I am trying to see what the singular points of the complex function $f(z)=\bar{z}^2-i\bar{z}$ are. Singular points is where the function fails to be continuous or is indeterminate. Just to be clear, I am getting that the function has singular points at $z=0$ and $z=-i$. I get that both are poles of order 1 of the function, as $f(z)=\bar{z}(\bar{z}-i)=\frac{|z|^2}{z}(\frac{|z|^2}{z}-i)$. Is this correct? I am starting to get myself confused.
Note: Singular points here include both zeroes and poles of $f(z)$. So my question is whether the points I found are zeroes or poles.
Definition of a pole used:
The singular point $z_0$ of an analytic function $f(z)$ is called a pole of order $n$ if $f(z)$ in the neighborhood of $z_0$ can be expressed as
$$f(z)=\frac{\phi(z)}{(z-z_0)^n} $$
where $\phi(z)$ is analytic in the neighborhood of $z_0$ and $\phi(z_0) \neq 0$.
Best Answer
The function is continuous on the whole complex plane and its zeros are at $z=0$ and $z=-i$.