# Wednesday January 15th

> See references:
> Taylor's Complex Analysis, Stein, Barry Simon (5 volume set), Hormander (technically a PDEs book, but mostly analysis)

> Good Paper: [Hormander 1955](https://projecteuclid.org/download/pdf_1/euclid.acta/1485892151)

We'll mostly be working from Simon Vol. 2A, most problems from from Stein's Complex.

## Topology and Algebra of $\CC$

To do analysis, we'll need the following notions:

1. Continuity of a complex-valued function $f: \Omega \to \Omega$

3. Complex-differentiability: For $\Omega \subset \CC$ open and $z_0 \in \Omega$, there exists $\varepsilon > 0$ such that $D_\varepsilon = \theset{z \suchthat \abs{z - z_0} < \varepsilon} \subset \Omega$, and $f$ is **holomorphic** (complex-differentiable) at $z_0$ iff $$\lim_{h\to 0} \frac 1 h (f(z_0 + h) - f(z_0))$$ exists; if so we denote it by $f'(z_0)$.

Example
: $f(z) = z$ is holomorphic, since $f(z+ h) - f(z) = z+h-z = h$, so $f'(z_0) = \frac h h = 1$ for all $z_0$.

Example
:   Given $f(z) = \bar z$, we have $f(z+h)-f(z) = \bar h$, so the ratio is $\frac{\bar h}{h}$ and the limit doesn't exist.

    Note that if $h\in \RR$, then $\bar h = h$ and the ratio is identically 1, while if $h$ is purely imaginary, then $\bar h = -h$ and the limit is identically $-1$.

We say $f$ is *holomorphic on an open set* $\Omega$ iff it is holomorphic at every point, and is holomorphic on a closed set $C$ iff there exists an open $\Omega \supset C$ such that $f$ is holomorphic on $\Omega$.

Fact
: If $f$ is holomorphic, writing $h = h_1 + ih_2$, then the following two limits exist and are equal:

  \begin{align*}
  \lim_{h_1 \to 0} \frac{f(x_0 + iy_0 + h_1) - f(x_0 + iy_0)}{h_1} = \dd{f}{x}(x_0, y_0) \\
  \lim_{h_2 \to 0} \frac{f(x_0 + iy_0 + ih_2) - f(x_0 + iy_0)}{ih_2} = \frac 1 i \dd{f}{y}(x_0, y_0) \\
  \implies \dd{f}{x} = \frac 1  i \dd{f}{y}
  .\end{align*}

  So if we write $f(z) = u(x, y) + i v(x, y)$, we have

  \begin{align*}
  \dd{u}{x} + i \dd{v}{x} \mid_{(x_0, y_0)} = \frac 1 i \qty{
  \dd{u}{y} + i \dd{v}{y}
  } \mid_{(x_0, y_0)}
  ,\end{align*}

  and equating real and imaginary parts yields **the Cauchy-Riemann equations**:

  \begin{align*}
  \dd{u}{x} + i \dd{v}{x} = -i \dd{u}{y} + \dd{v}{y} \\
  \iff \dd{u}{x} = \dd{v}{y} \quad\text{and}\quad \dd{u}{y} = - \dd{v}{x}
  .\end{align*}

The usual rules of derivatives apply:

1. $(\sum f)' = \sum f'$

Proof
: Direct.

2. $(\prod f)' =$ product rule
  
Proof
: Consider $(f(z+h)g(z+h) - f(z)g(z))/h$ and use continuity of $g$ at $z$.

3. Quotient rule
  
Proof
: Nice trick, write $$q = \frac f g$$ so $qg = f$, then $f' = q'g + qg'$ and $q' = \frac {f'} g - \frac{fg'}{g^2}$.

4. Chain rule

Proof
:   Use the fact that if $f'(g(z)) = a$, then $$f(z+h) - f(z) = a h + r(z, h),\quad \abs{r(z, h)} = o(\abs h) \to 0.$$

    Write $b = g'(z)$, then $$f(g(z + h)) = f(g(z) + b h + r_1 ) = f(g(z)) + f'(g(z))bh + r_2$$ by considering error terms, and so $$\frac 1 h (f(g(z+h)) - f(g(z))) \to f'(g(z)) g'(z)$$.