Let f:U→C function which is differentiable and U is open in C. Suppose f′(z0) exists where z0=a+ib∈ U⊂C. f(z)=u+iv where u:U→R and v:U→R. First take h=t∈R.f′(z0)=limt→0f(a+t+ib)−f(a+ib)tBreaking it we get limt→0u(a+t,b)−u(a,b)t+ilimt→0v(a+t,b)−v(a,b)t=∂u∂x|z0+i∂v∂x|z0
Now take h=it, t∈R
f′(z0)=limt→0f(a+ib+it)−f(a+ib)it
Breaking it down we get limt→0u(a,b+t)−u(a,b)it+ilimt→0v(a,b+t)−v(a,b)it=∂v∂y|z0−i∂u∂y|z0
Equating the two equations we get f is complex differentiable at z0 and ∂u∂x|z0=∂v∂y|z0∂v∂x|z0=−∂u∂y|z0These two equations are called Cauchy Riemann Equation
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