how to Show that sin(z)/z has removable singularity at 0

Question : Show that sin(z)/z has removable singularity at 0 Solution : Given function \[f(z)\ =\frac{sin\left(z\right)}{z}\] To that sin(z)/z has removable singularity at 0 First we shall find it’s Laurent series expansion \[f\left(z\right)=\frac{sin\left(z\right)}{z}\] \[=\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots..}{z}\] \[f\left(z\right)=1-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots.\] If the Laurent series expansion…

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