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

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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 of f(z) about z=z0 \[· · · an z – z0-n + · · · + a-1 z – z0-1 + a0 + a1(z – a) + a2z – z02 + · · · . \] Satisfies \[a_{-n}\ =\ 0\] for all n ∈ N then the point z = z0 is a removal singularity.

Clearly Laurent series expansion of sin(z)/z about z=0 is \[\frac{\sin{\left(z\right)}}{z}=\ 1-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots.. \] Satisfies \[a_{-n}\ =\ 0\] for all n ∈ N

Therefore z = 0 is a removal singularity of sin(z)/z


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