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:


 * {| style="background:none;"


 * $$f(z)\,$$
 * $$= \int f'(z)\,dz$$
 * $$= \int u_x (z, 0) - iu_y (z, 0)\,dz$$
 * $$= \int v_y (z, 0) + iv_x (z, 0)\,dz$$
 * }
 * $$= \int u_x (z, 0) - iu_y (z, 0)\,dz$$
 * $$= \int v_y (z, 0) + iv_x (z, 0)\,dz$$
 * }
 * $$= \int v_y (z, 0) + iv_x (z, 0)\,dz$$
 * }

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


 * {| style="background:none;"


 * $$f'(z)\,$$
 * $$= u_x (x, y) + iv_x (x, y)\,$$
 * $$= v_y (x, y) - iu_y (x, y)\,$$
 * }
 * $$= v_y (x, y) - iu_y (x, y)\,$$
 * }
 * }

And by the Cauchy-Riemann equations we arrive at


 * {| style="background:none;"


 * $$f'(z)\,$$
 * $$= u_x (x, y) - iu_y (x, y)\,$$
 * $$= v_y (x, y) + iv_x (x, y)\,$$
 * }
 * $$= v_y (x, y) + iv_x (x, y)\,$$
 * }
 * }

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,


 * {| style="background:none;"


 * $$u_x(x, y) - i u_y(x, y)\,$$
 * $$ = {F(x, y) + \overline{F(x, y)} \over{2}} - i{F(x, y) - \overline{F(x, y)} \over{-2i}}$$
 * $$ = {F(x, y)\over{2}} + {\overline{F(x, y)}\over{2}} + {F(x, y)\over{2}} - {\overline{F(x, y)}\over{2}}$$
 * $$ = F(x, y)\,$$
 * }
 * $$ = {F(x, y)\over{2}} + {\overline{F(x, y)}\over{2}} + {F(x, y)\over{2}} - {\overline{F(x, y)}\over{2}}$$
 * $$ = F(x, y)\,$$
 * }
 * $$ = F(x, y)\,$$
 * }

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


 * {| style="background:none;"


 * $$f(z)\,$$
 * $$= \int u_x (z, 0) - iu_y (z, 0)\,dz$$
 * $$= \int v_y (z, 0) + iv_x (z, 0)\,dz$$
 * }
 * $$= \int v_y (z, 0) + iv_x (z, 0)\,dz$$
 * }
 * }