Cauchy Riemannian Equations in Cartesian Form - (VTU 2020,2019,2018,2017)
The necessary condition for a complex function f(z)=u(x,y)+iv(x,y) is said to be analytic is that, there exists four first order continuous partial derivatives ∂u∂x,∂u∂y,∂v∂x,∂v∂y such that it satisfies the conditions ∂u∂x=∂v∂x ∂v∂x=−∂u∂y Proof: Let f(z)=u(x,y)+iv(x,y) be a complex function and is analytic at z=x+iy. According to definition of Analytic function we have f′(z)=limδz→zf(z+δz)−f(z)δz. Let δz=δx+iδy then f(z+δz)=u(x+δx,y+δy)+iv(x+δx,y+δy) ⟹f′(z)=[u(x+δx,y+δy)−u(x,y)]+i[v(x+δx,y+δy)−v(x,y)]δx+iδy ⟹f′(z)=limδz→0[u(x+δx,y+δy)−u(x,y)]δx+iδy+ilimδz→0[v(x+δx,y+δy)−v(x,y)]δx+iδy Case 01: If z approaches horizontally, i.e., δy=0 then δz=δx ⟹f′(z)=limδx→0[u(x+δx,y)−u(x,y)]δx+ilimδx→0[v(x+δx,y)−v(x,y)]δx ⟹f′(z)=∂u∂x+i∂v∂x⟶(1)Case 02: If z approaches vertically, i.e., δx=0 then δz=iδy ⟹f′(z)=limiδy→0[u(x,y+δy)−u(x,y)]iδy+limiδy→0[v(x,y+δy)−v(x,y)]δy ⟹f′(z)=−i∂u∂y+∂v∂y⟶(2) Comparing (1) and (2) we will get ∂u∂x=∂v∂x,∂v∂x=−∂u∂y
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