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

[Stumped1]

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!

 

[Prove It]

What you have done seems fine...

 

[chisigma]

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

$\chi$ $\sigma$

 

[alyafey22]

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.