Partial derivatives and complex numbers

  • Thread starter nmsurobert
  • Start date
  • #1
274
30

Homework Statement


show that the following functions are differentiable everywhere and then also find f'(z) and f''(z).
(a) f(z) = iz + 2

so f(z) = ix -y +2


then u(x,y) = 2-y, v(x,y) = x

Homework Equations


z=x+iy
z=u(x,y) +iv(x,y)
Cauchy-Riemann conditions says is differentiable everywhere if :
∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x

The Attempt at a Solution


so using the Cauchy-Riemann conditions i find that the function is differentiable everywhere. the part im stuck on is finding the first derivative.
f'(z) should be in the form of two partial derivatives right? because of the way the variables are set up.
so...
f'(z) =
∂z/∂x = ∂z/∂u(∂u/∂x) + ∂z/∂v(∂v/∂x)
∂z/∂y = ∂z/∂u/(∂u/∂y) + ∂z/∂v(∂v/∂y)

but where do i go from here? i can solve partials of u with respect to x or y but i dont know how to solve the partials of z with respect to u.

thank you!!!
 

Answers and Replies

  • #2
34,885
6,624

Homework Statement


show that the following functions are differentiable everywhere and then also find f'(z) and f''(z).
(a) f(z) = iz + 2

so f(z) = ix -y +2


then u(x,y) = 2-y, v(x,y) = x

Homework Equations


z=x+iy
z=u(x,y) +iv(x,y)
Cauchy-Riemann conditions says is differentiable everywhere if :
∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x

The Attempt at a Solution


so using the Cauchy-Riemann conditions i find that the function is differentiable everywhere. the part im stuck on is finding the first derivative.
f'(z) should be in the form of two partial derivatives right? because of the way the variables are set up.
so...
f'(z) =
∂z/∂x = ∂z/∂u(∂u/∂x) + ∂z/∂v(∂v/∂x)
∂z/∂y = ∂z/∂u/(∂u/∂y) + ∂z/∂v(∂v/∂y)

but where do i go from here? i can solve partials of u with respect to x or y but i dont know how to solve the partials of z with respect to u.

thank you!!!
Isn't f just a linear function of z?
 
  • #3
130
30
The derivative of the function of z does not consist of partial derivatives, you are looking for df/dz. The process to do this is to use limits as both Δx and Δy approach zero, where the numerator is analogous to the definition of the single variable derivative is divided by Δx + iΔy, analogous to h in single variable differentiation

[itex]\frac{d}{dz} = \lim_{Δx,Δy\to0} \frac{u(x+Δx,y+Δy) - u(x,y) + iv(x+Δx,y+Δy)) - iv(x,y)}{Δx+iΔy} [/itex]

If you set Δy = 0 first, and let Δx → 0, then
[itex]\frac{df}{dz} = \frac{∂u(x,y)}{∂x} + i\frac{∂v(x,y)}{∂x} . . . (1) [/itex]

If you set Δx = 0 first, and let Δy → 0, then
[itex]\frac{df}{dz} = -i\frac{∂u(x,y)}{∂y} + \frac{∂v(x,y)}{∂y} . . . (2) [/itex]

Both of these are ways to calculate df/dz, IF the real part of (1) is equivalent to the real part of (2), and the imaginary part of (1) is equivalent to the imaginary part of (2). This is where the Cauchy-Riemman condition comes from.
If we equate the real parts of equations (1) and (2), we have:
[itex] \frac{∂u(x,y)}{∂x} = \frac{∂v(x,y)}{∂y} [/itex]
If we equate the imaginary parts of equations (1) and (2), we have:
[itex] \frac{∂v(x,y)}{∂x} = -\frac{∂u(x,y)}{∂y} [/itex]
 
Last edited:
  • #4
34,885
6,624
[itex]\frac{d}{dz} = \lim_{Δx,Δy\to0} \frac{u(x+Δx,y+Δy) - u(x,y) + iv(x+Δx,y+Δy)) - iv(x,y)}{Δx+iΔy} [/itex]
d/dz is an operator, not a number or function. On the left side you should have ##\frac{df}{dz}## or something similar, indicating that you are taking the derivative of f with respect to z.
 
  • #5
130
30
d/dz is an operator, not a number or function. On the left side you should have ##\frac{df}{dz}## or something similar, indicating that you are taking the derivative of f with respect to z.

You're right, I entirely forgot the df on the first part.

It won't let me edit the post so just keep that in mind :]
 

Related Threads on Partial derivatives and complex numbers

Replies
1
Views
914
Replies
4
Views
993
  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
862
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
860
  • Last Post
Replies
1
Views
893
  • Last Post
Replies
5
Views
839
Top