Expanding a complex function in Taylor series Expand the function
$$ f(z) = \frac {2(z + 2)} {z^2 − 4z + 3} $$
in a Taylor series about the point $ z = 2 $ and find the circle C inside of which the series converges. Find a Laurent series that converges in the region outside of C.
I tried writing the denominator as $ (z-3)(z-1) $ to find the singularities $ z = 1,3 $ both simple. There exists a circle around $ z = 2 $ such that $f(z) $ is analytic so we can write it as a Taylor series. I got stuck trying to make the function look like a known Taylor series like $e^z$ or $\frac {1} {1-z}$
 A: Hint. Starting with a standard partial fraction decomposition
$$
\frac{2(z+2)}{z^2-4z+3}=\frac{3}{1-z}-\frac{5}{3-z}
$$
you may then set $u:=z-2$, that is $z=u+2$, obtaining
$$
\frac{3}{1-z}-\frac{5}{3-z}=-\frac{3}{1+u}-\frac{5}{1-u}
$$ with $u \to 0$ as $z \to 2$.
Can you take it from here?
A: Split them into partial fraction and upon solving you will get,
$  \frac {2(z+2)}{(z-1)(z-3)} = \frac {3} {(z-(-1))} + \frac {5} {z-3} $
taylor's series expansion is
$ \frac {1}{1-x} = \frac{1}{1-a} + \frac {x-a}{(1-a)^2} + \frac {(x-a)^2}{(1-a)^3} + ...$
so for $\frac {3}{(z-(-1)} $ , value of a=-1    i.e.,      $ \mathcal\ z ^\left(-1 \right) $ $\left[ \frac {1}{z-a} \right] $
the taylor's series for 
$\frac {3}{z-(-1)}$ = $ 3 \{  \frac{1}{2} + \frac{(x+1)}{4} +\frac {(x+1)^2}{8} + ...\}$  
similaryly for $\frac {5}{(z-3)} $ , value of a=3       i.e.,   $ \mathcal\ z ^\left(-1 \right) $ $\left[ \frac {1}{z-a} \right] $
the taylor's series for 
$\frac {5}{z-3}$ = $ 5 \{  \frac{-1}{2} + \frac{(x-3)}{4} +\frac {(x-3)^2}{8} + ...\}$  
