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Show that when $z\neq0$,

(a) $$\frac{e^z}{z^2}=\frac{1}{z^2}+\frac{1}{z}+\frac{1}{2!}+\frac{z}{3!}+\frac{z^2}{4!}+...$$

(b) $$\frac{\sin(z^2)}{z^4}=\frac{1}{z^2}-\frac{z^2}{3!}+\frac{z^6}{5!}-\frac{z^{10}}{7!}+...$$

I don't understand the Taylor series completely, nor how to use it to prove the series when $z\neq0$.

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These are Laurent series, not Taylor series, as you have poles at $0$. But start with the Taylor series for $e^z$ and $\sin(z^2)$.

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so i have that $e^z=\sum_{i=0}^\infty \frac{z^n}{z!}$ and $(|z|<\infty)$. do I use this somehow? – xfitpi Apr 24 '13 at 23:37
Yes, except you need to fix your typos :) Since you want $\dfrac{e^z}{z^2}$, there's something obvious to do. – Ted Shifrin Apr 25 '13 at 0:31
I'm stuck on how to incorporate the $z^2$ and how it changes the summation – xfitpi Apr 26 '13 at 0:28
Since these are Taylor series, you are working with real numbers, so you want to write $e^x = \Sigma_{i = 0}^{\infty} \frac{x^n}{n!}$ and $\sin x = \Sigma_{i = 0}^{\infty} (-1)^n \cdot \frac{x^{2n + 1}}{(2n+1)!} $ . For your first series, you seek $e^x \cdot \frac{1}{x^2} = \frac{1}{x^2} \cdot \Sigma_{i = 0}^{\infty} \frac{x^n}{n!}$ . How do you bring that $x^{-2}$ into the summation? How do you fix the trouble with the negative powers of x that are implied in the first couple terms? What do you do for $\sin (x^2) \cdot \frac{1}{x^4}$? – RecklessReckoner Apr 26 '13 at 14:06

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