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I am trying to find the Maclaurin series for $f(z)=\frac{z^2}{z^2+4}$. I use the fact that $\frac{1}{1+z}=\sum_{n=0}^\infty (-1)^nz^n$ and factored to get $\frac{z^2}{z^2+4}=\frac{z^2}{4}\Big(\frac{1}{\frac{z^2}{4}+1}\Big)$= $\sum_{n=0}^\infty {(-1)^n \frac{z^{2n+2}}{4^{n+1}}}$. I am not getting the correct terms though, did I do something wrong?

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No, you didn't do anything wrong. That's the correct series. –  Antonio Vargas Mar 26 '13 at 17:43

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up vote 4 down vote accepted

Your work is nice, but two things I'd add:

(1) The given series expansion is valid for

$$\left|\;\frac{z^2}{4}\;\right|<1\iff |z|^2<4\iff |z|<2\ldots$$

(2) You could also have done as well

$$\frac{z^2}{z^2+4}=1-\frac{4}{z^2+4}=1-\frac{1}{1+\left(\frac{z}{2}\right)^2}=\ldots$$

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