在(Bleft(0,r ight))
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}
ight|<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}
ight)=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:$$frac{z{3}+z{2}+2}{zleft(z{2}-1 ight){2}}$$
(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: