Unraveling the Equivalence: Chain Rule and Cauchy-Riemann Equations

[fredrick08]

Homework Statement

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

Homework Equations

CR equations: \partialu/\partialx=\partialv/\partialy and \partialu/\partialy=-\partialv/\partialx

The Attempt at a Solution

Ok the I am unsure by how i am meant to use the chain rule here? and instead of typing out the dirvative I am 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 doesn't 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?

 

[elibj123]

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:

\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}

 

[fredrick08]

ok then, but can i ask how do u do dr/dx and dtheta/dx? since r and theta are part of x?

 

[fredrick08]

do i rearrange?? coz r=x/cos(theta)?? then diffrentiate with respect to x?

 

[fredrick08]

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)

 

[fredrick08]

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(\theta)/sqrt(1-cos^2(\theta)/r^2))=1+(cos(\theta)/sqrt(1-sin^2(\theta)/r^2)), which clearly does no equal?