复变函数积分
在\(B\left(0,r\right)\)
The same statement holds for any \(a, b \in \mathbb{C}\). (Note that since \(f\) is an entire function, \(\int_{a}^{b} f(z) d z\) is independent of path, and hence well defined.
Proof:
Fix an arbitrary positive number \(R>\max \{|a|,|b|\} .\) It is easy to check that when \(|w|,\left|w_{0}\right|<R\) (each branch of ) \(\log \frac{z-w}{z-w_{0}}\) is a well defined holomorphic function of \(z\) on \(\{|z| \geq R\},\) so
\(g_{w_{0}}(w):=\frac{1}{2 \pi i} \int_{|z|=R} f(z) \log \frac{z-w}{z-w_{0}} d z\)
defines a holomorphic function of \(w\) on \(\{|w|<R\} .\) Moreover, \(g_{w_{0}}\left(w_{0}\right)=0\) and
\(g_{w_{0}}^{\prime}(w)=-\frac{1}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z-w} d z=-f(w)\)
That is to say,
\(g_{w_{0}}(w)=-\int_{w_{0}}^{w} f(z) d z\)
Letting \(w_{0}=a, w=b,\) the conclusion follows.
isolated singularity:$$\frac{z{3}+z+2}{z\left(z{2}-1\right){2}}$$
(z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}
Series of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=-1
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=0
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=1
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=\infty
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=-1
residue of z^3*cos(1/(z-2)) at infinity
分式线性变换: