Let's say I have a function $f(z)$ =
\begin{cases}
1/z^2&\text{if}\, z\neq 0\\
1&\text{if}\, z= 0\\
\end{cases}
And I want to find the domain in which this function is analytic, at first sight, I notice that the limit as $z$ approaches 0 for $1/z^2$ is equal to $\infty$ and the function itself evaluated at 0 is equal to 1, so this function is discontinuous at $z$ = $0$ and so it must be non differentiable at $z$ = $0$
Using DeMovire's theorem, $1/z^2$ = $1/r^2$ (cos($2\theta$) – sin($2\theta$))
When I use the Cauchy Riemann equations in polar form:
$\left( \frac{\partial u}{\partial r}\right) = \frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \ \ \ \ \ \text{and} \ \ \ \ \left(\frac{\partial v}{\partial r} \right) = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right)$
I find that the equations are satisfied at every point except at $r$ = $0$
Does this mean that this function is analytic in its whole domain except the origin? I just need some clarification, thanks.
Best Answer
Yes, but there is a simpler approach. Since the identity function is analytic, and since the product of two analytic functions is again analytic, the $z\mapsto z^2$ is analytic. Moreover, since $f$ analytic and without zeros $\implies\frac1f$ analytic, $z\mapsto\frac1{z^2}$ is analytic too (on $\mathbb{C}\setminus\{0\}$).