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.

  • $\begingroup$ You are correct. $g$ is holomorphic everywhere except at the poles $\pm a$ (which coincide if $a=0$). $\endgroup$
    – MPW
    Mar 5, 2019 at 16:35
  • $\begingroup$ “postive or negative $a$” doesn't make sense when $a$ is complex, but yes, the only points where $g$ is not defined are $\pm a$. $g$ is holomorphic on its whole domain. $\endgroup$ Mar 5, 2019 at 16:35

1 Answer 1


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

  • $\begingroup$ Oh alright it’s good I’m on the right track, the thing that made me unsure however is that the question asks for the ‘largest subset’ and then to illustrate this with a sketch. This makes me assume there are smaller subsets and that I’m unsure about how to sketch the full complex set, not including {a,-a}. That’s why I believed I was wrong. $\endgroup$
    – L G
    Mar 5, 2019 at 16:45
  • $\begingroup$ well $g$ is also holomorphic on every open non-empty subset of this one, which is probably why they ask for the largest one. $\endgroup$
    – tbrugere
    Mar 5, 2019 at 16:48
  • $\begingroup$ As to what kind of sketch they are asking for, I honestly have no idea… If you want a sketch of the full complex set, not including these two points, you could just draw two axes and these two points, but I’m not sure whether that’s what they want $\endgroup$
    – tbrugere
    Mar 5, 2019 at 16:52
  • $\begingroup$ Yeah that’s what I was thinking, it’s a bit confusing thanks a lot though, just curious the question does include ‘let $\gamma = /gamma (0;|a|+1)$ ‘ I assumed this path had nothing to do with the question but I’m probably wrong $\endgroup$
    – L G
    Mar 5, 2019 at 16:56
  • $\begingroup$ I don’t know that $\gamma(0; |a| + 1)$ notation. Is it perhaps the circle of center 0 and radius $|a| + 1$ ? $\endgroup$
    – tbrugere
    Mar 5, 2019 at 17:05

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