Finding Laurent Series with different range I am a bit confused about finding Laurent series. For example, f(z)=7/(z-2)(z+5) this function and and i am trying to find Laurent series of |z|<2, 2<|z|<5, |z|>5 this three cases.
I know how to do the second case, which is just combine the Laurent series of 1/(z-2) when |z|>2 & 1/(z+5) when |z|<5.
However, for the case |z|<2, i know how to do it with  1/(z-2), but what should i do with 1/(z+5)? i firstly put f(z) into partial fraction and then?
Same for the third case, do not know how to do  1/(z-2) when |z|>5.
Furthermore, if there is a situation that to find |z|>2 and |z|<5, what should i do?
Thank you so much for help !
 A: Start with partial fractions:
$$\frac{7}{(z-2)(z+5)}=\frac{1}{z-2}-\frac{1}{z+5}\ .$$
Let's assume that $2<|z|<5$, then I think you should be able to do the other cases for yourself.  For the second term,
$$\frac{1}{z+5}=\frac{1}{5}\frac{1}{1+(z/5)}=\frac{1}{5}\sum_{n=0}^\infty\Bigl(-\frac{z}{5}\Bigr)^n\ .$$
Note that this is valid because for the given $z$-values we have
$$\Bigl|\frac{z}{5}\Bigr|=\frac{|z|}{5}<1\ .$$
If we try to do the same for the other term,
$$\frac{1}{z-2}=-\frac{1}{2}\frac{1}{1-(z/2)}=-\frac{1}{2}\sum_{n=0}^\infty\Bigl(\frac{z}{2}\Bigr)^n$$
it is not correct since
$$\bigl|\frac{z}{2}\Bigr|=\frac{|z|}{2}>1$$
and so the series diverges.  So do it this way:
$$\frac{1}{z-2}=\frac{1}{z}\frac{1}{1-(2/z)}=\frac{1}{z}\sum_{n=0}^\infty\Bigl(\frac{2}{z}\Bigr)^n\ .$$
The series converges since
$$\Bigl|\frac{2}{z}\Bigr|=\frac{2}{|z|}<1\ .$$
Similarly, in the case $|z|<2$ we have
$$\frac{1}{z+5}=\frac{1}{5}\frac{1}{1+(z/5)}=\frac{1}{5}\sum_{n=0}^\infty\Bigl(-\frac{z}{5}\Bigr)^n\ ,$$
which is valid because
$$\Bigl|\frac{z}{5}\Bigr|=\frac{|z|}{5}<\frac{2}{5}<1\ .$$
See how you go with the rest of the problem.  Good luck!
