# Real/complex differentiable function

1. Dec 2, 2011

### Maybe_Memorie

1. The problem statement, all variables and given/known data

Is the following function real and complex-differentiable everywhere?

f(z) = Re(z)

2. Relevant equations

Cauchy-Riemann equations fy = ifx

3. 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) = fx(z)Δx + fy(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

fx(z)Δx = ∂f/∂x (Δx) = ∂x/∂x (Δx) = (Δx)
fy(z)Δy = 0

f(z+Δz) - f(z) = fx(z)Δx + fy(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.

fy = ifx

fy = 0
ifx = i

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

Is this correct?

2. Dec 2, 2011

### obafgkmrns

You've shown that Re(z)y ≠ iRe(z)x. The Cauchy-Riemann condition is fy = ifx. What's the logical conclusion? ;o)

3. Dec 2, 2011

### Maybe_Memorie

I'm sorry? What do you mean?

f(z) is Re(z).

4. Dec 2, 2011

### HallsofIvy

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

5. Dec 2, 2011

### Maybe_Memorie

So f is not complex differenctiable?