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On 30 Apr 2006 03:53:28 -0700, pindu12@xxxxxxxxxxxxx wrote:
>Hello there
>
> 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) ? )
No, 1/z does not have an antiderivative in the
entire domain C \ {0}.
Exactly what definition of ln(z) do you have
in mind? Whichever "branch" you choose, there
will be at least one point where it is not
even continuous, hence not differentiable.
>but any closed paths DO NOT have integral 0
>
>thank you
************************
David C. Ullrich
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