Why does this function residue equal 0?

Question

$$ f(z) = \frac{e^{2z}}{(z-1/2)^{2013}} $$ Why does this residue equal 0?

If I expand Laurent series, the right side will have $\dfrac{a_{2013}}{(z-1/2)^{2013}}$ $$ + \frac{a_{-2012}}{(z- 1/2)^{2012}}+\cdots \frac{a_{-1}}{(z-1/2)^{1}}+\cdots $$ and the left side will be equal to $$ \frac{\sum\frac{(z-1/2)^{2n}}{(2n!)}}{a_0 + a_1(z-1/2)^1 + a_2(z-1/2)^2+\cdots} $$ where a.-1 is the residue and is what we are looking for

where . indicate negative index

something about even and odd numbers but I can't seem to figure out the pattern

can someone please help me?

Answer 1

The residue isn't zero, though it is something very close to zero. Develop the actual series around $\;z=0.5\;$ :

$$2z=2\left(z-\frac12\right)+1\implies e^{2z}=\left(e^{z-\frac12}\right)^2\cdot e=e\,\sum_{n=0}^\infty\frac{\left(z-\frac12\right)^n}{n!}\implies$$

$$\frac{e^{2z}}{\left(z-\frac12\right)^{2013}}=\frac e{\left(z-\frac12\right)^{2013}}\left(1+\left(z-\frac12\right)+\frac{\left(z-\frac12\right)^2}{2!}+\ldots+\frac{\left(z-\frac12\right)^{2012}}{2012!}+\ldots\right)\implies$$

$$\text{the residue is}\;\;\frac e{2012!}$$

Answer 2

Timbuc was on the right track but neglected $2^n$. As was already shown, $2z=2(z-1/2)+1$ so we can write $$ \frac{\exp[2(z-1/2)]e}{(z-1/2)^{2013}} = \frac{e}{(z-1/2)^{2013}}\sum_{n=0}^{\infty}\frac{2^n(z-1/2)^n}{n!} $$ Therefore, the $a_{-1}$ terms occur at $n=2012$ so the residue is $$ a_{-1}=\frac{e2^{2012}}{2012!}\approx 9.05\times 10^{-5170} $$