Hey, Caleb!
You're right; I don't cover shit like this til' close to the end; and there I void all possible contours like this.
If you take:
\[
z \mapsto e^{-1/z}\\
\]
You will notice very similar behaviour; and that is where I cap the lambert reflection formula.
Where you are seeing here is more like:
\[
z \mapsto e^{-1/\sqrt{z}}\\
\]
Which has a branching singularity at \(0\) (coincidentally similarly at \(z = \infty\)). The rough draft I sent you is intended solely for singularities which aren't branching. Your graph above is clearly a branching singularity.
That's why I don't want to comment on this new question too much; but it deeply relates to "residue at infinity" as I'm talking about. It's just a branching residue rather than a meromorphic, or essential residue. I'm focusing on \(z \mapsto 1/z^k\) or \(z \mapsto e^{-1/z}\), I'm entirely ignoring \(z \mapsto e^{-1/\sqrt{z}}\). Not because I can't speak on it, but because I'm less confident, and more unsure what happens here.
Awesome job still!! I love these graphs! Keep em coming!
Sincere Regards, James
EDIT!
Also the graph you just posted is a Riemann surface defined off of a second order singularity. This means that the dimension of the surface is \(2\). This is just like \(z = y^2\) produces a Riemann Surface \(S\) in \(2\) complex dimension. So that, the "branching point" of the graph you just took, is only 2nd degree. It looks like \(z = y^2\). But the base singularity is essential... we can see this just how the function behaves near \(z=0\). It gets arbitrarily large the closer to \(z =0\) (which means it's essential). From there \(e^{-1/\sqrt{z}}\), is a second order Riemann Surface; or rather; there are two copies \(e^{\pm \frac{1}{y}}\) which are projections from the Riemann surface \(S\) in \(z\) to a singular complex variable \(y\).
I apologize if I've appeared a tad daft, Caleb. I'm just trying to do baby steps first; and build a slow theory. But I still believe I can answer your questions. Just remember that professional math is a marathon. And if you do a marathon, you'll come in 15th place. If you try to sprint your way through, you'll end up in last place...
You're right; I don't cover shit like this til' close to the end; and there I void all possible contours like this.
If you take:
\[
z \mapsto e^{-1/z}\\
\]
You will notice very similar behaviour; and that is where I cap the lambert reflection formula.
Where you are seeing here is more like:
\[
z \mapsto e^{-1/\sqrt{z}}\\
\]
Which has a branching singularity at \(0\) (coincidentally similarly at \(z = \infty\)). The rough draft I sent you is intended solely for singularities which aren't branching. Your graph above is clearly a branching singularity.
That's why I don't want to comment on this new question too much; but it deeply relates to "residue at infinity" as I'm talking about. It's just a branching residue rather than a meromorphic, or essential residue. I'm focusing on \(z \mapsto 1/z^k\) or \(z \mapsto e^{-1/z}\), I'm entirely ignoring \(z \mapsto e^{-1/\sqrt{z}}\). Not because I can't speak on it, but because I'm less confident, and more unsure what happens here.
Awesome job still!! I love these graphs! Keep em coming!
Sincere Regards, James
EDIT!
Also the graph you just posted is a Riemann surface defined off of a second order singularity. This means that the dimension of the surface is \(2\). This is just like \(z = y^2\) produces a Riemann Surface \(S\) in \(2\) complex dimension. So that, the "branching point" of the graph you just took, is only 2nd degree. It looks like \(z = y^2\). But the base singularity is essential... we can see this just how the function behaves near \(z=0\). It gets arbitrarily large the closer to \(z =0\) (which means it's essential). From there \(e^{-1/\sqrt{z}}\), is a second order Riemann Surface; or rather; there are two copies \(e^{\pm \frac{1}{y}}\) which are projections from the Riemann surface \(S\) in \(z\) to a singular complex variable \(y\).
I apologize if I've appeared a tad daft, Caleb. I'm just trying to do baby steps first; and build a slow theory. But I still believe I can answer your questions. Just remember that professional math is a marathon. And if you do a marathon, you'll come in 15th place. If you try to sprint your way through, you'll end up in last place...