How to show a function is analytic?

  • Thread starter Thread starter numberthree
  • Start date Start date
  • Tags Tags
    Function
Click For Summary
To show that a function like log(z-i) is analytic, it is essential to demonstrate its differentiability. This can be achieved by applying the Cauchy-Riemann equations, which require the partial derivatives of the real and imaginary components to be continuous and satisfy specific relationships. Alternatively, differentiating log(z-i) yields 1/(z-i), which is valid everywhere except at z=i, indicating the function is analytic in that region. Another method involves integrating the function over a closed contour and verifying that the integral equals zero. Overall, proving the existence of a derivative in a neighborhood confirms the function's analyticity.
numberthree
Messages
8
Reaction score
0
how to show a function is analytic??

I know that to show a function is analytic I need to prove its differentiable, but for a fuction like log(z-i), how could i show it is analytic?
 
Physics news on Phys.org


Use the "Cauchy-Riemann equations which should be mentioned early in any book on "functions of a complex variable". A function f(x+ iy)= u(x,y)+ iv(x,y) is analytic at z_0= x_0+ iy_0 if and only if the partial derivatives, \partial u/\partial x, \partial u/\partial y, \partial v/\partial x, and \partial v/\partial y are continuous at the point and
\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}
and
\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}
 


yes, i know wat u mean, but i don't know how to separate log(z-i) into u + iv form
 


numberthree said:
I know that to show a function is analytic I need to prove its differentiable, but for a fuction like log(z-i), how could i show it is analytic?

Why not just differentiate it and then show the derivative exists in a region surrounding a point then it is analytic in that region so:

\frac{d}{dz} \log(z-i)=\frac{1}{z-i}

and that derivative exists everywhere except at z=i.
 


Or you could integrate the function over a closed line and show the integral is zero.
 


use log(z) = log(|z|) + i (arg(z))
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K