# Solve this equation explicitly for $\beta$

Solve this equation explicitly for $\beta$:
$$T_{12}=\frac{x_{11}+x_{22}}{1+|\beta|^2}\beta$$
$x_{11}$ and $x_{22}$ are positive real numbers where $x_{11}>x_{22}$ and $T_{12}$ and $\beta$ are complex numbers. and $|\beta|<1$

Up to know I've tried this:

sol1 = Solve[
Subscript[T, 12] == (Subscript[x, 11] + Subscript[x, 22])/(1 + Abs[β]^2) β &&
{Subscript[x, 11], Subscript[x, 22]} ∈ Reals &&
{Subscript[T, 12], β} ∈ Complexes,
β
]


But it gives a complicated solution whereas the answer should be
$$\beta=\frac{T_{12}}{|T_{12}|}\sqrt{\frac{x_{22}}{x_{11}}}$$ according to the scientific article that I'm studying.

• sol1 = Block[{\$Assumptions = Subscript[x, 11] > Subscript[x, 22] > 0 && {Subscript[x, 11], Subscript[x, 22]} \[Element] Reals && Subscript[T, 12] \[Element] Complexes}, Solve[Subscript[T, 12] == ((Subscript[x, 11] + Subscript[x, 22]) \[Beta])/(1 + Abs[\[Beta]]^2), \[Beta], Complexes]] seems to work, but adding Abs[\[Beta]] < 1 in Solve takes too long. Commented Jan 10, 2016 at 22:14
• your answer cant be correct. the x's are a sum in the original, how can they appear otherwise in the result? Commented Jan 11, 2016 at 1:28

Edited to provide greater detail

eq = Subscript[T, 12] == (Subscript[x, 11] + Subscript[x, 22])/(1 + Abs[β]^2) β


First determine Abs[β]

Simplify[ComplexExpand[Abs[#] & /@ eq, {Subscript[T, 12], β},
TargetFunctions -> Abs], Subscript[x, 11] + Subscript[x, 22] > 0]
(* (1 + Abs[β]^2) Abs[Subscript[T, 12]] == Abs[β] (Subscript[x, 11] + Subscript[x, 22]) *)

Simplify[Solve[%, Abs[β]]] // Flatten
(* {Abs[β] -> (Subscript[x, 11] + Subscript[x, 22] - Sqrt[-4 Abs[Subscript[T, 12]]^2 +
(Subscript[x, 11] + Subscript[x, 22])^2])/(2 Abs[Subscript[T, 12]]),
Abs[β] -> (Subscript[x, 11] + Subscript[x, 22] + Sqrt[-4 Abs[Subscript[T, 12]]^2 +
(Subscript[x, 11] + Subscript[x, 22])^2])/(2 Abs[Subscript[T, 12]])}


and then β itself

Simplify[Solve[eq /. #, β] & /@ %] // Flatten
(* {β -> (Subscript[T, 12] (Subscript[x, 11] + Subscript[x, 22] -
Sqrt[-4 Abs[Subscript[T, 12]]^2 + (Subscript[x, 11] + Subscript[x, 22])^2]))
/(2 Abs[Subscript[T, 12]]^2),
β -> (Subscript[T, 12] (Subscript[x, 11] + Subscript[x, 22] +
Sqrt[-4 Abs[Subscript[T, 12]]^2 + (Subscript[x, 11] + Subscript[x, 22])^2]))
/(2 Abs[Subscript[T, 12]]^2)} *)