3.2 Exponential, Log, Multivalued Functions

Exponential Function

The exponential function is the solution of the differential equation

f^{'} (z) = f (z)

with $f (0) = 1$

f (z) = a_{0} + a_{1} z + \dots + a_{n} z^{n} + \dots

and

f^{'} (z) = a_{1} + a_{1} z + \dots + n a_{n - 1} z^{n - 1} + \dots

so we must have $n a_{n} = a_{n - 1}$ and $a_{0} = 1$ which means $a_{n} = \frac{1}{n!}$

\frac{1}{R} = \underset{n \to \infty}{lim sup} \sqrt[n]{\frac{1}{n!}} = 0 so R = \infty .

Properties

$e^{a + b} = e^{a} e^{b}$ for $a, b \in C$

(e^{z} e^{c - z})^{'} = e^{z} e^{c - z} + (- 1) e^{c - z} e^{z} = 0

Therefore, $e^{z} e^{c - z}$ is a constant. Let $z = 0$ , so

e^{z} e^{c - z} = e^{0} e^{c} = e^{c} .

Let $z = a$ and $c = a + b$ , then

e^{a + b} = e^{a} e^{b} .

e^{z} = e^{x + i y} = e^{x} e^{i y}

Remark. If $z = x + 0 i$ , we recover the real exponential function.

If $x > 0$ , then $e^{x} > 1$ , and if $x < 0$ , then $0 < e^{x} < 1$ .

Remark. $e^{\bar{z}} = \overset{―}{e^{z}}$ (use the series expansion of $e^{z}$ ).

Remark. If $z = 0 + i y$ , then $| e^{i y} |^{2} = e^{i y} e^{- i y} = 1$ .

In general $| e^{z} | = e^{z} \overset{―}{e^{z}} = e^{z} e^{\bar{z}}, e^{x} = | e^{z} |$

Therefore,

\begin{aligned} | e^{x + i y} |^{2} & = e^{x + i y} \cdot \overset{―}{e^{x + i y}} \\ = e^{x + i y} e^{x - i y} = e^{2 x} = (e^{x})^{2} \\ \Rightarrow | e^{x + i y} | = e^{x} . \end{aligned}

Periodicity

Definition: $f (z)$ has a period of $c$ if $f (z + c) = f (z)$ for all $z$ .

e^{i y} = \sum_{n = 0}^{\infty} \frac{(i y)^{n}}{n!} = 1 + \frac{i y}{1!} - \frac{y^{2}}{2!} - \frac{i y^{3}}{3!} + \frac{y^{4}}{4!} - \dots

i^{0} = 1, i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1, \dots

For even powers:

\sum_{k = 0}^{\infty} \frac{(i y)^{2 k}}{(2 k)!} = \sum_{k = 0}^{\infty} \frac{(- 1)^{k} y^{2 k}}{(2 k)!} = \cos y .

For odd powers:

\sum_{k = 0}^{\infty} \frac{(i y)^{2 k + 1}}{(2 k + 1)!} = i \sum_{k = 0}^{\infty} \frac{(- 1)^{k} y^{2 k + 1}}{(2 k + 1)!} = i \sin y .

Thus, $e^{i y} = \cos (y) + i \sin (y)$ and $y = \arg (e^{z})$

Note then that $e^{π i} = \cos (π) + i \sin (π) = - 1$

\begin{aligned} ⟹ e^{π i} + 1 = 0 \\ ⟹ e^{2 π i} = 1 \\ ⟹ e^{z + 2 π i} = e^{z} e^{2 π i} = e^{z} \end{aligned}

So the complex exponential function has a period of $2 π$

Complex Logarithm

Solving for the inverse of:

z = e^{w}

\begin{array}{r} | e^{w} | = e^{Re (w)} \end{array}

If $z \neq 0$ then $Re (w) = \log | z |$

\log z = \log (r e^{i θ}) = \log r + \log (e^{i θ}) = \log r + i θ

⟹ \log z = \log | z | + i \arg z

\begin{aligned} e^{\log z} & = e^{\log | z | + \arg z} \\ = | z | + e^{\arg z} \\ = | z | (\cos (θ) + i \sin (θ)) = z \end{aligned}

but,

\log e^{z} = \log e^{z + 2 π i} = z + 2 π i n

\log e^{z} = {z + 2 π n i, n \in Z}

The principal branch is:

