i need to calculate this integral [tex]f(z)=\frac{z}{e^{2\pi iz^2}-1}\\[/tex](adsbygoogle = window.adsbygoogle || []).push({});

in this area

[tex]\gamma _r=\left \{ |z|=r \right \},r>2[/tex]

i need to find the points which turn to zero in the denominator

and non zero in the numerator.

i got two such points

[tex]z=\pm \sqrt{n}[/tex]

by using this formula

[tex]res(\sqrt{a})=\frac{p(a)}{q(a)'}[/tex]

[tex]res(\sqrt{n})=\frac{1}{4\pi i}[/tex]

[tex]res(-\sqrt{n})=\frac{1}{4\pi i}[/tex]

the third point is z=0 but for it we have both numerator and denominator 0

i calculated the residium for it by [tex]res(f(x),a)=\lim_{x->a}(f(x)(x-a))[/tex] 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 dont know what point are inside the area

??

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# Homework Help: Calculating an integral threw residium question

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