The hardest and/or most important part of answering a question is making sure the real question is understood. The July 1, 2010 question Using fsolve with a dispersion relation posted to MaplePrimes seemed to be about obtaining a numeric solution of an equation. Turns out it was more a question about the behavior of an implicit function.
The equation
was given, along with the additional information that , and . Essentially, a solution for is sought in the particular case . The challenge was in the comments "On a sketch of vs , apparently on the lower branch, but there should be some on the upper branch" and "For this particular problem, I have obtained a negative . Am I attempting it the right way? Is there another way of attempting it?"
Making the substitutions for , and , the equations the user sent to fsolve were then and , where
When I tried the calculation, I got
In light of the comments about , we have to consider the possibility that there are multiple solutions, both real and complex. The real case is dispatched with the graph in Figure 1, a plot of against .

Figure 1 Graph of defined implicitly by

All real points satisfying lie in the first quadrant. There are no real values of that are negative. The little hitch near is not an aborted asymptote because = implies at this endpoint. Figure 2 zooms in on this end of the graph and poses two more questions.

Figure 2 A small portion of the graph of

What are the coordinates of the maximum shown in the graph, and between what two values of is not a function? The second question is equivalent to asking for the rightmost solution of . The solution is given by
The critical point that answers the first question is found by simultaneously solving and (computed implicitly) for . The result is
That there is only one real solution for when is clear from Figure 1 and the answers to these two questions. Just to be sure, a graph of is given in Figure 3, a figure that also shows is the only real solution of =0.

Figure 3

All that remains is to show there are no complex solutions of . We do this with Figure 4, a plot of with treated as a complex variable.

Figure 4 Graph of with treated as a complex variable

The magnitude of is zero at only one point, so that must be the real point .
Figure 2 and the concomitant calculations show that for , there are two values of that satisfy . For , there is only one solution of , and it is positive.
Note: The tinted cell and the four figures are tables containing hidden input. To see the input, use the Table Properties dialog to select "Show input."
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