The prove of the Laurent series formula

Question

I am trying to understand why at Laurent series we get that .

So, I need to understand why this formula is correct: .

Could someone help?

Answer 1

According to the residue theorem, $\int_{|z-z_0|=r}f(z)dz=2\pi i(\sum res(a_k,f))$ for finitely many singularities $a_k$ of $f$ in $|z-z_0|<r$. If $m=n$ then you get $\int_{|z-z_0|=r}{1\over z-z_0}dz$ with one simple pole at $z=z_0$ and $res(z_0,{1\over z-z_0})=\lim_\limits{z\to z_0}{z-z_0\over z-z_0}=1$, meaning $\int_{|z-z_0|=r}{1\over z-z_0}dz=2\pi i(res(z_0,{1\over z-z_0}))=2\pi i\cdot 1=2\pi i$.

If $m\ne n$ then either $n>m$, meaning $n\ge m+1$ meaning $n=m+1+k$, $k\in \Bbb{Z^+}$, ${(z-z_0)^{n}\over (z-z_0)^{m+1}}={(z-z_0)^{m+1+k}\over (z-z_0)^m+1}=(z-z_0)^k$ which has no singularities, meaning $\int_{|z-z_0|=r}f(z)dz=2\pi i(\sum res(a_k,f))=2\pi i\cdot(0)$.

Otherwise, $n<m$ and we have ${(z-z_0)^{n}\over (z-z_0)^{m+1}}={1\over (z-z_0)^k}$ for $k\ge 2$. Try to see what residue you get for $k\ge 2$.

Answer 2

You can easily compute $$ I = \int_{|z-z_0|=r} (z - z_0)^k \, dz $$ for $k = n - m - 1 \in \Bbb Z$.

If $k \ne -1$ then $(z-z_0)^k$ has the antiderivate $\frac{1}{k+1}(z-z_0)^{k+1}$ in $\Bbb C \setminus \{ z_0 \}$, and therefore $I = 0$.

If $k = -1$ then the parametrization $z = z_0 + r e^{it}$ of the curve immediately gives $$ I = \int_0^{2\pi} \frac{ire^{it}}{re^{it}} \, dt = 2 \pi i \, . $$