4
$\begingroup$

I am trying to evaluate the following integral, but don't know how to take the coefficient of $z$ out of the parenthesis to get it into the Cauchy integral form. Any help is appreciated.

$$ \oint_C \frac{\sin 2z}{(6z-\pi)^3}dz$$

$\endgroup$
4
$\begingroup$

Do you mean as in $$6^3\left(z-\frac{\pi}{6}\right)^3?$$

$\endgroup$
  • $\begingroup$ Hmm. I tried expanding it out and got into a mess, and didn't realize it was that simple. Thanks. $\endgroup$ – Joebevo Jul 17 '12 at 6:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.