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Let $$f(z)=\frac{\sin^2z}{z}, \quad z\neq0$$ with $z_0=0$ being an removable singularity of $f$ since $$\lim_{z \to 0}\, zf(z)=\lim_{z \to 0}\,\sin^2z=0$$ To find its Laurent series expansion within the annulus $|z|>0,$ does numerator have to be expanded in powers of $\frac1z$ ? Can $$\sin^2z=\bigg(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+ \dots\bigg)^2$$ be of use?

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    $\begingroup$ $z = 0$ is not an essential singularity of $f$. It is a removable singularity. And if you were to extend $f$ to cover $z = 0$, it would have a zero at $z = 0$. $\endgroup$ – Kenny Wong Aug 25 '18 at 13:54
  • $\begingroup$ Yes, removable is what I meant, confused the terms... $\endgroup$ – Jevaut Aug 25 '18 at 14:04
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Hint: Use $$\sin^2z=\dfrac{1-\cos2z}{2}$$ and the expansion $$\cos w=1-\dfrac{w^2}{2!}+\dfrac{w^4}{4!}+\cdots=\sum_{n=0}^\infty\dfrac{(-1)^nw^{2n}}{(2n)!}$$

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