update notations and fix typos
This commit is contained in:
Zheyuan Wu
@@ -12,16 +12,16 @@ We defined clircle to be a circle or a line.
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The circle equation is:
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Let $\zeta=u+iv$ be the center of the circle, $r$ be the radius of the circle.
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Let $z=u+iv$ be the center of the circle, $r$ be the radius of the circle.
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$$
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circle=\{z\in\mathbb{C}:|\zeta-c|=r\}
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circle=\{z\in\mathbb{C}:|z-c|=r\}
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$$
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This is:
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$$
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|\zeta|^2-c\overline{\zeta}-\overline{c}\zeta+|c|^2-r^2=0
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|z|^2-c\overline{z}-\overline{c}z+|c|^2-r^2=0
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$$
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If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
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@@ -29,7 +29,7 @@ If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps c
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We claim that a map is circle preserving if and only if for some $\alpha,\beta,\gamma,\delta\in\mathbb{R}$.
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$$
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\alpha|\zeta|^2+\beta Re(\zeta)+\gamma Im(\zeta)+\delta=0
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\alpha|z|^2+\beta Re(z)+\gamma Im(z)+\delta=0
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$$
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when $\alpha=0$, it is a line.
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@@ -38,7 +38,7 @@ when $\alpha\neq 0$, it is a circle.
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Proof:
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Let $w=u+iv=\frac{1}{\zeta}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
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Let $w=u+iv=\frac{1}{z}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
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Then the original equation becomes:
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@@ -48,13 +48,13 @@ $$
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Which is in the form of circle equation.
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EOP
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QED
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## Chapter 4 Elementary functions
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> $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$
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So, following the definition of $e^\zeta$, we have:
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So, following the definition of $e^z$, we have:
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$$
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\begin{aligned}
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@@ -65,51 +65,51 @@ e^{x+iy}&=e^xe^{iy} \\
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\end{aligned}
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$$
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### $e^\zeta$
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### $e^z$
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The exponential of $e^\zeta=x+iy$ is defined as:
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The exponential of $e^z=x+iy$ is defined as:
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$$
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e^\zeta=exp(\zeta)=e^x(\cos y+i\sin y)
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e^z=exp(z)=e^x(\cos y+i\sin y)
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$$
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So,
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$$
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|e^\zeta|=|e^x||\cos y+i\sin y|=e^x
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|e^z|=|e^x||\cos y+i\sin y|=e^x
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$$
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#### Theorem 4.3 $e^\zeta$ is holomorphic
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#### Theorem 4.3 $e^z$ is holomorphic
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$e^\zeta$ is holomorphic on $\mathbb{C}$.
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$e^z$ is holomorphic on $\mathbb{C}$.
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Proof:
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$$
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\begin{aligned}
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\frac{\partial}{\partial\zeta}e^\zeta&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
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\frac{\partial}{\partial z}e^z&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
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&=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
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&=0
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\end{aligned}
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$$
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EOP
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QED
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#### Theorem 4.4 $e^\zeta$ is periodic
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#### Theorem 4.4 $e^z$ is periodic
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$e^\zeta$ is periodic with period $2\pi i$.
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$e^z$ is periodic with period $2\pi i$.
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Proof:
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$$
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e^{\zeta+2\pi i}=e^\zeta e^{2\pi i}=e^\zeta\cdot 1=e^\zeta
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e^{z+2\pi i}=e^z e^{2\pi i}=e^z\cdot 1=e^z
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$$
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EOP
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QED
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#### Theorem 4.5 $e^\zeta$ as a map
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#### Theorem 4.5 $e^z$ as a map
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$e^\zeta$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
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$e^z$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
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$$
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e^{\pi i}+1=0
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@@ -119,36 +119,36 @@ This is a map from cartesian coordinates to polar coordinates, where $e^x$ is th
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This map attains every value in $\mathbb{C}\setminus\{0\}$.
