Larry Riddle, Agnes Scott College

Let \(\alpha = e^{\theta i}\). The branch tip for R^{k}(LR)^{∞} occurs at the point corresponding to the complex number
\[
z_k = \left(i+ir\alpha^{-1} + ir^2\alpha^{-2} + ir^3\alpha^{-3} \ldots + ir^k\alpha^{-k} \right)+ \left(ir^{k+1}\alpha^{-k+1} + ir^{k+2}\alpha^{-k} + \ldots\right).
\]
Then
\[
\begin{align}
x_k = -\text{Im}(z_k) &= 1 + r\sin(\theta) + r^2\sin(2\theta) + r^3\sin(3\theta) + \ldots + r^k\sin(k\theta) \\ \\
&\quad\quad+ r^{k+1}\sin((k-1)\theta) + r^{k+2}\sin(k\theta) + \ldots.
\end{align}
\]
The value of x_{m} is obtained in the same way to get
\[
\begin{align}
x_m = \text{Im}(z_k) &= 1 + r\sin(\theta) + r^2\sin(2\theta) + r^3\sin(3\theta) + \ldots + r^k\sin(k\theta) \\ \\
&\quad\quad+ r^{k+1}\sin((k+1)\theta) + r^{k+2}\sin((k+2)\theta)\\ \\
&\quad\quad+ \dots + r^m\sin(m\theta) + r^{m+1}\sin((m-1)\theta) + r^{m+2}\sin(m\theta) + \ldots.
\end{align}
\]
Subtracting the two expressions for the x-coordinates shows that
\[
\begin{align}
f_\theta (r) &= r^{k+1}\left(\sin((k+1)\theta) - \sin((k-1)\theta)\right) + r^{k+2}\left(\sin((k+2)\theta) - \sin(k\theta)\right) + \ldots \\ \\
&= 2\sin(\theta)\cos(k\theta)r^{k+1} + \left(\sin((k+2)\theta) - \sin(k\theta)\right)r^{k+2} + \ldots.
\end{align}
\]

Suppose kθ > 90°. Since 90° < kθ < 270° we have \(\cos(k\theta) < 0\) and \(\sin(\theta) > 0\), and so the coefficient of r^{k+1} will be negative, and in particular, non-zero. Therefore \(f_\theta (r)\) will behave like the function \({2\sin(\theta)\cos(k\theta)}r^{k+1}\) when r is close to 0.

If kθ = 90°, then the r^{k+1} term in the formula for \(f_\theta (r)\) will cancel.
Therefore the behavior of \(f_\theta (r)\) will be determined by the r^{k+2} term when r is small. Hence \(f_\theta (r)\) will behave like the function
\[
\left(\sin((k+2)\theta) - \sin(k\theta)\right)r^{k+2} = \left(\cos(2\theta) - 1\right) r^{k+2} .
\]
Again the coefficient will be negative. It is important to note that because m ≥ k+2, no term involving a power of r with exponent greater than or equal to m+1 will affect the r^{k+1} or r^{k+2} terms.

So in either case, when r is close to 0, the function \(f_\theta (r)\) will behave like Ar^{n} with A < 0. Thus the graph of \(f_\theta (r)\) will be decreasing and concave down for small values of r.

Example 1: θ = 80° (so k = 2 and m = 6, with kθ > 90°)

\[ f_\theta (r) = -1.85083r^3 - 0.98481r^4 - 0.34202r^5 + \ldots \]

Example 2: θ = 45° (so k = 2 and m = 10, with kθ = 90°)

\[ f_\theta (r) = -r^4 - \sqrt{2}r^5 - 2r^6 - \ldots \]