Let $f(z) =e^\frac1z + \frac{1}{z-1}$ and find the Laurent Series for $f(z)$ about zero in the regions $0<|z|<1$ and $|z|>1$.
I am somewhat confident in my working for the Laurent expansion but please correct me if I am wrong:
The commonly known expansion for $e^\frac1z$ is: $$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ which I believe is valid for all $|z|>0$ (and therefore valid for both cases in my question right?).
For $|z|>1$, we have that $\frac{1}{z-1}$ = $\frac{1}{z}$$\frac{1}{1-1/z}$ and $|\frac{1}{z}|<1$ holds for $|z|>1$ and therefore: $$\frac{1}{z-1}=\frac{1}{z}(1+\frac1z+\frac1{z^2}+\frac1{z^3}+\cdots)$$
So our answer for the Laurent expansion in the $|z|>1$ case is the sum of these two expansions, correct?
A slight tweak for the $0<|z|<1$ case, I obtained $\frac{1}{z-1}$ = $-1$$\frac{1}{1-z}$ and therefore the expansion is: $$\frac{1}{z-1}=-1(1+z+z^2+z^3+\cdots)$$ So adding this expansion to the same previous $e^\frac1z$ expansion gives us the Laurent Expansion for this $0<|z|<1$ case, right?
I would really appreciate verification of the steps of my calculations and perhaps any corrections of mistakes I may have made.