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

In summary, the integration of the complex conjugate of a function multiplied by the derivative of the function is purely imaginary. This can be proven by taking the real part of the integral expression, adding the complex conjugate, and using Cauchy's formula. The result is that the integral simplifies to 2i times the integral of u times dv.
  • #1
sbashrawi
55
0

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

 
Physics news on Phys.org
  • #2
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
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
 
  • #4
Take the real part of the integral expression by adding the complex conjugate.
 
  • #5
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
 
  • #6
integr( f + conj(f))*f'dz =

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

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

2i integr u dv
 
  • #7
Thank you very much
 

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

What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of functions of complex numbers. It is a powerful tool for solving problems in various areas of mathematics, physics, and engineering.

What is the integration of f(z)f' dz?

The integration of f(z)f' dz is a mathematical operation that involves finding the area under the curve of a function, where the function is multiplied by its derivative. In complex analysis, this operation is used to calculate the imaginary part of the integral of a complex function.

Why is the integration of f(z)f' dz important in complex analysis?

The integration of f(z)f' dz is important because it allows us to evaluate complex integrals and determine if the integral is purely imaginary. This information is useful in solving problems related to complex functions and their properties.

How do you show that the integration of f(z)f' dz is purely imaginary?

To show that the integration of f(z)f' dz is purely imaginary, we can use the Cauchy-Riemann equations. These equations relate the real and imaginary parts of a complex function and its derivative. By applying these equations to the integral, we can determine if the real part of the integral is equal to zero, which would make the integral purely imaginary.

Can you provide an example of the integration of f(z)f' dz being purely imaginary?

Yes, an example of the integration of f(z)f' dz being purely imaginary is the integral of eizsin(z)dz over the unit circle. By using the Cauchy-Riemann equations, we can show that the real part of this integral is equal to zero, making the integral purely imaginary.

Similar threads

Is f(z) =(1+z)/(1-z) a real function?
Replies
17
Views
2K
Classifying singularities of a function
Replies
2
Views
1K
Where is ##(z+1)Ln(z)## differentiable?
Replies
5
Views
1K
Show that the real part of a certain complex function is harmonic
Replies
7
Views
2K
When is an entire function a constant?
Replies
3
Views
2K
Complex Integration Along Given Path
Replies
3
Views
1K
Show that the given function is continuous
Replies
27
Views
2K
Derive an upper bound for |f(i)|
Replies
16
Views
1K
Understanding the Cauchy Integration Formula for Analytic Functions
Replies
4
Views
1K
Complex Analysis: Find Analytic Functions w/ |ƒ(z)-1| + |ƒ(z)+1| = 4
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top