Find the laurent series of the function

$$f(z)=\exp\left[\frac{\lambda}{2}(z-\frac{1}{z})\right]\quad \text{as}\quad \sum_{n=-\infty}^{n=\infty}C_{n}z^{n}\quad\text{ for}\;\;0<\left |z\right |<\infty$$

where $$C_{n}=\frac{1}{\pi}\int_{0}^{\pi}\cos(n\phi-\lambda \sin\phi)d\phi\quad n=0,\pm1,\pm2,\cdots$$

with $\lambda$ a given complex number and and taking the unit circle C given by $z=e^{i\phi}(-\pi\leq\phi\leq\pi)$ as contour in this region


1 Answer 1


Starting from the formula for the Laurent coefficients, we obtain $$ C_n=\frac 1 {2\pi i}\int_C\frac {f(z)} {z^{n+1}} \,dz=\frac 1 {2\pi i}\int_C \exp \left(\frac \lambda 2 \left(z- \frac 1 z \right)\right)z^{-n-1}\,dz=$$ $$ \frac 1 {2\pi i}\int_{-\pi}^\pi \exp \left(\frac \lambda 2 \left(e^{i\phi}- e{-i\phi} \right)\right)e^{-i\phi(n+1)}ie^{i\phi}\,d\phi=\frac 1 {2\pi} \int_{-\pi}^\pi \exp(i(\lambda \sin \phi -n\phi))\,d\phi.$$ Making use of the Euler formula $ e^{ix}=\cos x +i\sin x$ and the oddness of the sine function and the evenness of the cosine, we arrive to $$ C_n= \frac 1 {\pi} \int_0^\pi \cos(\lambda \sin \phi-n\phi)\,d\phi . $$

Using Maple, we obtain $$int(cos(n*phi-lambda*sin(phi)), phi = 0 .. Pi))/Pi$$ $$ assuming \,\,n::integer, n >= 0 $$ $${\it AngerJ} \left( n,\lambda \right) $$ and $$(int(cos(n*phi-lambda*sin(phi)), phi = 0 .. Pi))/Pi \,\, assuming \, n::integer,n<0 $$ $${\it AngerJ} \left( -n,-\lambda \right),$$ where ${\it AngerJ} \left( n,\lambda \right) $ is described here.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .