Complex Analysis: Theory and Applications · Problem 7.3.15

Taylor and Laurent Series — Howell & Matthews

Mathematics — Complex Analysis · laurent-series-fourier

Problem

Consider the real-valued function u(θ)=15−4cos⁡θu(\theta) = \dfrac{1}{5-4\cos\theta}. ((a)) Use the substitution cos⁡θ=12(z+1z)\cos\theta = \frac{1}{2}\left(z+\frac{1}{z}\right) and obtain u(θ)=f(z)=−z(z−2)(2z−1)=13(11−z2)−13(11−2z).u(\theta) = f(z) = \frac{-z}{(z-2)(2z-1)} = \frac{1}{3}\left(\frac{1}{1-\frac{z}{2}}\right) - \frac{1}{3}\left(\frac{1}{1-2z}\right). ((b)) Expand the function f(z)f(z) in part (a) in a Laurent series that is valid in the annulus A ⁣(0,12,2)A\!\left(0,\tfrac{1}{2},2\right). ((c)) Use the substitutions cos⁡(nθ)=12(zn+z−n)\cos(n\theta) = \frac{1}{2}(z^n + z^{-n}) in part (b) and obtain the Fourier series for u(θ)u(\theta), where u(θ)=13+13∑n=1∞2−n+1cos⁡(nθ)u(\theta) = \frac{1}{3} + \frac{1}{3}\displaystyle\sum_{n=1}^{\infty} 2^{-n+1}\cos(n\theta).

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