MHB Expand e^{\frac{z}{z-2}} Laurent Series: z=2

Stumped1
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expand $$e^{\frac{z}{z-2}}$$ in a Laurent series about $$z=2$$

I cannot start this.

my attempt so far has been

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

This is unlike the other problems I have worked. Seems I need to manipulate this equation some way before plugging into $$e^z$$'s maclaurin series

Thanks for any help with this!
 
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What you have done seems fine...
 
Stumped said:
expand $$e^{\frac{z}{z-2}}$$ in a Laurent series about $$z=2$$

I cannot start this.

my attempt so far has been

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

This is unlike the other problems I have worked. Seems I need to manipulate this equation some way before plugging into $$e^z$$'s maclaurin series

Thanks for any help with this!

Because is $\displaystyle \frac{z}{z-2} = 1 + \frac{2}{z-2}$ You obtain...

$\displaystyle e^{\frac{z}{z-2}} =e\ \{ 1 + \frac{2}{z- 2} + \frac{4}{2\ (z-2)^{2}} + ... + \frac{2^{n}}{n!\ (z-2)^{n}} + ...\}\ (1)$

Kind regards

$\chi$ $\sigma$
 
Note that $f$ must be written in the form

$$\sum_{n=-\infty}^\infty a_n (z-2)^n$$

where $a_n$'s are independent of $z$ as chisigma stated.
 
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