proof of the Cauchy-Riemann equations

Suppose that we have f:U-->C, a complex-valued function defined on some open subset U. Choose a point z₀ in U and think of f as mapping to R². What does it mean for f to have a derivative at z₀ in the sense of real functions? Well, first WLOG we assume that z₀=0 and f(z₀)=0 (this simplifies the algebra but makes no essential difference). The derivative exists at this point iff there exist a,b in C such that

f(z)=ax+by+k(z)z (*)

where k(z)--> 0 as z --> 0. (Note that a,b are just the partial derivatives of f with respect to x and y and we are writing z=x+iy, as usual.)

Looking at (*) we can rewrite it as

f(z)=(1/2)(a-ib)z + (1/2)(a+ib) bar(z) +k(z)z

where bar(z)=x-iy is the complex conjugate. (This just uses that z+bar(z)=2x and z-bar(z)=2iy.)

When z is nonzero we can divide both sides by it and we get

f(z)/z = (1/2)(a-ib) + (1/2)(a+ib)(bar(z)/z) +k(z)

Now f(z)/z has a limit at zero iff the coefficient (a+ib) of bar(z)/z is zero. This is because bar(z)/z is 1 if z is real but is -1 when z is imaginary.

If we write f=u+iv then a=du/dx + idv/dx and b=du/dy +idv/dy (at the point under consideration) and so we deduce that if f is holomorphic on U then then du/dx-dv/dy=0 and dv/dx+du/dy=0, as was required. Conversely, if f=u+iv for continuously differentiable real functions u(x,y) and v(x,y) satisfying these equations then f is holomorphic.