# Friday February 21st

## Singularities

Recall that there are three types of singularities:

- Removable
- Poles
- Essential

Recall that a function $g$ is holomorphic at $z_0$ iff
$$
\lim_{z\to z_0}  (z-z_0) g(z) = 0
$$

Theorem (3.2)
: An isolated singularity $z_0$ of $f$ is a pole $\iff \lim_{z\to z_0} f(z) = \infty$.

Theorem (3.3, Casorati-Weierstrass)
: If $f$ is holomorphic in $D_r(z_0) \setminus \theset{z_0}$ and has an essential singularity $z_0$, then there exists a radius $r$ such that $f(D_r(\theset{z_0})\setminus \theset{z_0})$ is dense in $\CC$.

Proof
:   Proceed by contradiction.
    Suppose there exists a $w\in \CC$ and a $\delta > 0$ such that
    $$
    D_\delta(w) \intersect f(D_r(\theset{z_0})\setminus \theset{z_0}) = \emptyset
    .$$

    If $z \in D_r(w)\setminus{z_0}$, then $\abs{f(z) - w} > \delta$.
    Define $g(z) = \frac{1}{f(z) - w}$ on $D_r(z_0) \setminus\theset{z_0}$; then $\abs{g(z)} < \frac{1}{\delta}$.

    > Note that this implies that $g(z)$ is holomorphic on $D_r(z_0) \setminus \theset{z_0}$.
    > $g(z)$ being holomorphic here follows from $f$ being holomorphic here.

    Then $g(z)$ has a removable singularity at $z = z_0$ by theorem 3.1.

    If $g(z_0) \neq 0$, then $f(z) - w$ is holmorphic at $z_0$, contradicting the fact that $z_0$ is an essential singularity.

    If instead $g(z_0) = 0$, then $z_0$ is a pole, again a contradiction.

> Note: revisit why this is a contradiction.


## Singularities at Infinity

The point $z=\infty$ can be one of three types of singularities:

1. *Removable* $\iff$ $f(z) = \sum_{k=-1}^\infty c_k \frac{1}{z^k}$.

    - I.e. only one positive exponent.

2. *Pole* $\iff$ $f(z) = \sum_{k = -\infty}^n c_k z^k$

    - I.e. there are finitely many positive exponents.

3. *Essential* $\iff$ $f(z) = \sum_{k=-\infty}^\infty c_k z^k$

    - There are infinitely many positive exponents.

Definition (Meromorphic)
:   A function $f$ is **meromorphic** on $\Omega$ iff there exists a sequence $\theset{z_i} \subset \Omega$ with no limit point in $\Omega$ such that

    1. $f$ is holomorphic on $\Omega \setminus\theset{z_i}$, and
    2. $f$ has poles at each $z_i$.

Theorem (3.4, Meromorphic Functions are Rational)
: $f$ is meromorphic on $\CP^1$ iff $f$ is a rational function.

Proof
:   \hfill
    $\implies$:
    By part 1 of the definition above, the point $z=0$ is either a pole or a removable singularity of the function $F(z) = f\qty{\frac 1 z}$.
    By part 2, $F$ has finitely many poles $\theset{z_k}_{k=1}^N$.
    So for each $k$, write
    $$
    f(z) = f_k(z) + g_k(z)
    $$
    where $f_k$ is the principal part and $g_k$ is holomorphic in a neighborhood of $z_k$.
    Then $f_k(z)$ is a polynomial in $\qty{\frac 1 {z-z_k}}$, say of degree $m_k$.
    But then
    $$
    F(z) \definedas f\qty{\frac 1 z} = \tilde f_\infty(z) + \tilde g_\infty(z)
    $$
    where $\tilde f_\infty(z)$ is a polynomial in $z$, and $\tilde g_\infty(z)$ is holomorphic near zero.
    Thus $\tilde f_\infty\qty{\frac 1 z}$ is a polynomial in $\frac 1 z$.

    Define $f_\infty(z) = \tilde f_\infty \qty{\frac 1 z}$ and
    $$
    H(z) = f(z) - f_\infty (z) - \sum_k f_k(z)
    .$$

    Then $H$ is entire and bounded and thus constant, and since $\lim_{z\to \infty} H(z) = 0$, $H$ is identically zero.
    Thus
    $$
    f(z) = f_\infty(z) + \sum_k f_k(z)
    $$

    \

    $\impliedby$:
    To be continued, uses the argument principle, Rouche's theorem, and Jordan's lemma.