Is the equality \sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b} true for complex numbers?

[gabel]

$$-1=\sqrt[3]{-1}=(-1)^{1/3} = (-1)^{2/6} = ((-1)^2)^{1/3}=1^{1/3} = 1$$

How can this be?

 

[HallsofIvy]

It can't be. The error is in thinking that $$\sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b}$$ for complex numbers. It is only true if the roots are real numbers.

 

[Tinyboss]

I think the assumption that 2*1/3=2/6 might be causing some trouble here, too. ;-)

 

[HallsofIvy]

I hadn't even noticed that! Though I suspect that was a typo.

 

[arildno]

It is a rather minor flaw, since the fundamental flaw is retained in the following "flawless" argument:
$$-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=((-1)^{2})^{\frac{1}{6}}=1$$

 

[Redbelly98]

gabel and arildno, welcome to PF.

I guess the explanation lies in the fact that -1 is one of the complex 6^th roots of 1.

 

[arildno]

gabel and arildno, welcome to PF.

Thank you for the warm welcome! :shy:

 

[yuiop]

gabel and arildno, welcome to PF.

How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel too

 

[arildno]

How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel too

Weclome to yuiop&gabel!

BTW, Redbelly is a newbie from 2008, so he hasn't been as tardy in welcoming me as you make it out to be..
In fact, you yourself has been even tardier than Redbelly..

 

[Redbelly98]

How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel too

Be sure to visit the recent https://www.physicsforums.com/showthread.php?t=465551"

Back on topic -- here is a version of this problem in "lowest terms", so to speak:

-1 = (-1)¹ = (-1)^2/2 = [(-1)²]^1/2 = 1^1/2 = 1​

 

[disregardthat]

It can't be. The error is in thinking that $$\sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b}$$ for complex numbers. It is only true if the roots are real numbers.

It's not true for real numbers, not the negative ones. And for rational exponentiation of complex numbers we usually define a branch, in which case the equality is true. For real numbers, rational exponentiation, x^(1/n), is usually defined as a real root of t^n-x = 0 (if such a root exists) in the complex domain, and the positive one is chosen if there are two such roots. In this case the equality does not apply generally, as the example shows.