Real/complex differentiable function

[Maybe_Memorie]

Homework Statement

Is the following function real and complex-differentiable everywhere?

f(z) = Re(z)

Homework Equations

Cauchy-Riemann equations f_y = if_x

The Attempt at a Solution

Let z = x + iy, z₁ = x - iy

Re(z) can be defined by Re(z) = (z + z₁)/2

A function is ℝ-differentiable if
f(z+Δz) - f(z) = f_x(z)Δx + f_y(z)Δy + Ω(Δz)
and
Ω(Δz)/Δz -> 0 as Δz -> 0.

Δz = Δx + iΔy


So, f(z+Δz) = (z + Δz + (z + Δz)₁)/2
= x + Δx

So f(z+Δz) - f(z) = Δx

f_x(z)Δx = ∂f/∂x (Δx) = ∂x/∂x (Δx) = (Δx)
f_y(z)Δy = 0

f(z+Δz) - f(z) = f_x(z)Δx + f_y(z)Δx + Ω(Δz)
So Δx = Δx + Ω(Δz)
Thus Ω(Δz) = 0 and evidently 0/Δz -> 0 as Δz -> 0.

So f(z) is ℝ-differentiable everywhere.


I know from a theorem that if f is ℝ-differentiable and satisfies the Cauchy-Riemann equations then it is ℂ-differentiable.


f_y = if_x

f_y = 0
if_x = i

So f does not satisfy the equations, hence not ℂ-differentiable.


Is this correct?

 

[obafgkmrns]

You've shown that Re(z)_y ≠ iRe(z)_x. The Cauchy-Riemann condition is f_y = if_x. What's the logical conclusion? ;o)

 

[Maybe_Memorie]

You've shown that Re(z)_y ≠ iRe(z)_x. The Cauchy-Riemann condition is f_y = if_x. What's the logical conclusion? ;o)

I'm sorry? What do you mean?

f(z) is Re(z).

 

[HallsofIvy]

His point is that Re(z) does NOT satisfy the Cauchy-Riemann equations.

 

[Maybe_Memorie]

So f is not complex differenctiable?