$$ g(z) = \frac{ e^{2 \pi i z}}{ (1 - e^{2 \pi i z})^2} $$ has a double pole at $ z = 0$. So $z^2 g(z) $ is holomorphic at $ z = 0$. So this can be expanded as a series $\Sigma a_n z^n$, which convergent absolutely and uniformly. Thus $$a_1 = \frac{d}{dz} z^2 g(z) |_{z=0}. $$ But $ \frac{d}{dz} z^2 g(z) $ has a single pole at $z=0$.
( since $$ \frac{d}{dz} z^2 g(z) = \frac{2ze^{2 \pi i z} (2 \pi i z e^{2 \pi i z} + (1 + \pi i z)(1 - e^{2 \pi i z}))}{(1 - e^{2 \pi i z})^3} $$)
Which is wrong? Please help.
P.S. My text says $a_1 = 0, a_2 = -1/12.$
Write $w=2\pi i z$. Your function is then $$\frac{e^w}{(1-e^w)^2}=\frac1{(1-e^w)(e^{-w}-1)} =\frac{1}{2-e^w-e^{-w}}=\frac1{2-2\cosh w} =\frac1{-w^2-w^4/12-\cdots}$$ where the series has only even powers of $w$.
Therefore $$\frac{e^w}{(1-e^w)^2} =-\frac1{w^2}\left(1+\frac{w^2}{12}+\cdots\right)^{-1} =-\frac1{w^2}\left(1-\frac{w^2}{12}+b_4w^4+b_6w^6+\cdots\right)$$ for some $b_{2k}$. (In general these can be expressed in terms of Bernoulli numbers).
Putting this back in terms of $z$ gives $$\frac1{4\pi^2z^2}\left(1+\frac{\pi^2 z^2}{3}+16\pi^4 b_4z^2 -64\pi^6 b_6z^6+\cdots\right)$$ etc.