Let 
 denote the change in argument of a function 
 around a closed loop 
.  Also let 
 denote
the number of Roots of 
 in 
 and 
 denote the number of Poles of 
 in
.  Then
![\begin{displaymath}[\arg f(z)]= {1\over 2\pi}(N-P).
\end{displaymath}](v_300.gif)  | 
(1) | 
 
To find 
 in a given region 
, break 
 into paths and find 
 for each path.  On a circular
Arc 
  | 
(2) | 
 
let 
 be a Polynomial 
 of degree 
.  Then
Plugging in 
 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}](v_307.gif)  | 
(4) | 
 
![\begin{displaymath}
\lim_{R\to \infty} {P(Re^{i\theta})\over Re^{i\theta n}} = \hbox{[constant]},
\end{displaymath}](v_308.gif)  | 
(5) | 
 
so
![\begin{displaymath}
\left[{ P(Re^{i\theta})\over Re^{i\theta n}}\right]= 0,
\end{displaymath}](v_309.gif)  | 
(6) | 
 
and
![\begin{displaymath}[\arg P(z)]= [\arg e^{i\theta n}] = n(\theta_2-\theta_1).
\end{displaymath}](v_310.gif)  | 
(7) | 
 
For a Real segment 
,
![\begin{displaymath}[\arg f(x)]= \tan^{-1}\left[{0\over f(x)}\right]= 0.
\end{displaymath}](v_312.gif)  | 
(8) | 
 
For an Imaginary segment 
,
![\begin{displaymath}[\arg f(iy)]= \left\{{\tan^{-1} {\Im[P(iy)]\over \Re[P(iy)]}}\right\}_{\theta_1}^{\theta_2}.
\end{displaymath}](v_314.gif)  | 
(9) | 
 
Note that the Argument must change continuously, so ``jumps'' occur across inverse tangent asymptotes.
© 1996-9 Eric W. Weisstein 
1999-05-26