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Variation of Argument

Let $[\arg f(z)]$ denote the change in argument of a function $f(z)$ around a closed loop $\gamma$. Also let $N$ denote the number of Roots of $f(z)$ in $\gamma$ and $P$ denote the number of Poles of $f(z)$ in $\gamma$. Then

\begin{displaymath}[\arg f(z)]= {1\over 2\pi}(N-P).
\end{displaymath} (1)

To find $[\arg f(z)]$ in a given region $R$, break $R$ into paths and find $[\arg f(z)]$ for each path. On a circular Arc
z = Re^{i\theta},
\end{displaymath} (2)

let $f(z)$ be a Polynomial $P(z)$ of degree $n$. Then
$\displaystyle {[}\arg P(z)]$ $\textstyle =$ $\displaystyle \left[{\arg\left({z^n {P(z)\over z^n}}\right)}\right]$  
  $\textstyle =$ $\displaystyle [\arg z^n] + \left[{\arg \left({P(z)\over z^n}\right)}\right].$ (3)

Plugging in $z = Re^{i\theta}$ gives
\begin{displaymath}[\arg P(z)]= [\arg Re^{i\theta n}] + \left[{\arg { P(Re^{i\theta})\over Re^{i\theta n}}}\right]
\end{displaymath} (4)

\lim_{R\to \infty} {P(Re^{i\theta})\over Re^{i\theta n}} = \hbox{[constant]},
\end{displaymath} (5)

\left[{ P(Re^{i\theta})\over Re^{i\theta n}}\right]= 0,
\end{displaymath} (6)

\begin{displaymath}[\arg P(z)]= [\arg e^{i\theta n}] = n(\theta_2-\theta_1).
\end{displaymath} (7)

For a Real segment $z = x$,
\begin{displaymath}[\arg f(x)]= \tan^{-1}\left[{0\over f(x)}\right]= 0.
\end{displaymath} (8)

For an Imaginary segment $z = iy$,
\begin{displaymath}[\arg f(iy)]= \left\{{\tan^{-1} {\Im[P(iy)]\over \Re[P(iy)]}}\right\}_{\theta_1}^{\theta_2}.
\end{displaymath} (9)

Note that the Argument must change continuously, so ``jumps'' occur across inverse tangent asymptotes.

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© 1996-9 Eric W. Weisstein