consequently, R1 = and R2 = , and 3 2 3 2 the annulus becomes {z ∈ C | 1 1 < |x| < }. 3 2 2) If |z| < R1 = f (z) = 1 , then 3 5 5z = · 6 6z 2 − z − 1 z z− 1 2 z+ 1 3 1 1 − 5 1 5 3 · 1 = · 2 · + · 1 1 1 1 6 1 1 6 + z− − − z+ 2 3 2 3 2 3 +∞ = 1 1 1 1 {(−3)n − 2n } z n . com 31 Complex Funktions Examples c-5 3) If Laurent series 1 1 < |z| < , then 3 2 f (z) = 1 1 1 1 + =− + · 2z − 1 1 + 3z 1 − 2z 3z +∞ = − 2n z n + n=0 +∞ 1 =− 1 1+ 3z 2n z n + n=0 1 3z +∞ n=0 (−1)n · 1 1 · n n z 3 +∞ (−1)n−1 1 · n.

If a−1 = 0, then +∞ f (z) = a−2 1 a−2 = 2 · n z z n=2 1 1 1− z = 1 a−2 · , z z−1 |z| > 1. 3 Find all Laurent series from 0, which are solution of the diﬀerential equation z 3 f (z) + 3z 2 + z f (z) + f (z) = 0, and the domain of convergence. Express the Laurent series by elementary functions. First method. Inspection. It is possible in some cases to solve a diﬀerential equation by inspection. Here we have an example of such an equation. This is, however, not always possible. We rearrange the equation in the following way, 0 = z 3 f (z) + (3z + z) f (z) + f (z) = z 3 f (z) + 3z 2 f (z) + {z f (z) + f (z)} d d d = z 3 f (z) + {z f (z)} = z 3 f (z) + z f (z) .

### Laurent Series, c-5 by Mejlbro L.

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