User:Selfworm/Math2

We are interested in studying how complex analytic functions behave when limited to the real line. So, let $\mathbb {D}$ be a domain that contains some part of the real line and let $f(x)\,$ be a complex analytic function from the domain$\mathbb {D}$ into $\mathbb {C}$. Since $f\,$ is analytic it has harmonic conjugates, call them $u(x,y)\,$ and $v(x,y)\,$. So,

$f(z)=u(x,y)+iv(x,y)=u(z,0)+iv(z,0)\,$

An important result follows:

$f(z)\,$	$=\int f'(z)\,dz$
	$=\int u_{x}(z,0)-iu_{y}(z,0)\,dz$
	$=\int v_{y}(z,0)+iv_{x}(z,0)\,dz$

We will derive the above result by using the conjugate of $f'\,(x)$. The domain of $f\,$ and ${\overline {f}}$ need not be the same so we will let ${\overline {\mathbb {D} }}=\{z:{\overline {z}}\in \mathbb {D} \}$, and $\mathbb {D^{*}} =\mathbb {D} \bigcap {\overline {\mathbb {D} }}$. In this way both $f\,$ and ${\overline {f}}$ will be defined on $\mathbb {D^{*}}$. Now since $f\,$ is analytic on $\mathbb {D^{*}}$ we have that

$f'(z)\,$	$=u_{x}(x,y)+iv_{x}(x,y)\,$
	$=v_{y}(x,y)-iu_{y}(x,y)\,$

And by the Cauchy-Riemann equations we arrive at

$f'(z)\,$	$=u_{x}(x,y)-iu_{y}(x,y)\,$
	$=v_{y}(x,y)+iv_{x}(x,y)\,$

Since we are interested in the real line we will let $y=0\,$. Consequently $z=x+yi=x+0i=x\,$.
To simplify notation we let $F=f'\,$ and we note that,

${\overline {F(z)}}=u_{x}(x,-y)+iu_{y}(x,-y)$

Now,

$u_{x}(x,y)={F(x,y)+{\overline {F(x,y)}} \over {2}}$ and $u_{y}(x,y)={F(x,y)-{\overline {F(x,y)}} \over {-2i}}$

and so,

$u_{x}(x,y)-iu_{y}(x,y)\,$	$={F(x,y)+{\overline {F(x,y)}} \over {2}}-i{F(x,y)-{\overline {F(x,y)}} \over {-2i}}$
	$={F(x,y) \over {2}}+{{\overline {F(x,y)}} \over {2}}+{F(x,y) \over {2}}-{{\overline {F(x,y)}} \over {2}}$
	$=F(x,y)\,$

And so we have proved that $f'(z)=F(z)=u_{x}(x,y)-iu_{y}(x,y)\,$ and integrating both sides and using the $y=0\,$ substitution we get the desired result .

$f(z)\,$	$=\int u_{x}(z,0)-iu_{y}(z,0)\,dz$
	$=\int v_{y}(z,0)+iv_{x}(z,0)\,dz$