5.0 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 z=u+iv be the center of the circle, r be the radius of the circle.
circle=\{z\in\mathbb{C}:|z-c|=r\}
This is:
|z|^2-c\overline{z}-\overline{c}z+|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|z|^2+\beta Re(z)+\gamma Im(z)+\delta=0
when \alpha=0, it is a line.
when \alpha\neq 0, it is a circle.
Proof
Let w=u+iv=\frac{1}{z}, 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.
Chapter 4 Elementary functions
e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}
So, following the definition of e^z, 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^z
The exponential of e^z=x+iy is defined as:
e^z=exp(z)=e^x(\cos y+i\sin y)
So,
|e^z|=|e^x||\cos y+i\sin y|=e^x
Theorem 4.3 e^z is holomorphic
e^z is holomorphic on \mathbb{C}.
Proof
\begin{aligned}
\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) \\
&=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
&=0
\end{aligned}
Theorem 4.4 e^z is periodic
e^z is periodic with period 2\pi i.
Proof
e^{z+2\pi i}=e^z e^{2\pi i}=e^z\cdot 1=e^z
Theorem 4.5 e^z as a map
e^z 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 z and \sin z
\cos z=\frac{1}{2}(e^{iz}+e^{-iz})
\sin z=\frac{1}{2i}(e^{iz}-e^{-iz})
\cosh z=\frac{1}{2}(e^z+e^{-z})
\sinh z=\frac{1}{2}(e^z-e^{-z})
From this definition, we can see that \cos z and \sin z are no longer bounded in the complex plane.
And this definition is still compatible with the previous definition of \cos and \sin when z is real.
Moreover,
\cosh(iz)=\cos z
\sinh(iz)=i\sin z
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 of Branch of \arg z and \log z
Let G be an open connected subset of \mathbb{C}\setminus\{0\}.
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).
A branch of \log(z) in G is a continuous function \beta, such that e^{\beta(z)}=z.
Note: G has a branch of \arg(z) if and only if it has a branch of \log(z).
Proof
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).
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).
If G=\mathbb{C}\setminus\{0\}, then not branch of \arg(z) exists.
Corollary of 4.10
Suppose \alpha_1 and \alpha_2 are two branches of \arg(z) in G.
Then,
\alpha_1(z)-\alpha_2(z)=2k\pi
for some k\in\mathbb{Z}.
Suppose l_1 and l_2 are two branches of \log(z) in G.
Then,
l_1(z)-l_2(z)=2k\pi i
for some k\in\mathbb{Z}.
Theorem 4.11
\log(z) is holomorphic on \mathbb{C}\setminus\{0\}.
Proof (continue on next lecture)
Method 1: Use polar coordinates. (See in homework)
Method 2: Use the fact that \log(z) is the inverse of e^z.
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.