Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?

Question

In general we know $A^{bc} = A^{cb}$ for integer $A$. I want to extend this to the case $A=-1$.

For integers $a,b$ I guess the above relation holds, \begin{align} (-1)^{2\cdot3} = ((-1)^2)^3 = 1 = ((-1)^3)^2 .\end{align}

But if we include the fraction

\begin{align} (-1)^{-\frac{1}{2}} &= ((-1)^{-1})^{\frac{1}{2}} = (-1)^{\frac{1}{2}} = i \\ &= ((-1)^{\frac{1}{2}})^{-1}=\frac{1}{i} = -i \end{align} this does not hold any more.

Is something wrong with my computation?

The problem here is that complex powers with non-integer exponent are ill-defined. — Crostul, Dec 15 '15 at 09:11
$-1$ is an integer for starters, second powers are not commutative for non-integer exponents. — Zelos Malum, Dec 15 '15 at 09:12
On base of what do you say $(-1)^{\frac12}=i$? Why not $(-1)^{\frac12}=-i$? We have $i^2=-1=(-i)^2$. — drhab, Dec 15 '15 at 09:13
But can we say mod function make it hold true?? As integers are also mods?? — Archis Welankar, Dec 15 '15 at 09:17
@ArchisWelankar Well, similar statement about absolute values holds, but that's not a good way to put it. — Abstraction, Dec 15 '15 at 09:19
If we do not stick to principal roots then we have $(-1)^{\frac{1}{2}}=\pm i$, then the results holds atleast for the example you have given — Shahid M Shah, Dec 15 '15 at 09:20
What i mean its just finding out$\sqrt{a^2+b^2}$ not much than that — Archis Welankar, Dec 15 '15 at 09:35

score 3 · Accepted Answer · 2015-12-15T09:32:49.637

3

In complex numbers, $$(a^b)^c=a^{bc}$$ doesn't hold. You just found a counterexample.

This is due to the $2k\pi$ undeterminacy of the argument. When you take the square root, it becomes a $k\pi$ i.e. a sign indeterminacy.

edited Dec 15 '15 at 09:32

answered Dec 15 '15 at 09:17

Please explain this downvote. – Dec 15 '15 at 09:22
@Arthur: I should have added "in general". – Dec 15 '15 at 09:30

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

Of course $(-1)^{ab} = (-1)^{ba}$, since multiplication is commutative. (No matter how you define exponentiation for complex numbers: fixed branch or multivalued.)

What isn't true is that $(-1)^{ab} = ((-1)^a)^b$ in general. This has been asked and answered many times on this site.

For similar questions, see for example

just to mention a few.

score 1 · Answer 3 · answered Dec 15 '15 at 09:26

For complex number $z$ and integer $a$, $z^a$ is defined and equals $z*z*...*z$ ($a$ times) or ${1\over z}*{1\over z}*...*{1\over z}$ ($-a$ times) for negative $a$.

For rational $a={p \over q}$, $z^a$ can be said to have $q$ values, with one value called "main" or principal value. For rational $a,b$, $(z^a)^b$ and $(z^b)^a$ have the same set of values, but may have different principal values. In your case (simplifying things a bit): $$((-1)^{-1})^{1 \over 2} = (-1)^{1 \over 2} = \{i, -i\} \\ ((-1)^{1 \over 2})^{-1} = (\{i, -i\})^{-1} = \{-i, i\}$$

When $a$ is irrational, $z^a$ is usually said to be undefined (one could say it has infinitely many values).

Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?

3 Answers3

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