Taking the derivative of complex functions

In summary, based on the Cauchy-Riemann theorem, it is determined that the function f(z) = 2x + ixy^2 is not differentiable at any point. This is because it does not satisfy both conditions of the theorem simultaneously, as one condition requires y = 0 while the other requires xy = 1. Therefore, the derivative of f(z) cannot be calculated. Additional resources are suggested for understanding the Cauchy-Riemann equations and complex derivatives.
  • #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!
 
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  • #2
MaestroBach said:
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
MaestroBach said:
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
PeroK said:
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
Mark44 said:
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
MaestroBach said:
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.

Try:

http://mathworld.wolfram.com/Cauchy-RiemannEquations.html

Or:

http://www1.spms.ntu.edu.sg/~ydchong/teaching/06_complex_derivatives.pdf
 

Related to Taking the derivative of complex functions

1. What is the definition of a derivative for complex functions?

The derivative of a complex function is defined as the limit of the difference quotient, where the difference between two points on the function approaches zero. It represents the rate of change of the function at a specific point.

2. How is the derivative of a complex function calculated?

The derivative of a complex function is calculated using the same rules as for real-valued functions, with the additional consideration of the complex conjugate. The derivative of a complex function is given by the limit of the difference quotient, which involves taking the derivative of both the real and imaginary parts of the function separately.

3. Can the derivative of a complex function be expressed as a single number?

No, the derivative of a complex function is a complex number, as it involves both real and imaginary components. It can also be written in terms of a vector, with the real and imaginary parts representing the x and y components respectively.

4. What is the geometric interpretation of the derivative of a complex function?

The geometric interpretation of the derivative of a complex function is that it represents the slope or tangent of the function at a specific point on the complex plane. It can also be thought of as the direction and magnitude of the instantaneous rate of change of the function.

5. What are some applications of taking the derivative of complex functions?

The derivative of complex functions has various applications in mathematics, physics, and engineering. It is used in the study of complex analysis, as well as in the fields of optics, electromagnetism, and fluid dynamics. It is also used in signal processing and control theory to analyze and manipulate complex signals and systems.

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