# Wednesday February 12th

## Singularities 

Let $f(z)$ be holomorphic on $\Omega$, then we have Cauchy's integral formula:

\begin{align*}
f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z} ~d\xi
.\end{align*}

*Example:*
Note that $f(z) = \frac 1 z$ is holomorphic on $\CC\setminus 0$.

Let $\Omega$ be an open set containing a disk $D$ and $\Omega\setminus p$ be a punctured domain.

Definition (Isolated Singularities)
:   We say $f$ has an *isolated singularity* at $p$ iff $f$ is defined and holomorphic on some deleted neighborhood of $p$.


    Classification of singularities:

    1. **Removable**: $\abs{f(z)}$ is bounded on some $D_r(p) \setminus p$.
        
        *Example*: $f(z) = \sin(z)/z$.


    2. **Poles**: $\lim_{z\to p} \abs{f(z)} = \infty$.
      
        *Example*: $f_n(z) = \frac{1}{z^n}$ at $p=0$


    3. **Essential**: neither 1 nor 2.

        *Example*: $f(z) = e^{\frac 1 z}$ at $z=0$.

Note that for singularities at $\infty$, we can just make the change of variables $z\mapsto \frac 1 z$.
Defining $F(z) = f(\frac 1 z)$, the singularities at 0 of $f$ correspond to singularities at infinity for $F$.

## Spherical Projection

We can solve for a spherical projection map $S^2 \to \CC$.
Let $(0,0,1)$ be the North pole of the sphere; then to map to $(x, y, 0)$ on the plane we can take the parameterization $\ell: (tx, ty, 1-t)$.
This yields

\begin{align*}
t \mapsto \qty{ \frac{2\Re(z)}{1 + \abs z^2} , \frac{2\Im(z)}{1 + \abs z^2}, 1 - \frac{2}{1 + \abs z^2}}
.\end{align*}

![Image](figures/2020-02-12-14:05.png)\

From this we can induce a spherical metric:

\begin{align*}
\phi(z_1, z_2) &= \frac{z \abs{z_1 - z_2}} { \sqrt{\abs z_1^2 + 1} \sqrt{\abs z_2^2 + 1}  }
.\end{align*}


Proposition (Continuous Extension Over Removable Singularities)
:   Let $p$ be a removable singularity of $f$.
    Then

    1. $\lim_{z\to p} f(z)$ exists.

    2. The function 
    \begin{align*}
    \tilde f(x) = \begin{cases}f(z) & z\neq p \\ \lim_{z\to p} f(z) & z=p  \end{cases}
    .\end{align*}
    is holomorphic on $D_r(p)$.

Example
:   Consider $$\frac{\sin(z)}{z} \converges{z\to 0}\to 1.$$

Proof (of Proposition)
:   Take $p=0$ and consider $g(z) = z^2 f(z)$.
    We can verify directly that $g$ satisfies the Cauchy-Riemann equations on $D_r(0)$.
    Then $g$ is holomorphic on $D_r(0)$ and vanishes to order 2 at $z=0$,
    and $$f(z) = \frac{g(z)}{z^2}$$ is holomorphic on $D_r(0)$.

    If $f(z)$ has a pole at $z_0$, then $\lim_{z\to z_0} \abs{f(z)} \to \infty$ by definition, iff $\lim_{z\to z_0} \frac{1}{\abs{f(z)}} = 0$ and thus the reciprocal has a zero at $z=z+0$.
    If $z_0$ is a zero of a nontrivial holomorphic function $f$, then $z_0$ is isolated, i.e. there exists a punctured disc $D_r(z_0)\setminus z_0$ on which $f$ is nonzero.

Theorem (???)
:   If $f$ is holomorphic in a connected domain $\Omega$ with a zero $z_0$, then there exists a non-vanishing holomorphic function $g(z)$ and some $n\in \NN$ such that $$f(z) = (z-z_0)^n g(z)$$.

Proof
:   Since $f$ is holomorphic, expand its power series $f(z) = \sum a_k (z-z_0)^k$.
    Since $f(z_0) = 0$, we have $a_0 = 0$.
    Choose the smallest $n$ such that $a_n \neq 0$, so 
    \begin{align*}
    f(z) 
    &= a_n(z-z_0)^n + a_{n+1}(z-z_0)^{n+1} + \cdots \\
    &= (z-z_0)^n \qty{a_n + \cdots} \\
    &\definedas (z-z_0)^n g(z)
    .\end{align*}

    Then $g(z_0) \neq 0$, so by continuity there exists an $r$ such that $\abs{g(z)} \geq \abs{a_n}/2$.

Definition (Pole)
:   A function $f$ defined on a deleted neighborhood of $z_0$ has a **pole** at $z_0$ if the function $F = \frac 1 f$ with $F(z_0) \definedas 0$ is holomorphic in a full neighborhood of $z_0$.