character of f

 

Thm 1

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$ .

$f'(z) = 0$ on $\Omega$

$\Rightarrow$  $f$ is a constant function.

 

  Let $f(z) = u(x, y) + iv(x, y)$

  Since $f'(z) = \dfrac{\partial f}{\partial x} = -i \dfrac{\partial f}{\partial y} =0$,

  $\dfrac{\partial u}{\partial x} = 0$,     $\dfrac{\partial u}{\partial y} = 0$

  $\dfrac{\partial v}{\partial x} = 0$,     $\dfrac{\partial v}{\partial y} = 0$

  $\therefore$  $f$ is constant function

 

 

 

Thm 2

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$ .

Let $f$ be an analytic function and $c \in \mathbb{R}$

$\left|f(z) \right| = c$ on $\Omega$

$\Rightarrow$  $f(z)$ is a constant function.

 

  Let $f(z) = u(x, y) + iv(x, y)$

  Since $\left|f \right| = \left|u + iv \right| = c$,

  $u^{2} + v^{2} = c^{2}$

 

  Through the partial differentiation with respect to x and y, we can get

  $u \cdot \dfrac{\partial u}{\partial x} + v \cdot \dfrac{\partial v}{\partial x} = 0$

  $u \cdot \dfrac{\partial u}{\partial y} + v \cdot \dfrac{\partial v}{\partial y} = 0$

 

  By Cor 1 and Thm 2,

  C-R equation of $f$ is satisfied at all points.

 

  Therefore

  $u \cdot \dfrac{\partial u}{\partial x} - v \cdot \dfrac{\partial u}{\partial y} = 0$

  $u \cdot \dfrac{\partial u}{\partial y} + v \cdot \dfrac{\partial u}{\partial x} = 0$

 

  If you solve the above system of equations,

  $\dfrac{\partial u}{\partial x} = 0$,     $\dfrac{\partial u}{\partial y} = 0$

 

  $\therefore$  $u$ is a constant function

  Similarly, we can show $v$ is a constant function.

 

  $\therefore$  $f$ is a constant function.

 

 

 

Thm 3

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$ .

Let $f$ be an analytic function and $l$ be a curve.

$f(z) \in l$ on $\Omega$

$\Rightarrow$  $f$ is a constant function.

 

  By 이중적분의 변수변환,

  $\displaystyle \iint_{f(\Omega)} \; \mathrm{d}A = \iint_{\Omega} \left| \frac{\partial (u,v)}{\partial (x, y)} \right| \; \mathrm{d}x \mathrm{d}y$

 

  By defenition,

  $\left| \dfrac{\partial (u,v)}{\partial (x, y)} \right| = \left|\det \begin{pmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \\
\end{pmatrix} \right| = \left| \dfrac{\partial u}{\partial x} \dfrac{\partial v}{\partial y} - \dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x} \right|$

 

  By Cor 1 and Thm 2,

  C-R equation of $f$ is satisfied at all points.

 

  Therefore

  $\left| \dfrac{\partial (u,v)}{\partial (x, y)} \right| = (\dfrac{\partial u}{\partial x})^{2} + (\dfrac{\partial v}{\partial x})^{2} = (\dfrac{\partial u}{\partial y})^{2} + (\dfrac{\partial v}{\partial y})^{2}$

 

  Since $\displaystyle \iint_{f(\Omega)} \; \mathrm{d}A = 0$ and $(\dfrac{\partial u}{\partial x})^{2} + (\dfrac{\partial v}{\partial x})^{2} \geq 0$,

  $(\dfrac{\partial u}{\partial x})^{2} + (\dfrac{\partial v}{\partial x})^{2} = 0$

 

  $\therefore$  $\dfrac{\partial u}{\partial x} = 0$,  $\dfrac{\partial v}{\partial x} = 0$

 

  Similarly, we can get

  $\dfrac{\partial u}{\partial y} = 0$,  $\dfrac{\partial v}{\partial y} = 0$  

 

  $\therefore$  $f$ is a constant function.

 

 

 

Thm 4

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$ be an analytic function.

Define $\phi : \mathbb{C} \to \mathbb{R}^{2}$ by

$\phi(a + ib) = \begin{pmatrix}
a \\
b \end{pmatrix}$

$\Rightarrow$  For $\alpha \in \mathbb{C}$,  $\phi( f' \times \alpha ) = D(\phi \circ f \circ \phi^{-1}) \times \phi(\alpha)$

 

  By Cor 1,

  $f' = \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x}$

 

  By Cor 1 and Thm 2,

  C-R equation of $f$ is satisfied at all points.

 

  Let $\alpha = h + ik$

  $f' \times \alpha = (\dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x}) (h+ik) $
                 $= (\dfrac{\partial u}{\partial x}h - \dfrac{\partial v}{\partial x}k) + i (\dfrac{\partial v}{\partial x}h + \dfrac{\partial u}{\partial x}k)$

                 $= (\dfrac{\partial u}{\partial x}h + \dfrac{\partial u}{\partial y}k) + i (\dfrac{\partial v}{\partial x}h + \dfrac{\partial v}{\partial y}k)$

 

  Therefore

  $\phi(f' \times \alpha) = \begin{pmatrix}
\dfrac{\partial u}{\partial x}h + \dfrac{\partial u}{\partial y}k \\
\dfrac{\partial v}{\partial x}h + \dfrac{\partial v}{\partial y}k \end{pmatrix} = \begin{pmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \\
\end{pmatrix} \begin{pmatrix}
h \\
k \end{pmatrix}$

 

  $\therefore$  $\phi( f' \times \alpha ) = D(\phi \circ f \circ \phi^{-1}) \times \phi(\alpha)$