- #1
Ed Quanta
- 297
- 0
Ok, I am told in a complex analysis book that the gradient squared of u is equal to the gradient squared of v which is equal to 0.
We know the derivate of w exists, and w(z)=u(x,y) + iv(x,y)
Thus the Cauchy Riemann conditions must hold. (When I use d assume that it refers to a partial derivative)
So du/dx=dv/dy and du/dy=-dv/dx by Cauchy Riemann
We know the gradient squared of u is equal to d^2u/dx^2 + d^2u/dy^2 is equal to d^2v/dy^2 - d^2v/dx^2
We know the gradient squared of v is equal to d^2v/dx^2 + d^2v/dy^2 which is equal to d^2u/dx^2 - d^2u/dy^2
Am I correct in assuming this? And if I am, I don't see how the gradient squared of u is equal to the gradient squared of v. Any suggestions?
We know the derivate of w exists, and w(z)=u(x,y) + iv(x,y)
Thus the Cauchy Riemann conditions must hold. (When I use d assume that it refers to a partial derivative)
So du/dx=dv/dy and du/dy=-dv/dx by Cauchy Riemann
We know the gradient squared of u is equal to d^2u/dx^2 + d^2u/dy^2 is equal to d^2v/dy^2 - d^2v/dx^2
We know the gradient squared of v is equal to d^2v/dx^2 + d^2v/dy^2 which is equal to d^2u/dx^2 - d^2u/dy^2
Am I correct in assuming this? And if I am, I don't see how the gradient squared of u is equal to the gradient squared of v. Any suggestions?