Say I have a smooth odd function $f(z):\mathbb{C} \to \mathbb{C}$. Then is the function $$ g = \begin{cases} \frac{f(z)}{z} & \text{ if } z \neq 0 \\ \frac{d f}{dz} (0) & \text{ if } z = 0\\ \end{cases} $$ also smooth?
The limit of $f(z)/z$ as $z\to 0$ is just $f^{\prime}(0)$ by L'Hospital's rule, so that's where I get the $g(0)$. That also shows $g(z)$ is continuous. From this, I can also show that it is differentiable, but how do I show it is smooth? (I tried taking multiple derivatives, but that got messy really quickly) I know that away from zero, $f(z)/z$ is smooth, that $f(z)/z$ is even, that the derivative of an even function is odd (so is $0$ at $z=0$) and the derivative of an odd function is even. I don't see how this wouldn't be smooth but there might be some weird case. And I certainly don't know how to prove it at any rate.
Or, if my piecewise function above isn't the correct one, but there is another way to extend $f(z)/z$ over $0$ I'd be very appreciative.