Laurent Series for $\frac{\cos(z)}{z^2}$ centered at $0$

Question

How would you prove that this is entire if f(z) is (cos(z)-1) if z doesn't equal zero and is -1/2 if z equals zero?

Answer 1

It looks like you're asking about the series for the "repaired version" of $$f(z)=\frac{\cos(z)-1}{z^2}$$ centered at the origin. By "repaired version", I mean that while $f(z)$ is clearly not defined at $z=0$, we can extend it to a function that is defined at $z=0$ in a very natural way.

To see this, note that the series of $\cos(z)-1$ is $$\sum_{j=1}^\infty\frac{(-1)^jz^{2j}}{(2j)!}=z^2\sum_{j=1}^\infty\frac{(-1)^jz^{2j-2}}{(2j)!},$$ which converges everywhere, so the function $$g(z):=\sum_{j=1}^\infty\frac{(-1)^jz^{2j-2}}{(2j)!}$$ converges everywhere. Moreover, since $$z^2f(z)=\cos(z)-1=z^2\sum_{j=1}^\infty\frac{(-1)^jz^{2j-2}}{(2j)!}=z^2g(z)$$ for all $z\ne 0$, then $f$ and $g$ agree everywhere but at the origin (where $f$ is undefined). Thus $g$ is an extension of $f$, removing the singularity at $z=0$ to make an entire function. You can also readily check that $g(0)=-\frac12.$