- π \leq \arg (z) \leq π

Properties

\begin{aligned} \log (z_{1} z_{2}) & = \log (z_{1}) + \log (z_{2}) \end{aligned}

\begin{aligned} \log (r_{1} e^{i (θ_{1})} r_{2} e^{i (θ_{2})}) & = \log (r_{1} r_{2} e^{i (θ_{1} + θ_{2})}) \\ = \log | r_{1} r_{2} | + i (θ_{1} + θ_{2} + 2 π n_{1} + 2 π n_{2}) \\ = \log | r_{1} | + i (θ_{1} + 2 π n_{1}) + \log | r_{2} | + i (θ_{2} + 2 π n_{2}) \\ = \log (z_{1}) + \log (z_{2}) \end{aligned}

For $a^{b}$ where $a, b \in C$
if $b \in Z$ :

\begin{aligned} a^{b} & = e^{b (\ln | a | + i (\arg (a) + 2 π k)} \\ = e^{b (\ln | a | + i \arg (a))} e^{i 2 π k b} \\ = e^{b (\ln | a | + i \arg (a))} \cdot 1 \\ = e^{(\ln | a |^{b} + i \arg (a))} \end{aligned}

if $b \in Q$ : $b = \frac{p}{q}$ in lowest terms:

\begin{aligned} a^{p / q} & = e^{\frac{p}{q} (\ln | a | + i (\arg a + 2 π k))} \\ = e^{\frac{p}{q} (\ln | a | + i \arg a)} e^{i 2 π \frac{p}{q} k} \end{aligned}

which has $q$ distinct branches

\begin{aligned} a^{b} & = e^{b (\ln | a | + i (\arg a + 2 π k))} \\ = e^{b (\ln | a | + i \arg a)} e^{i 2 π b k} \end{aligned}

Which has infinite solutions.

Multi-valued Functions

$f (z)$ is analytic in an arbitrary set of points $A$ if it is the restriction of a function which is analytic in some open neighbourhood containing $A$ .

Let $f (z) = \sqrt{z}$
Recall that if $z = r (\cos θ + i \sin θ)$ then there are two answers:

{\begin{cases} \sqrt{r} (\cos \frac{θ}{2} + i \sin \frac{θ}{2}) \\ - \sqrt{r} (\cos \frac{θ}{2} + i \sin \frac{θ}{2}) \end{cases}

z = r e^{i θ}, \sqrt{z} = \pm \sqrt{r} e^{i θ / 2}

Principle Branch

To make $\sqrt{z}$ single values we restrict $- π < θ < π$
This means the image has no negative real component.

Set the principal argument $\arg \in (- π, π]$ .

Define $\sqrt{z} = \sqrt{| z |} e^{i \arg (z) / 2}$ .

Above the cut.
$z_{ε}^{+} = e^{i (π - ε)}, ε ↓ 0$

| z_{ε}^{+} | = 1, \arg (z_{ε}^{+}) = π - ε

$\sqrt{z_{ε}^{+}} = 1 \cdot e^{i (π - ε) / 2} = e^{i (π / 2 - ε / 2)} ⟶ e^{i π / 2} = i$
Below the cut.
$z_{ε}^{-} = e^{i (- π + ε)}, ε ↓ 0$
$| z_{ε}^{-} | = 1, \arg (z_{ε}^{-}) = - π + ε$
$\sqrt{z_{ε}^{-}} = 1 \cdot e^{i (- π + ε) / 2} = e^{- i (π / 2 - ε / 2)} ⟶ e^{- i π / 2} = - i$

Thus $lim_{\begin{matrix} z \to - 1 \\ above \end{matrix}} \sqrt{z} = i, lim_{\begin{matrix} z \to - 1 \\ below \end{matrix}} \sqrt{z} = - i$
so $\sqrt{z}$ has a jump on $(- \infty, 0]$

Branch

A branch of $f$ is a single valued function $F : U \to C$ so that $U$ is open and connected, $F (z)$ has only a single value and $F (z)$ is analytic on $U$