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I want to find Laurent series of the complex function $$f(z) = \cos\left(\frac{z^2-4z}{(z-2)^2}\right)$$ at $z_0=2$. I will be thankful for any hints.

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Hint. You can substitute

$$\frac{z^2-4z}{(z-2)^2}=1-\frac{4}{(z-2)^2}$$

into the series for $\cos z$ (which is convergent for all $z\in\mathbf{C}$): $$\sum_{n\geqslant 0}\frac{(-1)^n}{(2n)!}z^{2n}.$$

This will give the Laurent series centered at $z=2$.

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  • $\begingroup$ Thanks, that answers my question. $\endgroup$ – Detrin May 11 at 18:52
  • $\begingroup$ @Detrin glad I could be of help. $\endgroup$ – rae306 May 11 at 18:57

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