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#### Definition 4.6-8 $\cos\zeta$ and $\sin\zeta$
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#### Definition 4.6-8 $\cos z$ and $\sin z$
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$$
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\cos\zeta=\frac{1}{2}(e^{i\zeta}+e^{-i\zeta})
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\cos z=\frac{1}{2}(e^{iz}+e^{-iz})
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$$
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$$
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\sin\zeta=\frac{1}{2i}(e^{i\zeta}-e^{-i\zeta})
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\sin z=\frac{1}{2i}(e^{iz}-e^{-iz})
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$$
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$$
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\cosh\zeta=\frac{1}{2}(e^\zeta+e^{-\zeta})
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\cosh z=\frac{1}{2}(e^z+e^{-z})
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$$
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$$
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\sinh\zeta=\frac{1}{2}(e^\zeta-e^{-\zeta})
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\sinh z=\frac{1}{2}(e^z-e^{-z})
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$$
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From this definition, we can see that $\cos\zeta$ and $\sin\zeta$ are no longer bounded.
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From this definition, we can see that $\cos z$ and $\sin z$ are no longer bounded in the complex plane.
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And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $\zeta$ is real.
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And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $z$ is real.
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Moreover,
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$$
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\cosh(i\zeta)=\cos\zeta
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\cosh(iz)=\cos z
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$$
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$$
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\sinh(i\zeta)=i\sin\zeta
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\sinh(iz)=i\sin z
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$$
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### Logarithm
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@@ -175,37 +175,58 @@ If $y\in(-\pi,\pi]$, then $\log a=b$ means $e^b=a$ and $Im(b)\in(-\pi,\pi]$.
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If $a=re^{i\theta}$, then $\log a=\log r+i(\theta_0+2k\pi)$.
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#### Definition 4.10
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#### Definition 4.10 of Branch of $\arg z$ and $\log z$
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Let $G$ be an open connected subset of $\mathbb{C}\setminus\{0\}$.
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A branch of $\arg(\zeta)$ in $G$ is a continuous function $\alpha$, such that $\alpha(\zeta)$ is a value of $\arg(\zeta)$.
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A branch of $\arg(z)$ in $G$ is a continuous function $\alpha:G\to G$, such that $\alpha(z)$ is a value of $\arg(z)$.
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A branch of $\log(\zeta)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(\zeta)}=\zeta$.
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A branch of $\log(z)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(z)}=z$.
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Note: $G$ has a branch of $\arg(\zeta)$ if and only if it has a branch of $\log(\zeta)$.
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Note: $G$ has a branch of $\arg(z)$ if and only if it has a branch of $\log(z)$.
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If $G=\mathbb{C}\setminus\{0\}$, then not branch of $\arg(\zeta)$ exists.
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Proof:
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Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(\zeta)$ in $G$.
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Suppose there exists $\alpha(z)$ such that $\forall z\in G$, $\alpha(z)\in G$, then $l(z)=\ln|z|+i\alpha(z)$ is a branch of $\log(z)$.
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Suppose there exists $l(z)$ such that $\forall z\in G$, $l(z)\in G$, then $\alpha(z)=Im(z)$ is a branch of $\arg(z)$.
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QED
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If $G=\mathbb{C}\setminus\{0\}$, then not branch of $\arg(z)$ exists.
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#### Corollary of 4.10
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Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(z)$ in $G$.
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Then,
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$$
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\alpha_1(\zeta)-\alpha_2(\zeta)=2k\pi i
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\alpha_1(z)-\alpha_2(z)=2k\pi
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$$
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for some $k\in\mathbb{Z}$.
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Suppose $l_1$ and $l_2$ are two branches of $\log(z)$ in $G$.
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Then,
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$$
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l_1(z)-l_2(z)=2k\pi i
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$$
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for some $k\in\mathbb{Z}$.
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#### Theorem 4.11
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$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
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$\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
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Proof:
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Method 1: Use polar coordinates. (See in homework)
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Method 2: Use the fact that $\log(\zeta)$ is the inverse of $e^\zeta$.
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Method 2: Use the fact that $\log(z)$ is the inverse of $e^z$.
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Suppose $h=s+it$, $e^h=e^s(\cos t+i\sin t)$, $e^h-1=e^s(\cos t-1)+i\sin t$. So
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