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!
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!