Another Cauchy-Goursat Question....?

if i have the integral:

$\displaystyle \int_{|z|=1} \frac{\sin(z)}{(z^{2}-25)(z^{2}+9)}dz$

to prove that it equals zero, would i be correct in splitting it up into:

$\displaystyle \frac{\sin(z)}{340(z-5)}+ \frac{ i \sin(z)}{204(z-3i)}- \frac{i \sin(z)}{204(z+3i)}- \frac{\sin(z)}{340(z+5)}$

and then pointing out that the numerator is "entire", and the singularities don`t lie within the connected domain (contour) and, THUS, by the Cauchy-Goursat Theorm the integral takes the value zero? (since each part does?)

(i can`t exactly do any more than that - can i?)

I can`t just say "the numerator is entire, the denominator`s singularities aren`t inside the circle therefore = 0!!

Thanks!

if i have the integral:

$\displaystyle \int_{|z|=1} \frac{\sin(z)}{(z^{2}-25)(z^{2}+9)}dz$

to prove that it equals zero, would i be correct in splitting it up into:

$\displaystyle \frac{\sin(z)}{340(z-5)}+ \frac{ i \sin(z)}{204(z-3i)}- \frac{i \sin(z)}{204(z+3i)}- \frac{\sin(z)}{340(z+5)}$

and then pointing out that the numerator is "entire", and the singularities don`t lie within the connected domain (contour) and, THUS, by the Cauchy-Goursat Theorm the integral takes the value zero? (since each part does?)

(i can`t exactly do any more than that - can i?)

I can`t just say "the numerator is entire, the denominator`s singularities aren`t inside the circle therefore = 0!!

Thanks!