# Another complex question

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 $$\partial$$u/$$\partial$$r=1/r($$\partial$$v/$$\partial$$$$\theta$$) and $$\partial$$v/$$\partial$$r=-1/r($$\partial$$u/$$\partial$$$$\theta$$) are equivelent to the cauchy riemann equations.

## Homework Equations

CR equations: $$\partial$$u/$$\partial$$x=$$\partial$$v/$$\partial$$y and $$\partial$$u/$$\partial$$y=-$$\partial$$v/$$\partial$$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

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???