Complex Analysis: Show Integration of f(z)f' dz is Purely Imaginary

[sbashrawi]

Homework Statement

Let gama be a closed curve and f be analytic function. Show that the integration of f(z)f' dz is puerly imaginary

Homework Equations

The Attempt at a Solution

 

[Eynstone]

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.

 

[sbashrawi]

Thank you very much

I am sorry the true statement 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

 

[Count Iblis]

Take the real part of the integral expression by adding the complex conjugate.

 

[sbashrawi]

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

 

[Count Iblis]

integr( f + conj(f))*f'dz =

integr (2Re(f))*f'dz =

integr[(2 u) (du + i dv)] =

2i integr u dv

 

[sbashrawi]

Thank you very much