Singularities of a Complex Function with Complex Conjugate

Question

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 $\phi(z_0) \neq 0$.

Answer 1

The function is continuous on the whole complex plane and its zeros are at $z=0$ and $z=-i$.

Answer 2

The function $f$ is conjugate analytic everywhere hence it is continuos, harmonic and real analytic everywhere but it is not complex analytic anywhere; however, it is complex differentiable at $-\frac{i}{2}$ with derivative $0$ since $\partial_{\bar z}f=2\bar z-i$ cancels there (and for fun one can easily show that the limit at $-\frac{i}{2}$ of $\frac{f(z)-f(-\frac{i}{2})}{z+\frac{i}{2}}$ is the limit of $\frac{\bar{(z+\frac{i}{2})^2}}{z+\frac{i}{2}}$ so it is indeed zero!)