Cauchy–Riemann equations
Simple example

Suppose that z = x + i y {\displaystyle z=x+iy}. The complex-valued function f (z) = z 2 {\displaystyle f(z)=z^{2}} is differentiable at any point z in the complex plane. f (z) = (x + i y) 2 = x 2 − y 2 + 2 i x y {\displaystyle f(z)=(x+iy)^{2}=x^{2}-y^{2}+2ixy} The real part u (x, y) {\displaystyle u(x,y)} and the imaginary part v (x, y) {\displaystyle v(x,y)} are u (x, y) = x 2 − y 2 v (x, y) = 2 x y {\displaystyle {\begin{aligned}u(x,y)&=x^{2}-y^{2}\\v(x,y)&=2xy\end{aligned}}} and their partial derivatives are u x = 2 x; u y = − 2 y; v x = 2 y; v y = 2 x {\displaystyle u_{x}=2x;\quad u_{y}=-2y;\quad v_{x}=2y;\quad v_{y}=2x}

We see that indeed the Cauchy–Riemann equations are satisfied, u x = v y {\displaystyle u_{x}=v_{y}} and u y = − v x {\displaystyle u_{y}=-v_{x}}.
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