# 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.