A result of {}
from Solve
indicates that Solve
thinks the solution set is empty. We can see that this is so, if we rationalize the left-hand side of the equation:
lhs = 8/(π u (-I a + γ) Sqrt[1 + (h^2 v^2 u^2)/(-I a + γ)^2]) +
(16 (-I a + γ) (1 - Sqrt[1 + (h^2 v^2 u^2)/(-I a + γ)^2]))/(h^2 π v^2 u^3);
conjugates = lhs /. {{e_ :> e}, {r : Sqrt[_] :> -r, r : 1/Sqrt[_] :> -r}};
Times @@ conjugates // Expand // Simplify
(* -(64/(π^2 u^2 (-a^2 + h^2 u^2 v^2 - 2 I a γ + γ^2))) *)
Since the numerator is a constant 64
, the equation lhs == 0
cannot be satisfied.
One might put the terms together and set the numerator equal to zero. But this yields a result that makes lhs
undefined:
Solve[Numerator@Together@eqn == 0, u]
lhs /. %
(*
{{u -> 0}}
{Indeterminate}
*)
Whether or not this can be considered a valid solution depends on whether clearing the denominator can be justified in the problem in which the equation arose.
Reduce
in place of solve and it resulted in "False." $\endgroup$ – Fred Kline Oct 24 '16 at 6:06x
(to then take the limitx->0
)solx = Reduce[expr == x, u];
returns a message...A likely reason for this is that the solution set depends on branch \ cuts of Wolfram Language functions
. However one can find a solutionu==0
either by taking the limit or by looking atSeries[expr, {u, 0, 2}]
. $\endgroup$ – b.gates.you.know.what Oct 24 '16 at 6:18