- #1
MaestroBach
- 40
- 3
- Homework Statement
- Determine where f(z) = 2x + ixy^2 is differentiable and find its derivative in such locations.
- Relevant Equations
- Cauchy Riemann: Ux = Vy and Vx = -Uy
So just based on the cauchy riemann theorem, I think:
Ux = 2 = Vy = 2xy, so f(z) is differentiable on xy = 1, and also that Vx = y^2 = -Uy = 0. That doesn't make sense to me because if 0 = y^2, then y = 0, yet that wouldn't satisfy xy = 1, would it?
Furthermore, I'm not sure how I would calculate the derivative. If f(z) = something with z in it, like z^2, then I would definitely know how. However, I'm not sure how to deal with my function when it's in terms of x and y.
Can I use ∂/∂z = (1/2)(∂/∂x - i∂/∂y)?
I also saw a different thing where f' = Ux + iVx but I swear that doesn't work for a couple simple equations I was testing it on.
It seemed like f' = Ux + Vx worked though.Would appreciate any help!
Ux = 2 = Vy = 2xy, so f(z) is differentiable on xy = 1, and also that Vx = y^2 = -Uy = 0. That doesn't make sense to me because if 0 = y^2, then y = 0, yet that wouldn't satisfy xy = 1, would it?
Furthermore, I'm not sure how I would calculate the derivative. If f(z) = something with z in it, like z^2, then I would definitely know how. However, I'm not sure how to deal with my function when it's in terms of x and y.
Can I use ∂/∂z = (1/2)(∂/∂x - i∂/∂y)?
I also saw a different thing where f' = Ux + iVx but I swear that doesn't work for a couple simple equations I was testing it on.
It seemed like f' = Ux + Vx worked though.Would appreciate any help!