Another complex question

  • Thread starter fredrick08
  • Start date
  • #1
376
0

Homework Statement


f(z)=u(r,theta)+iy(r,theta)... where x=rcos(theta) and y=rsin(theta), use chain rule to show that [tex]\partial[/tex]u/[tex]\partial[/tex]r=1/r([tex]\partial[/tex]v/[tex]\partial[/tex][tex]\theta[/tex]) and [tex]\partial[/tex]v/[tex]\partial[/tex]r=-1/r([tex]\partial[/tex]u/[tex]\partial[/tex][tex]\theta[/tex]) are equivelent to the cauchy riemann equations.


Homework Equations


CR equations: [tex]\partial[/tex]u/[tex]\partial[/tex]x=[tex]\partial[/tex]v/[tex]\partial[/tex]y and [tex]\partial[/tex]u/[tex]\partial[/tex]y=-[tex]\partial[/tex]v/[tex]\partial[/tex]x


The Attempt at a Solution


Ok the im unsure by how i am meant to use the chain rule here? and instead of typing out the dirvative im goin to just write i.e d/dx..

i did, dz/dr=dz/dx*dx/dr=1(cos(theta) and dz/dtheta=dz/dy*dy/dtheta=rcos(theta)... but that doesnt make sense... its the same as the provided equations without the 1/r.. but if i do the CR equations i get, du/dx=1 and du/dy=0????
 

Answers and Replies

  • #2
240
2
Your use of the chain rule is incorrect here.

Since we're talking about functions of several variables, the chain rule must consider all the possible derivatives. So for example:

[tex]\frac{\partial u}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial u}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial u}{\partial \theta}[/tex]
 
  • #3
376
0
ok then, but can i ask how do u do dr/dx and dtheta/dx???? since r and theta are part of x?
 
  • #4
376
0
do i rearrange?? coz r=x/cos(theta)?? then diffrentiate with respect to x?
 
  • #5
376
0
ok so if i do the chain rule that elib gave... i get (1/cos(theta))*cos(theta)+-1/(r*sqrt(1-x^2/r^2)*-rsin(theta).... but when i do dv/dy i kind of get similar answer, but instead of the x^2 its y^2, and the last term is cos(theta) and not sin(theta)
 
  • #6
376
0
ok for the x^2 and y^2 values, i subbed in there respecful values x=rcos(theta) y=rsin(theta) then i get...
1+(sin([tex]\theta[/tex])/sqrt(1-cos^2([tex]\theta[/tex])/r^2))=1+(cos([tex]\theta[/tex])/sqrt(1-sin^2([tex]\theta[/tex])/r^2)), which clearly does no equal???
 

Related Threads on Another complex question

Replies
1
Views
715
Replies
7
Views
818
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Top