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    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!
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    (Original post by Hasufel)
    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!
    To be perfectly honest, I would be inclined to say precisely that "the numerator is entire and the zeros of the denominator lie outside the contour". I can't really see the purpose of such a question, but such is life!
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    (Original post by Hasufel)
    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!
    Just tell whoever asked it that it is a stupid question and that they should come up with examples that illustrate purpose rather than function.


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    Just to point out, if you're doing partial fractions and you start seeing coefficients like 204, 340 - chances are that isn't the intended method. However, as the other two pointed out, this is practically a trivial problem if you're allowed to use Cauchy's Thorem, and there isn't any need to split it up to use CT. To be honest, this is the kind of level of difficulty you'd expect at A-Level if CT was covered. As with your previous questions, really you just need to show that the function is holomorphic inside, and on, the contour by showing that any singularity lies outside (which is straightforward here since the singularities clearly all have modulus >1) and the function is holomorphic everywhere else (which you rarely need to explicitely show in a complex analysis course - it's clear with this given that sin(z) and the denominator are holomorphic, so the product of the two will be whereever the denominator is non-zero).
    • Thread Starter
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    Thanks for the input, guys! - to be honest, i thought it might just be a straitforward: "numerstor..blah, denominator..blah" => each =0

    - Just like to be thorough, though!

    cheers! +rep
 
 
 
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