4.5 KiB
Math416 Lecture 6
Review
Linear Fractional Transformations
Transformations of the form f(z)=\frac{az+b}{cz+d},$a,b,c,d\in\mathbb{C}$ and ad-bc\neq 0 are called linear fractional transformations.
Theorem 3.8 Preservation of clircles
We defined clircle to be a circle or a line.
The circle equation is:
Let \zeta=u+iv be the center of the circle, r be the radius of the circle.
circle=\{z\in\mathbb{C}:|\zeta-c|=r\}
This is:
|\zeta|^2-c\overline{\zeta}-\overline{c}\zeta+|c|^2-r^2=0
If \phi is a non-constant linear fractional transformation, then \phi maps clircles to clircles.
We claim that a map is circle preserving if and only if for some \alpha,\beta,\gamma,\delta\in\mathbb{R}.
\alpha|\zeta|^2+\beta Re(\zeta)+\gamma Im(\zeta)+\delta=0
when \alpha=0, it is a line.
when \alpha\neq 0, it is a circle.
Proof:
Let w=u+iv=\frac{1}{\zeta}, so \frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}.
Then the original equation becomes:
\alpha\left(\frac{u}{u^2+v^2}\right)^2+\beta\left(\frac{u}{u^2+v^2}\right)+\gamma\left(-\frac{v}{u^2+v^2}\right)+\delta=0
Which is in the form of circle equation.
EOP
Chapter 4 Elementary functions
e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}
So, following the definition of e^\zeta, we have:
\begin{aligned}
e^{x+iy}&=e^xe^{iy} \\
&=e^x\left(\sum_{n=0}^{\infty}\frac{(iy)^n}{n!}\right) \\
&=e^x\left(\sum_{n=0}^{\infty}\frac{(-1)^ny^n}{n!}\right) \\
&=e^x(\cos y+i\sin y)
\end{aligned}
e^\zeta
The exponential of e^\zeta=x+iy is defined as:
e^\zeta=exp(\zeta)=e^x(\cos y+i\sin y)
So,
|e^\zeta|=|e^x||\cos y+i\sin y|=e^x
Theorem 4.3 e^\zeta is holomorphic
e^\zeta is holomorphic on \mathbb{C}.
Proof:
\begin{aligned}
\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) \\
&=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
&=0
\end{aligned}
EOP
Theorem 4.4 e^\zeta is periodic
e^\zeta is periodic with period 2\pi i.
Proof:
e^{\zeta+2\pi i}=e^\zeta e^{2\pi i}=e^\zeta\cdot 1=e^\zeta
EOP
Theorem 4.5 e^\zeta as a map
e^\zeta is a map from \mathbb{C} to \mathbb{C} with period 2\pi i.
e^{\pi i}+1=0
This is a map from cartesian coordinates to polar coordinates, where e^x is the radius and y is the angle.
This map attains every value in \mathbb{C}\setminus\{0\}.
Definition 4.6-8 \cos\zeta and \sin\zeta
\cos\zeta=\frac{1}{2}(e^{i\zeta}+e^{-i\zeta})
\sin\zeta=\frac{1}{2i}(e^{i\zeta}-e^{-i\zeta})
\cosh\zeta=\frac{1}{2}(e^\zeta+e^{-\zeta})
\sinh\zeta=\frac{1}{2}(e^\zeta-e^{-\zeta})
From this definition, we can see that \cos\zeta and \sin\zeta are no longer bounded.
And this definition is still compatible with the previous definition of \cos and \sin when \zeta is real.
Moreover,
\cosh(i\zeta)=\cos\zeta
\sinh(i\zeta)=i\sin\zeta
Logarithm
Definition 4.9 Logarithm
A logarithm of a is any b such that e^b=a.
If a=0, then no logarithm exists.
If a\neq 0, then there exists infinitely many logarithms of a.
Let a=re^{i\theta}, b=x+iy be a logarithm of a.
Then,
e^{x+iy}=re^{i\theta}
Since logarithm is not unique, we can always add 2k\pi i to the angle.
If y\in(-\pi,\pi], then \log a=b means e^b=a and Im(b)\in(-\pi,\pi].
If a=re^{i\theta}, then \log a=\log r+i(\theta_0+2k\pi).
Definition 4.10
Let G be an open connected subset of \mathbb{C}\setminus\{0\}.
A branch of \arg(\zeta) in G is a continuous function \alpha, such that \alpha(\zeta) is a value of \arg(\zeta).
A branch of \log(\zeta) in G is a continuous function \beta, such that e^{\beta(\zeta)}=\zeta.
Note: G has a branch of \arg(\zeta) if and only if it has a branch of \log(\zeta).
If G=\mathbb{C}\setminus\{0\}, then not branch of \arg(\zeta) exists.
Suppose \alpha_1 and \alpha_2 are two branches of \arg(\zeta) in G.
Then,
\alpha_1(\zeta)-\alpha_2(\zeta)=2k\pi i
for some k\in\mathbb{Z}.
Theorem 4.11
\log(\zeta) is holomorphic on \mathbb{C}\setminus\{0\}.
Proof:
Method 1: Use polar coordinates. (See in homework)
Method 2: Use the fact that \log(\zeta) is the inverse of e^\zeta.
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
\begin{aligned}
\frac{e^h-1}{h}&=\frac{(s+it)e^s(\cos t-1)+i\sin t}{s^2+t^2} \\
&=\frac{e^s(\cos t-1)}{s^2+t^2}+i\frac{\sin t}{s^2+t^2}
\end{aligned}
Continue next time.