277 lines
10 KiB
Markdown
277 lines
10 KiB
Markdown
# Math416 Lecture 4
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## Review
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### Derivative of a complex function
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$$
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\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)
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$$
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$$
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\frac{\partial f}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)
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$$
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### Angle between two curves
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Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
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The angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
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### Cauchy-Riemann equations
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$$
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\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)
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$$
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## Continue on last lecture
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### Theorem of conformality
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Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
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If $f'(z_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the same as the angle between the vectors $f'(z_0)\gamma_1'(t_0)$ and $f'(z_0)\gamma_2'(t_0)$.
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### Lemma of function of a curve and angle
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If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=z_0$ for some $t_0\in I$.
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Then,
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$$
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(f\circ \gamma)'(t_0)=f'(\gamma(t_0))\gamma'(t_0).
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$$
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> Looks like the chain rule.
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<details>
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<summary>Proof</summary>
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We want to show that
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$$
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\lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0}=f'(\gamma(t_0))\gamma'(t_0).
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$$
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> Notation:
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>
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> A function $g(h)$ is $O(h)$ if $\exists C>0$ such that $|g(h)|\leq C|h|$ for all $h$ in a neighborhood of $0$.
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>
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> A function $g(h)$ is $o(h)$ if $\lim_{h\to 0}\frac{g(h)}{h}=0$.
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>
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> $f$ is differentiable if and only if $f(z+h)=f(z)+f'(z)h+\frac{1}{2}h^2f''(z)+o(h^3)$ as $h\to 0$. (By Taylor expansion)
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Since $f$ is holomorphic at $\gamma(t_0)=z_0$, we have
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$$
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f(z_0)=f(z_0)+(z-z_0)f'(z_0)+o(z-z_0)
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$$
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> This result comes from Taylor Expansion of the derivative of the function around the point $z_0$
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and
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$$
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f(\gamma(t_0))=f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))
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$$
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So,
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$$
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\begin{aligned}
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\lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0}
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&=\lim_{t\to t_0}\frac{\left[f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))\right]-f(\gamma(t_0))}{t-t_0} \\
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&=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))}{t-t_0} \\
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&=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))}{t-t_0} +\lim_{t\to t_0}\frac{o(\gamma(t)-\gamma(t_0))}{t-t_0} \\
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&=f'(\gamma(t_0))\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0} +0\\
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&=f'(\gamma(t_0))\gamma'(t_0)
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\end{aligned}
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$$
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</details>
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#### Definition 2.12 (Conformal function)
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A function $f:G\to \mathbb{C}$ is called conformal if it preserves the angle between two curves.
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#### Theorem 2.13 (Conformal function)
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If $f:G\to \mathbb{C}$ is conformal at $z_0\in G$, then $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
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Example:
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$$
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f(z)=z^2
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$$
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is not conformal at $z=0$ because $f'(0)=0$.
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#### Lemma of conformal function
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Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial z}(z_0)$, $b=\frac{\partial f}{\partial \overline{z}}(z_0)$.
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Let $\gamma(t_0)=z_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
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<details>
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<summary>Proof</summary>
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$f=u+iv$, $u,v$ are real differentiable.
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$$
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a=\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
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$$
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$$
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b=\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
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$$
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$$
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\gamma'(t_0)=\frac{d\alpha}{dt}+i\frac{d\beta}{dt}
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$$
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$$
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\overline{\gamma'(t_0)}=\frac{d\beta}{dt}-i\frac{d\alpha}{dt}
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$$
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$$
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\begin{aligned}
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(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial z}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{z}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
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&=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\alpha}{dt}+i\frac{d\beta}{dt}\right)\\
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&+\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\beta}{dt}-i\frac{d\alpha}{dt}\right) \\
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&=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\alpha}{dt}-\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\beta}{dt}\right]\\
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&+i\left[\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\alpha}{dt}+\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\beta}{dt}\right] \\
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&=\left[a+b\right]\frac{d\alpha}{dt}+i\left[a-b\right]\frac{d\beta}{dt} \\
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&=\left[u_x+iv_x\right]\frac{d\alpha}{dt}+i\left[v_y-iu_y\right]\frac{d\beta}{dt} \\
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&=a\gamma'(t_0)+b\overline{\gamma'(t_0)}
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\end{aligned}
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$$
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</details>
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#### Theorem of differentiability
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Let $f:G\to \mathbb{C}$ be a function defined on an open set $G\subset \mathbb{C}$ that is both holomorphic and (real) differentiable, where $f=u+iv$ with $u,v$ real differentiable functions.
