Flawed reasoning between closed contour integral and Cauchy's residue theorem Thanks for any help in advance.
I'm trying to justify the answer to :
$$\oint_V \frac{e^{3z}}{z-\ln2} \,dz$$
over the square of vertices $\pm$$1$$\pm$$i$.
By the Cauchy's integral theorem, 
as there is a simple pole at $z = \ln  2$, so 
$$\oint_V \frac{f(z)}{z-z_{0}} \,dz$$ 
becomes $2\pi  i\cdot e^{3ln2} = 16\pi i$
However, if you expand $e^{3z}$, we have a fraction of 
$$ \frac{1 + (3z) + \frac{(3z)^{2}}{2} + ...}{z-\ln2} $$
which gives us a residue of $1$ at $z = \ln 2$. 
Then, by the residue theorem 
$$\oint_V f(s) \,ds$$ $ = 2\pi \cdot $ (sum of residues) $= 2\pi i \cdot (1)$
which gives us an answer of $ 2\pi i$
What is the flaw in my reasoning? Have i applied the residue theorem wrongly?

You are expanding $e^{3z}$ around $z=0$. You should expand it around $z=\ln2$.
