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I need some help with this exercise: I need to obtain the power series development of this function: $$f(z)=\frac{\cos(z+1)}{(z^2-1)z}$$ Centered in $z_0=-1$ and valid in $z_1=\frac{1}{2}-i$

I know how to develop the power series of $\cos$ or $\frac{1}{1-\text{anything}}$ but this exercise is too difficult, I suppose.

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Hint: Express $\cos(z+1)$ as a power series in powers of $(z+1)$. Decompose $\frac{1}{(z^2-1)z}$ into a sum of partial fractions $\frac{A}{z+1}$, $\frac{B}{z-1}$, $\frac{C}{z}$. Express $\frac{1}{z}=-\frac{1}{1-(z+1)}$ as a power series in $(z+1)$. Similarly express $\frac{1}{z-1}=-\frac{1}{2}\frac{1}{1-\frac{z+1}{2}}$ as a series in $(z+1)$...

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Hint:$

Find the Laurent Series of $\cos(z+1)$ (Is centered in $z_0=-1$) and

$\frac{1}{z(z^2-1)} = \frac{1}{z(z-1)(z+1)} = \frac{1}{[(z+1) - 1][(z+1) - 2](z+1)}$

Now $\frac{1}{(z+1)-1}$ and $\frac{1}{(z+1) -2}$ has a laurent serie that you must know and the term $\frac{1}{z+1}$ also is easy to work.

Now you can use the Cauchy Formula for Products of Series.

http://en.wikipedia.org/wiki/Cauchy_product

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you can use the tree poles to calculate your series -1,0,1 XE around the zero $$\frac{x^2}{2 (x+1)}+\frac{x^2 \cos (2)}{2 (x-1)}-\frac{1}{2} x \cos (1)-\frac{\cos (1)}{x}+\sin (1)$$ or around the x=1 $$\frac{x^2}{8 (x+1)}+\frac{3 x^2 \sin (2)}{4 (x-1)}+\frac{5 x^2 \cos (2)}{8 (x-1)}-\frac{x}{4 (x+1)}+\frac{1}{8 (x+1)}-\frac{2 x \sin (2)}{x-1}+\frac{5 \sin (2)}{4 (x-1)}-x \cos (1)-\frac{2 x \cos (2)}{x-1}+\frac{15 \cos (2)}{8 (x-1)}-\frac{\cos (1)}{x}+2 \cos (1)$$ compare the solution of the taylor series around the zero $$\frac{5}{6} x^2 \sin (1)-\frac{1}{2} x \cos (1)-\frac{\cos (1)}{x}+\sin (1)$$ enter image description here

the gree one is the taylor series therst color is the function with the other teo series.

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