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\centerline{{\it Class 13 }{\bf (Wednesday February 11)}}
 
\begin{description}
\item[SUMMARY] Complex Logarithms and the debut of branch cuts
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\item[CURRENT READING] Brown \& Curchill, pages 75-81
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\item[NEXT READING] Brown \& Curchill pages 81-85
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{\bf The Complex Logarithm $\log z$}\\
Let us define $w = \log z$ as the inverse of $z = e^w$

But we know that $\exp [ \ln |z| + i(\theta + 2n \pi)] = z$, where $n \in Z$,
from our knowledge of the exponential function.

So we can define $$\log z = \ln |z| + i \arg z = \ln |z| + i \Arg z + 2 n \pi i = \ln r + i \theta$$
where $r = |z|$ as usual, and $\theta$ is the argument of $z$

If we only use the principal value of the argument, then we define the principal value of $\log z$ as $\Log z$, where
$$\Log z =  \ln |z| + i\Arg z = \Log |z| + i \Arg z$$

\underline{\bf Examples}\\
Compute $\Log (-2) and \Log(2), \log(-2)$, and $\log(2)$, $\Log(-4)$ and $\log(-4)$

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{\bf Logarithmic Identities}\\
$z=e^{\log z}$ but $\log e^z = z + 2k\pi i$ (Is this a surprise?)

$\log(z_1 z_2) = \log z_1 + \log z_2$

$\log {\displaystyle \left( \frac{z_1}{z_2} \right)} = \log z_1 - \log z_2$

However these do not neccessarily apply to the principal branch of the logarithm, written as $\Log z$.

\newpage
{\bf $\Log z$: the Principal Branch of $\log z$}\\
$\Log z$ is a single-valued function and is analytic in the domain $D^*$ consisting
of all points of the complex plane {\it except for those lying on the nonpositive real axis}, where
$$\frac{d}{d z} \Log z = \frac{1}{z}$$

Sketch the set $D^*$ and convince yourself that it is an open connected set.\\
(Ask yourself: Is every point in the set an interior point?)
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The set of points $\Re z \le 0 \cap \Im z = 0$ is a line of discontinuities known
as a {\bf branch cut}. 
By putting in a branch cut we say that we ``construct $\Log z$ from $\log z$.'' Why can we not evaluate $\log z$ along the entire positive $x$-axis?\\

{\bf Analyticity of $\Log z$}\\
We can use a version of the Cauchy-Riemann Equations in polar coordinates to help us investigate the analyticity of $\Log z$\\

Why don't we investigate the analyticity of $\log z$?\\

If $x = r \cos \theta$ and $y = r \sin \theta$ one can trewrite $f(z) = u(x,y) + i v(x,y)$ into $f = u(r,\theta) + i v(r,\theta)$ in that case, the CREs become:
$$u_r = \frac{1}{r}v_{\theta}, \qquad v_r = - \frac{1}{r}u_{\theta}$$
and the expression for the derivative $f'(z) = u_x + i v_x$ can be re-written $$f'(z) = e^{-i \theta}(u_r + i v_r)$$

\underline{\sc Groupwork}\\
Using this information, show that $\Log z$ is analytic and that $\frac{d}{d z} \Log z = \frac{1}{z}$.\\
(HINT: You will need to write down $u(r,\theta)$ and $v(r,\theta)$ for $\Log z$)

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\newpage
{\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 $f'(i)$?

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{\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
$$\log z = \ln |z| + i \theta, \qquad \mbox{ where } \alpha < \theta < \alpha + 2 \pi$$
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 $\log z$

\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 if 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
$$ z^c = \exp( c \log z)$$
 
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 Examples}\\
Compute the following:\\
(a) $(1+i)^{1-i}=$
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(b) $(0.5-i)^3=$
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(c) $(-1)^{2/3}=$
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\underline{\bf Differentiating}\\
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 the principal branch)
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If $g(z) = z^(1-i)$, Find $f'(1+i)$ (Use the principal branch)
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