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Maybe_Memorie
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Homework Statement
Is the following function real and complex-differentiable everywhere?
f(z) = Re(z)
Homework Equations
Cauchy-Riemann equations fy = ifx
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?