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Question: Does is(...) assume real?

Question:Does is(...) assume real?

C_R 2867 Maple
assuming abs signum
November 05 2023
3

Here is an example without assumptions that should return false for complex numbers (IMO)

prp := cos(z)/abs(cos(z)) - abs(cos(z))/cos(z) = 0:
is(prp);
                              true

For comparision: simplify requires restriction to the real domain

simplify(prp);
(simplify(prp) assuming real);
                          2                    
                 -|cos(z)|  sec(z) + cos(z)    
                 -------------------------- = 0
                          |cos(z)|             

                             0 = 0

This only makes sense if is(...) assumes real or I am wrong with complex numbers and simplify should simplify this expression without assumptions.

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