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On 30 Apr 2006 03:53:28 -0700, 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
>
what is 'cts' supposed to mean ? continuous ?
>
>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
>
Apparently you don't realize what an 'explosive'
subject you attack by just writing down ln(z) for
complex z ! This expression is a priori *not* well defined
for complex z; one may define so-called principal
values for ln(z) that extend the usual definition
for real (strictly) positive z to any complex non-zero z
(so that the expression becomes well defined again)
*But* you must then accept one of two things:
1) ln(z) will remain *undefined* for negative real z
2) ln(z) will not even be continuous at negative real z's
(so in both cases the function ln(z) will not be an
anti-derivative of 1/z in the whole domain C\{0})
Using any "uniformization" of ln(z) defined in this domain
(i.e. getting rid of the problem of multiple values
at the same place by choosing between these), continuous
in its interior and continuous 'from above' at neg. reals
will result in a function whose limit 'from below' at one of
these reals will differ from its value at the same place
by 2*pi*sqrt(-1). This complex number p (like 'period':
it is a generator of the periods of e^z) verifies e^p=1.
(Above and below refer here to the usual drawing of
the complex plane with a real 'axis' going horizontally
from left to right, and sqrt(-1) being above this axis.)
And this must be so *because* (1) => (3)
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