Re: complex analysis Q

pindu12@xxxxxxxxxxxxx wrote:

i have a theorem in my notes

suppose f:D--->Complex number where D is a connected domain
suppose f cts then the following are equivalent

1) f has an anti derivative F in D s.t F ' = f
2) integral f(z) dz only depends on end points
3) integral f(z) dz =0 for any CLOSED path

well i see (2) <=> (3)
but i dont see if (1) is true then (3) is
take for example f(z)=1/z on the domain C/{0}
it is cts, and has an antiderivative (ln(z) ? )
but any closed paths DO NOT have integral 0

No, f(z) = 1/z has no antiderivative in C\{0}. What's ln(z)?

Note that when a function _f_ has an antiderivative F, then, for
every path _g_ defined on an interval [a,b], the integral of _f_ along
_g_ is equal to F(g(b)) - F(g(a)). This proves that 1) => 2). On the
other hand, if 2) holds then fix some _w_ in D and define F(z) as the
integral of _f_ along some path which goes from _w_ to _z_; this
makes sense because you're assuming 2). Well, it turns out that F is an
antiderivative of _f_. So, 2) => 1).

Best regards,

Jose Carlos Santos
.