2 unsolved questions:
- Prove that all the singular points of rational function $f(z)=\frac{P_1(z)}{P_2(z)}$, where $P_{1,2}$ are polynomials, are either removable singularities or poles.
- Consider function $f(z)$, and it's derivative $f'(z)$. Given $z_0$ is simple pole of $f'(z)$, characterize the point $z_0$ with respect to $f(z)$.
Thanks in advance!
We can just prove that the limit when z tends to the singularity of $f$ exists and is in the compactification of $\mathbb{C}$ (i.e., infinity is an acceptable value for our limit). This is because singularities of complex functions are characterized by how the limit looks in the singularity. If the limit exists and is finite, then the singularity is removable. If the limit is infinite, then the singularity is a pole. If the limit does not exist, we got an essential singularity. (All converse propositions hold, so this is a great characterization). So, the proof just comes from the fact that polynomials are holomorphic over $\mathbb{C}$, so they admit a Taylor expansion around our singularity $z_0$. We get then that $\lim_{z \to z_0}{f(z)}=\infty$
2. Since every complex derivable function is differentiable infinite times, $f(z_0)$ can't be continous at $z_0$