Weird simplification of complex value and its conjugate?

Question

Consider the following simplification

(Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] // FullSimplify

(Sqrt[Conjugate[A]] Sign[A])/Sqrt[A]

I am a bit confused by this output. I would have expected Mathematica to return just 1, since

$$A=|A|e^{i(\phi+2\pi n)}~~~,~~~A^*=|A|e^{-i(\phi+2\pi n)}~~~,~~~n\in\mathbb{Z}\,,$$

obviously leads to

$$\frac{\sqrt{A}\sqrt{A^*}}{\sqrt{AA^*}}=\frac{|A|\sqrt{e^{i(\phi+2\pi n)}}\sqrt{e^{-i(\phi+2\pi n)}}}{|A|}=e^{i\frac{\phi+2\pi n}{2} -i\frac{\phi+2\pi n}{2}}=1$$

on all branches $n\in \mathbb{Z}$.

Instead, we get the above output, where I'm not even sure what Sign of a complex number is supposed to mean. What is going on? Is there a way to make Mathematica simplify this properly?

EDIT:

Just to convince everyone that there is nothing special going on for Arg[A] >= Pi, see the following plot

Plot3D[ReIm[(Sqrt[x Exp[I \[Phi]]] Sqrt[x Exp[-I \[Phi]]])/Sqrt[  x Exp[I \[Phi]] x Exp[-I \[Phi]]]], {x, 0, 50}, {\[Phi], 0, 20 \[Pi]}]

Real part is always 1, imaginary part is always 0.

EDIT2:

It seems that the trouble of $\sqrt{(-1)\cdot(-1)}$ vs $\sqrt{-1}\cdot \sqrt{-1}$ is addressed even in the relevant wikipedia article. There it is pointed out that $\sqrt{z^*}\neq \sqrt{z}^*$ when the principal square root function is considered. I guess Mathematica uses exactly that principal function version, which explains why it does not return what I expected.

Personally, I would have preferred if on the field of complex numbers we had non-identical $(-1)=e^{i \pi}$ and $(-1)^*=e^{-i \pi}$, which would resolve the issue and make $\sqrt{z^*}= \sqrt{z}^*$ true. But that's not standard, so I guess I can't expect it from Mathematica either.

Answer 1

Since Sqrt is a function, it must be defined with a branch cut (this is not a shortcoming of Mathematica, just a result of requiring functions to be single-valued). The cut used by Mathematica means that phases $\phi$ of complex arguments are moved into the interval $-\pi < \phi\le \pi$ before applying the rule $\sqrt{\exp(i \phi)} = \exp(i \phi/2)$.

In your derivation, this rule is applied without moving the argument to the correct branch. This is a misunderstanding based on the fact that in mathematics the interpretation of the symbol $\sqrt{\ldots}$ isn't as strict as in Mathematica. You therefore have to adjust your expectation to correspond to the meaning of Sqrt, or replace the function Sqrt by something else.

I assume you don't want to throw out Sqrt from the code, so here is what you could do to get the expected result:

Simplify[(Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] /. A -> Abs[A] Exp[I ϕ], 
Assumptions -> {-π < ϕ < π}]

1

What I did is replace A by its AbsArg representation and added the assumption about the phase ϕ being in the open interval excluding the boundaries.

Answer 2

It evidently depends on the value of A. Try

(Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] // 
  PowerExpand[#, 
    Assumptions -> A \[Element] Complexes] & // FullSimplify

Piecewise[{{-1, Arg[A] >= Pi}}, 1]

So it's 1 if Arg[A] <= Pi

Answer 3

(Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] /. A -> -1
(*  -1  *)

because it's

I * I / 1

Answer 4

In addition to all the answers about complex numbers, Mathematica makes no assumptions that A is a number. It could just as easily be a matrix, for example:

In[13]:= (Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] /. A -> {{1, I}, {-I, -1}}

Out[13]= {{1, 1}, {1, -1}}

In[14]:= (Sqrt[Conjugate[A]] Sign[A])/Sqrt[A] /. A -> {{1, I}, {-I, -1}}

Out[14]= {{1, 1}, {1, -1}}

in which case returning a scalar would be incorrect.

Answer 5

A bit long for a comment. Consider:

a = Sqrt[A] /. A -> -1
(* I *)

Correct. Sqrt[-1] is I by definition.

b = Sqrt[Conjugate[A]] /. A -> -1
(* I *)

Correct. Conjugate[-1] is -1, and Sqrt[-1] is I by definition.

c = Sqrt[A Conjugate[A]] /. A -> -1
(* 1 *)

Correct. Sqrt[1] is unambiguously 1.

(a b)/c
(* -1 *)

Correct.

(Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] /. A -> -1
(* -1 *)

Now, you consider this wrong, but it's the same as (a b)/c. Mathematica is thus consistent, which is a crucial thing to a computer algebra system. To get the answer you want, Mathematica would have to use a different branch cut for different instances of Sqrt in your expression. How is it supposed to know that this is what you intend?