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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.

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  • $\begingroup$ 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
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Remember that

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

So that

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

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