Let, fz = u + iv , is an analytic function. ⇒Refz = u ⇒ ⇒Refz2 = u2 ∎ ∂ ∂x Refz2 = ∂ ∂x u2 = 2 u ux ⇒ ∂2 ∂x2 Refz2 = ∂2 ∂x2 u2 = 2 u uxx + ux ux = 2 uuxx + ux2 -----1 ∎ ∂ ∂y Refz2 = ∂ ∂y u2 = 2 u uy ⇒ ∂2 ∂y2 Refz2 = ∂2 ∂y2 u2 = 2 u uyy + uy uy = 2 uuyy + uy2 -----2 Taking 1 + 2, We have ∂2 ∂x2 + ∂2 ∂y2 Refz2 = 2 uuxx + ux2 + 2 uuyy + uy2 ∂2 ∂x2 + ∂2 ∂y2 Refz2 = 2u uxx + uyy + 2 ux2 + uy2 ∂2 ∂x2 + ∂2 ∂y2 Refz2 = 2f'z2 Because 1 fz = u + iv is an analytic function, then u is a harmonic function. So, uxx + uyy = 0. 2 fz = u + iv ⇒f'z = ux + ivx = ux - iuy by CR - equations. 2 fz = u + iv ⇒f'z = ux2 + uy2 2 fz = u + iv ⇒f'z2 = ux2 + uy2