Suppose that we have *f:U-->***C**,
a complex-valued function defined on some open subset *U*.
Choose a point *z _{0}* in

*U*and think of

*f*as mapping to

**R**

^{2}. What does it mean for

*f*to have a derivative at

*z*in the sense of real functions? Well, first WLOG we assume that

_{0}*z*and

_{0}=0*f(z*(this simplifies the algebra but makes no essential difference). The derivative exists at this point iff there exist

_{0})=0*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.