Why doesn't $z^n\cdot\left(\frac{a+b}{z}\right)^n = (a+b)^n$ always hold?

Question

When I entered is(z^n*((a+b)/z)^n = (a+b)^n); into Maple, the output was false and I guess Maple assumes that $a,b,n$ and $z$ can be any number in $ℂ$.

I thought $$z^n\cdot\left(\frac{a+b}{z}\right)^n$$ was clearly $$\left(z\cdot\frac{a+b}{z}\right)^n$$ and therefore $$(a+b)^n$$

Why does this not apply?

Edit:
The output is still false when typing assume(z <> 0): is(z^n*((a+b)/z)^n = (a+b)^n);

Edit 2:
assume(z >= 0): is(z^n*((a+b)/z)^n = (a+b)^n); returns true.

Edit 3:
assume(z = 0): is(z^n*((a+b)/z)^n = (a+b)^n); returns false.

Isn't this a contradiction?

Edit 4:
assume(n in ℕ): is(z^n*((a+b)/z)^n = (a+b)^n); returns false.

For $a<0$ and $z<0$, Maple fails to evaluate the expression, although it is true. This means Maple returned false instead of admitting that it is unable to determine the result.
This is a flaw.

@Mattos I tried that. Maple throws a division-by-zero error. — xoxox, Mar 06 '16 at 11:05
@Mattos That cannot be since the expression $;\frac{a+b}z;$ is given, so it is assumed, I suppose, that automatically $;z\neq0;$ — DonAntonio, Mar 06 '16 at 11:05
Have you typed it correctly? I really doubt. If yes then, yes! It's definitely a contradiction. — Sufyan Naeem, Mar 06 '16 at 12:38
@Mankind Good point, I added that case to my answer. I think it suffices when $n$ is nonnegative. — xoxox, Mar 06 '16 at 14:31

score 8 · Accepted Answer · answered Mar 06 '16 at 11:35

8

In complex numbers, $(ab)^c=a^cb^c$ does not necessarily hold.

$$((-1)(-1))^{1/2}=1\ne(-1)^{1/2}(-1)^{1/2}=i^2=-1.$$

answered Mar 06 '16 at 11:35

2

I don't know, but since when does $(-1)^{1/2} = i$? – user49685 Mar 06 '16 at 12:26
1

@user49685 https://en.wikipedia.org/wiki/Imaginary_unit – xoxox Mar 06 '16 at 12:37
1

In complex numbers, $a^c$ is not even defined unambiguously. – Najib Idrissi Mar 06 '16 at 13:13
2

Writing $(-1)^{1/2}=i$ is wrong to begin with, unless you specifically tell what branch of the square root you're using. – egreg Mar 06 '16 at 13:36
@user49685: since when does $(ab)^c=a^cb^c$ hold ? – Mar 06 '16 at 13:43

Therkel · Answer 2 · 2016-03-07T09:42:02.307

It looks like Maple has trouble evaluating the left hand side of your equation. You can help Maple by asking it to expand (or simplify) the left-hand-side of your expression before using the is command from the assume facility.

eqn := z^n*((a+b)/z)^n = (a+b)^n;
assume(z::real,n::integer);

As we saw in your question, is(eqn); returns false but

is(simplify(eqn));

returns true. I suppose this arises from limitations of the Assume Facility; as referred to on the help page, this may be documented in [Corless, Monagan] (I was unable to find an online copy of it).

Side note

If $n$ is any complex number, we cannot say anything in general (non-uniqueness) but applying the assumption that $(a+b)/z>0$ seems to be enough for the Assume Facility (with a helping hand from the simplify command):

restart;
eqn := z^n*((a+b)/z)^n = (a+b)^n;

assume((a+b)/z>1);
is(simplify(eqn));

returns true although $z^n$ is ambiguous for $z,n \in \mathbb C$. This seems to be a result of the way Maple defines the complex exponential to have a unique solution, see Section 5.1 The Complex Exponential Function (document can only be opened in Maple).

_{[Corless, Monagan] Corless, Robert, and Monagan, Michael. "Simplification and the Assume Facility." Maple Technical Newsletter, Vol. 1 No. 1. Birkhauser, 1994.}

Why doesn't $z^n\cdot\left(\frac{a+b}{z}\right)^n = (a+b)^n$ always hold?

2 Answers2

Side note

Linked