Originally Posted by

**0-)** Thank you Laurent!

I think your working is slightly wrong, all the $\displaystyle \sqrt{b}$'s should be $\displaystyle b$.

So $\displaystyle \displaystyle \frac{z^2}{z^2+b^2} = 1-\frac{b}{2i}\left(\frac{1}{z-ib}-\frac{1}{z+ib}\right)$

Yes, there're mistakes in my post, but (for me) it was on the left-hand side: it should have been

$\displaystyle \frac{z^2}{z^2+b}=1-\frac{b}{z^2+b}= 1-\frac{b}{(z+i\sqrt{b})(z-i\sqrt{b})}=$ $\displaystyle 1-\frac{b}{2i\sqrt{b}}\left(\frac{1}{z-i\sqrt{b}}-\frac{1}{z+i\sqrt{b}}\right)$.

From this, I think I am right in saying that if I is the same integral in my previous post but with $\displaystyle b^2$ instead of $\displaystyle b$ then

$\displaystyle |b| \leq |a|\implies I=0$

Can someone confirm this for me?

No, this is the *contrary*. Or you can say that your former integral is 0 if $\displaystyle |b|>|a|^2$. Indeed, it is zero if the poles $\displaystyle \pm\sqrt{b}$ are outside the circle of integration.