Find the Laurent series expansion in powers of $z$ of
$f(z) = \frac{\cos(z^2)}{z^3}$
valid in the region $|z| > 0$

My Instinct is to make use of the fact that $\cos(z^2) = \frac{1}{2}(e^{z^2 i\theta} + e^{-z^2 i\theta} ) $. But I am a bit lost and have never seen a Laurent series with trigonometric functions in before this.

  • $\begingroup$ Start with the Taylor series for $\cos(z) = 1 - z^2/2! + z^4/4! - z^6/6! + \ldots$. Take $z\to z^2$ and divide the whole thing by $z^3$. $\endgroup$ – Winther May 15 '18 at 14:39
  • $\begingroup$ Your formula for $cos$ contains a mysterious $\theta$. $\endgroup$ – Torsten Schoeneberg May 15 '18 at 14:42

Remember that

$$ \cos z = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{(2k)!}z^{2k} $$

So that

$$ \frac{\cos z^2}{z^3} = \frac{1}{z^3}\sum_{k = 0}^{+\infty} \frac{(-1)^k}{(2k)!}z^{4k} $$


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.