@Alessandro Where $\gamma$ is a contour around $0$ (where you assume your pole is). Now sub in $w = 1/z$ so that $\gamma$ becomes a contour around infinity.
It's like real analysis, except functions are actually nice rather than trying to backstab you with counterexamples to every plausible sounding assertion
Suppose
$$z\frac{\cos z}{\sin z}= \sum_{-\infty}^{n} a_nz^n $$
the laurent series of $f(z)= z\frac{cosz}{sinz} $ on the ring π<|z|<2π.Find the $a_n$.
Now i know $a_n= \frac{1}{2Ï€i} \int\frac{f(z)}{z^{n+1}}dz$ so for $$n=0$$i plug in the $f$ and i try to use the residues theorem but i dont ...
@Studentmath in general integral operators are compact and self adjoint if they satisfy some symmetry condition, then you can apply the spectral theorem, if you're actually asking for a specific orthonormal eigendecomposition im not gonna work it out
Also it's weird that this is the only quarter of the 3 in which we actually finished everything we intended to, last quarter we lost some calendar days and in first quarter we were trying to cover so much
Like I'm willing to take certain things on faith that things could be made rigorous so that we could focus more on the ideas. Part of it is that I completely glaze over at epsilon detail in analysis
@Zee Religion is when you have faith based on unverifiable personal/spiritual experience, this faith of mine is, I know this has been done rigorously and I'd rather get to cool things than spend a week staring at Greek symbols
Suppose
$$z\frac{\cos z}{\sin z}= \sum_{-\infty}^{n} a_nz^n $$
the laurent series of $f(z)= z\frac{cosz}{sinz} $ on the ring π<|z|<2π.Find the $a_n$.
Now i know $a_n= \frac{1}{2Ï€i} \int\frac{f(z)}{z^{n+1}}dz$ so for $$n=0$$i plug in the $f$ and i try to use the residues theorem but i dont ...
