How can I find some values for which an expression is real?

Question

I've a complicated expression which depends on several variables. I want to find some values for those variables such that the complicated expression is real. How can I do that?

The obvious command to use is FindInstance but for some reason this command (used with a simplified but still illustrative function below) does not work:

FindInstance[x + y - Log[z + a](b + c) ∈ Reals, {x,y,z,a,b,c}]

The error returned is "FindInstance::nsmet: The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist." However it's fairly obvious what a possible instance is: $x=y=0, z=a=b=c=1$; in fact as long as $z+a > 0$ the condition is met.

Answer 1

If you are willing to assume that the parameters are real valued:

FindInstance[x + y - Log[z + a] (b + c) \[Element] Reals, {x, y, z, a, b, c} \[Element] Reals]

Or you can get a fuller account:

Reduce[x + y - Log[z + a] (b + c) \[Element] Reals, {x, y, z, a, b, c} \[Element] Reals]

Answer 2

Look at FunctionDomain

FunctionDomain[x + y - Log[z + a] (b + c), {x, y, z, a, b, c}, Reals]

(* a + z > 0 *)

Consequently, given that all variables are real the expression is real provided that a + z > 0. Then

solns = FindInstance[a + z > 0, {x, y, z, a, b, c}, Reals, 5]

(* {{x -> -(14/5), y -> -(31/10), z -> -(117/5), a -> 73, b -> -(21/5), 
  c -> -3}, {x -> -(1/5), y -> 12/5, z -> 99/5, a -> 49, b -> -(1/5), 
  c -> 47/10}, {x -> 5/2, y -> -(23/10), z -> 43/5, a -> 39, b -> 7/10, 
  c -> 1/2}, {x -> 39/10, y -> -(18/5), z -> -(69/5), a -> 80, b -> 23/5, 
  c -> -(3/10)}, {x -> 23/5, y -> 9/2, z -> 171/10, a -> 49, b -> -(18/5), 
  c -> -(1/5)}} *)

Verifying,

And @@ (x + y - Log[z + a] (b + c) ∈ Reals /. solns)

(* True *)