# Wednesday February 26th

## Argument Principle and Application

Let $f$ be holomorphic in $\Omega$ which is open, simple, and connected.
Then $f(z_0) = 0$ implies there exists an integer $m$ such that $f(z) = (z-z_0)^m g(z)$ where $g(z_0) \neq 0$.

Let $N_\Omega(f)$ be the number of zeros of $f$ inside $\Omega$, and $N_\Omega(f, a)$ be the number of zeros of $f-a$ in $\Omega$.
Writing $f = f_1 f_2$ where $f_1 = (z-z_0)^m$ and $f_2 = g(z)$, we have

\begin{align*}
\frac {f'} f
&= \frac {f_1'} {f_1} + \frac {f_2'} {f_2} \\
&= \frac m {z-z_0} + \frac {g'} g
.\end{align*}

Now integrating both sides yields

\begin{align*}
\frac 1 {2\pi i } \int_{D_r(z_0)} \frac{f'(z)}{f(z)} ~dz = m
,\end{align*}

so the integral of this function counts the number of zeros of $f$ in $D_r(z_0)$.

Proposition (Argument Principle)
: Let $f$ be holomorphic in a neighborhood of $\bar{D_r(z_0)}$ and suppose that $f$ is non-vanishing on all of $\bd D_r(z_0)$.
  Then

  \begin{align*}
  \frac 1 {2\pi i } \int_{D_r(z_0)} \frac{f'(z)}{f(z)} ~dz = N_{D_r(z_0)}(f)
  .\end{align*}

  More generally, if $q(z)$ is another holomorphic function in a neighborhood of $\bar{D_r(z_0)}$ and $z_1, \cdots, z_k$ are the distinct zeros of $f$ in $D_r(z_0)$ with orders $m_1, \cdots, m_k$, then

  \begin{align*}
  \frac 1 {2\pi i } \int_{D_r(z_0)} q(z) \frac{f'(z)}{f(z)} ~dz
  = \sum_{j=1}^k q(z_k) m_k
  .\end{align*}

Proof
: Write $f(z) = \prod_{j=1}^k (z-z_j)^{m_j} g(z)$.
  By Leibniz's rule, if $h = f_1 \cdots f_\ell$, then

  \begin{align*}
  \frac {h'} h
  &= \sum_{j=1}^\ell \frac {f_j'} {f_j} \\
  \implies q\frac {f'} f &= 1\frac {g'} g + \sum_{j=1}^k \frac{m_j q}{z-z_j}
  .\end{align*}

  Since $\frac {g'} g$ is holomorphic in the closed disc, integrating both sides yields the desired formula.

Note that if we replace $f$ by a family $f_t$ of continuous functions, an integer-valued continuous function must be constant.

Corollary
: Let $f_t(z)$ for $0\leq t \leq 1$ be a family of holomorphic functions on $D_{r+\eps}(z_0)$ for some $\eps > 0$.)
  Suppose $f_t(z)$ is continuous for all $z$ in this disc, uniformly in $z$, and for all $t$, $f_t(z)$ is nonvanishing on the boundary.

  Then the following integral is independent of $t$:

  \begin{align*}
    \frac 1 {2\pi i } \int_{D_r(z_0)} \frac{f_t'(z)}{f_t(z)} ~dz
  .\end{align*}

Theorem (Rouché's Theorem)
:   Let $f, g$ be holomorphic in a neighborhood of $\bar{D_r(z_0)}$ and suppose that on $\bd D_r(z_0)$ we have

    \begin{align*}\label{star}
    \abs{f(z) - g(z)} < \abs{f(z)} + \abs{g(z)}
    .\end{align*}

    Then $f$ and $g$ are nonvanishing on $\bd D_r(z_0)$ and
    \begin{equation}
    N_{D_r(z_0)} (f) =
    N_{D_r(z_0)} (g)
    .\end{equation}

Proof
:  If $f(w) = 0$ for some $w\in \bd D_r(z_0)$, then $\abs{-g(w)} = \abs{g(w)}$, but this contradicts condition \ref{star}.

  So let $t\in [0, 1]$ with $f_t(z) = (1-t)f(z) + t g(z)$.
  Then (claim) $f_t$ is nonvanishing on the boundary, so we can apply the previous corollary.

  Suppose otherwise that there exists $w$ on the boundary such that $f_t(w) = 0$ for some $t$, so $(1-t)f(w) + tg(w) = 0$.
  Then rearranging terms yields

  \begin{align*}
  f(w) &= t(g(w) - f(w)) \\
  g(w) &= (1-t)(g(w) - f(w))
  .\end{align*}

  But then

  \begin{align*}
  \abs{f(w) + g(w)}
  &= t\abs{g(w) - f(w)} + (1-t) \abs{g(w) - f(w)} \\
  &= \abs{g(w) - f(w)}
  ,\end{align*}
  which contradicts condition \ref{star}

  By the corollary, the integral is continuous in $t$ and integer-valued, and thus constant.

Corollary (Fundamental Theorem of Algebra)
: Let $p(z) = \sum_{j=1}^n a_j z^j$ be a polynomial of degree $n$, so $a_n \neq 0$.
  Let $f(z) = a_n z^n$ and $g(z) = p(z)$.
  If
  $$
  \abs{z} > \frac{\abs a_0 + \cdots + \abs a_m}{\abs a_n} > 1
  $$
  then
  \begin{align*}
  \abs{f(z) - g(z)}
  &= \abs{ a_0 + \cdots + a_{n-1}z^{n-1} }\\
  &\leq \abs{z}^{n-1} \qty{ \abs a_0 + \cdots + \abs a_{n-1} } \\
  &< \abs{a_n} \abs{z}^n \\
  &= \abs{f} \\
  &\leq \abs{f} + \abs{g}
  .\end{align*}


Note that this is useful because it tells you where the zeros are, namely in the disc $\abs{z} < \frac{\sum \abs a_i}{\abs a_n}$.

Example
: Let $p(z) = 9 - 8 z + 20z^2$, then all of the zeros are in a disc of radius $r = \frac 7 4$.

> Qual alert: problems about power series, Rouché's, linear mapping, integration.

Example
: Let $f(z) = z^9 - 2z^6 + z^2 -8z - 2$.

  How many zeros are in the unit disc?
  Take $g(z) = -8z$, the largest term.
  Then $\abs{f(z) - g(z)} \leq 1+2+1+2 = 6 < \abs{f} + \abs{g} = 8$,
  so condition \ref{star} is satisfied.
  Thus they both have the same number of zeros, but $g$ has exactly one zero.

  What about $\abs{z} = 2$?
  Then set $g(z) = z^9$, then check $\abs{f(z) - z^9} \leq 150 < 152$, so all 9 zeros lie in this disc.

Exercise
: Let $g(z) = z^4 - 4z - 5$, how many zeros are in $\abs{z} \leq 1$?
  Note the root on the boundary.