Simple complex function question

Click For Summary
SUMMARY

The discussion focuses on finding the complex function f(z) = u(x,y) + iv(x,y) given the real part U = x^2 - 2xy - y^2 and checking for analyticity using the Cauchy-Riemann equations. The partial derivatives u_x and u_y are calculated as 2x - 2y and -(2x + 2y), respectively. From these, the corresponding v_y and v_x are derived, leading to the integration of v yielding v = 2xy - y^2 + x^2 + c, where h(x) is identified as a function of x. The discussion clarifies that h(x) can be treated as a constant of integration in the context of differentiation with respect to y.

PREREQUISITES
  • Understanding of complex functions and their components
  • Familiarity with Cauchy-Riemann equations
  • Knowledge of partial derivatives and integration techniques
  • Basic concepts of analyticity in complex analysis
NEXT STEPS
  • Study the implications of the Cauchy-Riemann equations in complex analysis
  • Explore the concept of analyticity and its significance in complex functions
  • Learn about the integration of functions of multiple variables
  • Investigate the role of constants of integration in differential equations
USEFUL FOR

Students and professionals in mathematics, particularly those studying complex analysis, as well as educators seeking to clarify concepts related to complex functions and the Cauchy-Riemann equations.

John O' Meara
Messages
325
Reaction score
0
Find f(z)= u(x,y) + iv(x,y), given U [tex]= x^2 - 2xy - y^2 \\[/tex] and check for analyticity.
We have to find v(x,y) as follows:
[tex]u_x = v_y[/tex] and [tex]u_y = -v_x[/tex] Cauchy-Riemann equations
[tex]u_x = 2x - 2y[/tex] and [tex]u_y = -(2x+2y) \\[/tex].
Therefore[tex]v_y = 2x - 2y[/tex]...(i)
and [tex]v_x = 2x + 2y \\[/tex] ...(ii), integrating (i) with respect to y and then differentiating it with respect to x , we v=[tex]2xy - y^2 +[/tex]h(x) and [tex]v_x = 2y + \frac{dh}{dx} \\[/tex] on comparision with (ii) [tex]\frac{dh}{dx} = 2x[/tex] therefore h(x)= [tex]x^2+c \\[/tex] Therefore [tex]v = 2xy - y^2 +x^2+c\\[/tex]. Question. How can h(x) be a constant of integration, I thought the constant of integration could only be a pure number?
 
Last edited:
Physics news on Phys.org
If you are integrating and differentiating wrt y, then sure, h(x) can be considered a constant of integration. It's derivative wrt y is 0. Of course, it can't wrt x. It's all relative.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K