| First read the statement 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. | 
