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Taking the derivative of complex functions

  • #1

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!
 

Answers and Replies

  • #2
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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
It's been many years since I had a course on complex analysis, but both of the Cauchy-Riemann conditions have to be satisfied on some open subset of ##\mathbb C## for the function to be complex differentiable, not just one of those conditions.

Also, your notation is a little hard to follow. You can write a partial derivative using a BBCode subscript, like this: Ux, or using LaTeX, like this: ##V_y##. There are links to tutorials on each of these techniques at the lower left corner of the page.
MaestroBach said:
, 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?
No. x is undefined if y = 0.
MaestroBach said:
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!
 
  • #3
PeroK
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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?

Would appreciate any help!
You have two conditions to satisfy. The first is that ##y = 0##. The second is that ##xy = 1##. This means that you cannot satisfy both conditions at any point. The function, therefore, is not differentiabe at any point.
 
  • #4
You have two conditions to satisfy. The first is that ##y = 0##. The second is that ##xy = 1##. This means that you cannot satisfy both conditions at any point. The function, therefore, is not differentiabe at any point.

Appreciate it!

Sorry to bother you but any thoughts on how I would have taken the derivatives, had they existed? I still am not sure how to handle derivatives of f(z) when written in terms of x and y.
 
  • #5
It's been many years since I had a course on complex analysis, but both of the Cauchy-Riemann conditions have to be satisfied on some open subset of ##\mathbb C## for the function to be complex differentiable, not just one of those conditions.

Also, your notation is a little hard to follow. You can write a partial derivative using a BBCode subscript, like this: Ux, or using LaTeX, like this: ##V_y##. There are links to tutorials on each of these techniques at the lower left corner of the page.
No. x is undefined if y = 0.

Yeah I was actually trying to figure out how to use LaTeX here, but something tripped me up a little so I didn't use it.
 
  • #6
PeroK
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