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?