Calculating an integral threw residium question

[nhrock3]

i need to calculate this integral $$f(z)=\frac{z}{e^{2\pi iz^2}-1}\\$$

in this area
$$\gamma _r=\left \{ |z|=r \right \},r>2$$

i need to find the points which turn to zero in the denominator
and non zero in the numerator.
i got two such points
$$z=\pm \sqrt{n}$$
by using this formula
$$res(\sqrt{a})=\frac{p(a)}{q(a)'}$$
$$res(\sqrt{n})=\frac{1}{4\pi i}$$
$$res(-\sqrt{n})=\frac{1}{4\pi i}$$

the third point is z=0 but for it we have both numerator and denominator 0
i calculated the residium for it by $$res(f(x),a)=\lim_{x->a}(f(x)(x-a))$$ formula
but then
my prof says some stuff that involves the area
he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
for each point have a residiu and i need to sum the residiums inside.
but here the area is not defined
its not like (by radius 3)

i don't know what point are inside the area

??

 

[nhrock3]

anyone?>?>

 

[nhrock3]

sorry there is a mistake
the area is
$$\gamma _r=\left \{ |z|=r \right \},n<r^2<n+1$$

and the integral is from plus to minus infinity