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Then, $f$ is conformal at every point $z_0\in G$ if and only if $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
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<details>
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<summary>Proof</summary>
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We prove the equivalence in two parts.
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($\implies$) Suppose that $f$ is conformal at $z_0$. By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $z_0$. In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $z_0$, $Df(z_0)$, acts as a similarity transformation. Any similarity in $\mathbb{R}^2$ can be written as a rotation combined with a scaling; in particular, its matrix representation has the form
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$$
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\begin{pmatrix}
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A & -B \\
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B & A
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\end{pmatrix},
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$$
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for some real numbers $A$ and $B$. This is exactly the matrix corresponding to multiplication by the complex number $a=A+iB$. Therefore, the Cauchy-Riemann equations must hold at $z_0$, implying that $f$ is holomorphic at $z_0$. Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(z_0)=a\neq 0$.
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($\impliedby$) Now suppose that $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$. Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $z_0$ is
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$$
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f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|),
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$$
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for small $h\in\mathbb{C}$. Multiplication by the nonzero complex number $f'(z_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=z_0$, we have
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$$
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(f\circ\gamma)'(t_0)=f'(z_0)\gamma'(t_0),
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$$
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and the angle between any two tangent vectors at $z_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $z_0$.
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For further illustration, consider the special case when $f$ is an affine map.
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Case 1: Suppose
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$$
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f(z)=az+b\overline{z}.
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$$
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The Wirtinger derivatives of $f$ are
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$$
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\frac{\partial f}{\partial z}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{z}}=b.
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$$
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For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{z}}=b=0$. Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$. If $b\neq 0$, then the map mixes $z$ and $\overline{z}$, and one can check that the linearization maps the real axis $\mathbb{R}$ into the set $\{(a+b)t\}$, which does not uniformly scale and rotate all directions. Thus, $f$ fails to be conformal when $b\neq 0$.
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Case 2: For a general holomorphic function, the lemma of conformal functions shows that if
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$$
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(f\circ \gamma)'(t_0)=f'(z_0)\gamma'(t_0)
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$$
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for any differentiable curve $\gamma$ through $z_0$, then the effect of $f$ near $z_0$ is exactly given by multiplication by $f'(z_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $z_0$.
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</details>
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### Harmonic function
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Let $\Omega$ be a domain in $\mathbb{C}$. A function $u:\Omega\to \mathbb{R}$
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> A domain is a connected open set.
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Say $g:\Omega\to \mathbb{R} \text{ or } \mathbb{C}$ is harmonic if it satisfies the Laplace equation
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$$
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\Delta g=\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}=0.
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$$
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#### Theorem of harmonic conjugate
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Let $f=u+iv$ be holomorphic function on domain $\Omega\subset \mathbb{C}$. Then $u$ and $v$ are harmonic functions on $\Omega$.
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<details>
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<summary>Proof</summary>
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$$
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\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.
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$$
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Using the Cauchy-Riemann equations, we have
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$$
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\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x\partial y}, \quad \frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y\partial x}.
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$$
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So,
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$$
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\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.
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$$
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</details>
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If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$.
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Example:
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$$
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u(x,y)=x^2-y^2
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$$
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is harmonic on $\mathbb{C}$.
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To find a harmonic conjugate of $u$ on $\mathbb{C}$, we need to find a function $v$ such that
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$$
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\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=2y, \quad \frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2x.
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$$
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Integrating, we get
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$$
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v(x,y)=2xy+G(y)
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$$
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$$
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\frac{\partial v}{\partial y}=2x+G'(y)=2x
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$$
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So,
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$$
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G'(y)=0 \implies G(y)=C
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$$
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$$
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v(x,y)=2xy+C
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$$
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is a harmonic conjugate of $u$ on $\mathbb{C}$.
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Combine $u$ and $v$ to get $f(x,y)=x^2-y^2+2xyi+C=(x+iy)^2+C=z^2+C$, which is holomorphic on $\mathbb{C}$.
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