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\centerline{{\it  Class 14} {\bf (Friday February 13)}}
 
\begin{description}
\item[SUMMARY] Complex Exponents $z^c$ and $c^z$ and More on Branch Cuts
\vspace{-12pt}
\item[CURRENT READING] Brown \& Curchill, pages 75-81
\vspace{-12pt}
\item[NEXT READING] Brown \& Curchill pages 81-85
\end{description}

{\bf Branching: Producing Single-valued functions from Multiple-valued ones}\\
A single-valued function $F(z)$ is said to be a {\it branch} of a multiple-valued
function $f(z)$ in a domain $D$ if $F(z)$ is single-valued and analytic in $D$ and has the property that for each $z \in D$, the value $F(z)$ is one of the values of $f(z)$\\
 
Branch cuts do not have to be along the $x$-axis, they can be any line which when removed from the domain of definition of the multi-valued function, produce a 
single valued function.\\
 
\underline{\bf Example}\\
Determine the domain of analyticity for the function $f(z) = \Log(3 z -i)$ and
compute $f'(z)$ What is $f(i)$? What about $f'(i)$?\\
 
(HINT: think of it as $\Log(w)$.  Where in the $w$-plane is $\Log(w)$ not analytic?)

\vfill
\vfill
 
{\bf Other branches of $\log z$}\\
One can define other analytic branches of $\log z$ by choosing different branch 
cuts.\\
The usual way to do this is to make the branch cut along $\theta = \alpha$ starting at the origin, so that
$$\log z = \ln |z| + i \theta, \qquad \mbox{ where } \alpha < \theta \le \alpha + 
2 \pi$$
These branches of $\log z$ can be denoted ${\cal L_{\alpha}}$, where $\theta = \alpha$ is the branch cut. \\
 
A {\bf branch point} of a function $f$ is a point which is common to all branch 
cuts of $f$. So, 0 is a branch point of $\log z$
 
\newpage
\underline{\sc Groupwork}\\
Now consider $\Log(z^2+1)$. This is a much more complicated function.\\
How does it differ from $\log(z^2+1)$ or ${\cal L}_{\alpha}(z^2+1)$?\\
What are the equations that its branch cut will satisfy? \\
Sketch the branch cut for $\Log(z^2+1)$ on the axes below.

\vfill
\underline{\bf Roots of Complex Numbers (Reprise)}\\
From previous identities about complex logarithms and the complex exponential, we can show (for $z \neq 0$) that
$$z^n =  \exp(n \log z), \qquad \qquad \mbox{ as long as } n \in Z$$
 
Similarly, $$z^{1/n} = \exp \left( \frac{1}{n} \log z \right)$$
 
But 
\begin{eqnarray*}
\exp \left( \frac{1}{n} \log z \right) &  = & \exp \left( \frac{1}{n} [ \ln |z|+ i ( \Arg z + 2 k \pi)] \right) \qquad (k \in Z) \nonumber \\
 & = & |z|^{1/n} \exp \left(i [\frac{\Arg z}{n} + \frac{2 k \pi}{n}] \right) \nonumber \\
 & = & |z|^{1/n} \exp {\left( \frac{i \theta + 2 k \pi i}{n} \right) }\qquad \qquad \mbox{where } \theta = \Arg z \nonumber
\end{eqnarray*}
But you should recognize the right hand side as the familiar formula for finding
 the root of a complex number, where $k$ is restricted to $0, 1, 2, \ldots, n-1$
. Why would we do that? [HINT: how many distinct values does $\exp(2 k \pi i/n)$
 have when $k$ can be any integer and $n$ is fixed?] What do you think happens i
f we try and raise a complex number to something besides an integer or rational 
number? How will we deal with {\bf complex exponents}?
 
\newpage
{\bf Complex Exponents}\\

If $z \neq 0$ and $c \in {\cal C}$, the function $z^c$ is defined as
$$ z^c = \exp( \log z^c) = \exp (c \log z) $$

Since $\log z$ is a multi-valued function, $z^c$ will have multiple values. 
How many values depends on the nature of $c$.
$$
z^c = \left\{
\begin{array}{cll}
z^{n/m}  & \mbox{if $c$ is rational, i.e. $n/m$} & \mbox{finite number of values (m)}\\
z^n & \mbox{if $c=n$, where $n$ is an integer} & \mbox{single value}\\
z^c & \mbox{all other complex numbers} & \mbox{infinite  number of values}\\
\end{array}
\right.
$$
\underline{\bf Example}\\
Show that $i^i$ is purely real.
\vfill
\underline{\sc Groupwork}\\
Compute the following:\\
(a) $(0.5-\frac{\sqrt{3}}{2}i)^3=$
\vfill
(b) $(-1)^{2/3}=$
\vfill
(c) $(1+i)^{1-i}=$
\vfill
\newpage
{\bf Derivatives of $z^c$ and $c^z$}\\
If you choose a branch of $z^c$ which is analytic on an open set, then
$$\frac{d}{d z}(z^c) = c z^{c-1}$$
where the branch of the log used in evaluating $z^c$ is the same branch used in evaluating
$z^{c-1}$

Similarly, we can define the {\it complex exponential function with base $c$}
$$c^z = \exp(z \log c)$$

If a single value of $c$ is chosen, then $c^z$ is an entire function such that
$$\frac{d}{d z}(c^z) = \frac{d}{d z} \exp( z \log c) = c^z \log c$$

\underline{\bf Examples}\\
If $f(z) = (1+i)^z$, Find $f'(1-i)$ (Use ${\cal L}_{0}$)
\vfill
If $g(z) = z^{(1-i)}$, Find $f'(1+i)$ (Use ${\cal L}_{0}$)
\vfill

\newpage
{\bf Branch cuts}\\
A {\bf branch cut} is a set of points on which we must select a particular value of a multi-valued function in order to construct a single-valued function. For example, the set of points $\Im z = 0 \cap \Re z \le -$ is the branch cut on which we select the argument function to have the singular value of $\pi$.\

Let us look at $\Log(3 z -i)$ again. First, let's consider $\Log(w)$. \\
Where in the $w$-plane is $\Log(w)$ not analytic?\\[24pt]

What are the set of points where Re$(w) < 0$ and Im $w = 0$ if $w = 3 z - i$?\\
Draw these points on the axes below.

\vspace{2.5in}


\newpage
{\bf Branch chasing}\\
Other functions have branch cuts, too, of course. We usually have to {\it choose} the branch cut so that our function in question is analytic at a required point.\\

\underline{\sc Groupwork}\\
1: Define a branch of $f(z) = (z^2-1)^{1/2}$ so that it is analytic on the exterior of the unit circle $|z| > 1$

(a) First write down the {\it principal branch} of $f(z)$. Where are its branch cuts? [What equations/inequalities do you have to solve?]
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(b) Sketch the branch cuts on the axes below.
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(c) What if we redefine $(z^2-1)^{1/2} = z(1-1/z^2)^{1/2}$? Where are the branch cuts of the principal branch of our new function? Sketch them below
\vspace{3in}

(d) Write down the function which solves our problem \#1
\vfill

The problem of ``branch chasing'' for complicated functions can be extremely
tedious but it is necessary if we want them to be analytic. Thankfully, it
is seldom necessary for our elementary applications.


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