Question

Give the first four terms of the Laurent's expansion in the stated region: (sin(z) - z)/(z^2*cos (z)) in the region 0< abs(z) < pi*2

Answer 1

*

Answer 2

Ain't Laurent's series (if not stated otherwise) around (0,0) ?

Answer 3

around zero .... wolf calculates as
http://www.wolframalpha.com/input/?i=laurent+expansion+%28sin%28z%29+-+z%29%2F%28z^2*cos+%28z%29%29+at+z%3D0

Answer 4

Consider me a deserter on this one. Can do, but the efficiency will be not excellent.

Answer 5

Let \( z = re^{i\theta}\) then \( dz = r i e^{i\theta }\)
Let \( a_n \) be the coefficients of Laurent series
\[ \huge a_n = \frac1{2i\pi}\int _\gamma \frac{f(z)}{(z-0)^n} dz \\
\huge= \frac1{2i\pi}\int _\gamma \frac{\sin (r {e^{i \theta}) - r {e^{i \theta}}}}{r^{n+2}e^{i\theta {(n+2)}} \cos (re^{i\theta})} r ie^{i \theta}d\theta\]

Try to evaluate these using contour integral. I'm not sure if i should put \( r = 1 \) or \(r = 2 \pi \) since there are singularities in \( {\pi \over 2}\) and \( {3\pi \over 2}\).

I don't know any simpler method.

Answer 6

Woops, the limit of integration should be
\[ \huge \int_0^{2\pi}\]

Answer 7

also try this
\[ f(z) = \frac{\sin z - z}{z^2 \cos z}=\frac{-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}{z^2(1-\frac{z^2}{2!}+\frac{z^4}{‌​4!}-\ldots)}=\frac{-\frac{z}{3!}+\frac{z^3}{5!}-\ldots}{(1-\frac{z^2}{2!}+\frac{z‌​^4}{4!}-\ldots)} \]