Let gama be a closed curve and f be analytic function. Show that the integration of f(z)f' dz is puerly imaginary
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#2
Eynstone
336
0
Welcome sbashrawi,
As f(z)f'(z) = (1/2) d/dz f^2 , Cauchy's formula shows that what you claim is invalid unless gamma encircles some poles of f with real residues at them.
#3
sbashrawi
55
0
Thank you very much
I am sorry the true statment is that :
integration of ( conjugate of f ) * f' *dz is purely imaginary.
I tried to prove it using the winding number but I couldn't
#4
Count Iblis
1,851
7
Take the real part of the integral expression by adding the complex conjugate.
#5
sbashrawi
55
0
Hi
here is what I did:
integ(conj(f)*f'dz) = integr( f + conj(f))*f'dz
which implies
integ [ conj(f) - 2 Re(f)] * f' dz = 0
letting f = u + iv , then the expression will be
integ[ -u -iv] * f' dz = 0
then I couldn't find how it is purely imaginary from this step