I’m struggling a lot with this question:
—- Fix $a \in \mathbb{C}$ where $g(z) = \frac{1}{z^2-a^2}$, Let $\gamma = \gamma (0;|a|+1)$. Find the largest subset of $\mathbb{C}$ where g(z) is holomorphic. Illustrate this with a sketch. —- I understand that a holomorphic function is a function is differentiable in the complex plane, I believe that the only time g(z) would not be holomorphic is when the function is invalid e.g $\frac{1}{0}$, however this leaves me with just assuming the only points g(z) is not holomorphic is where $z^2 = a^2$ however this then leaves just the subset $\mathbb{C}$ not including positive or negative a, which doesn’t seem right. Any help help would be fab.
You are right : The function $g$ is holomorphic on $\mathbb{C} - \{a, -a\}$ as you said. when $z$ is in this set, you can say
$g(z) = - \frac{1}{a^2} \frac{1}{1 - z^2}$
which you can develop into a power series near $z$ using the development of $\frac{1}{1 + z}$
(which does prove $g$ is holomorphic on $z$)
You could also calculate the derivative $g'(z)$ using the formula $\lim_{z' \rightarrow z} \frac{g(z') - g(z)}{z' - z}$ which will converge for each z in this set