I'm confused that (-1)^0.2 =1 or -1 and the other 4 answers that are complex numbers

Well that is quite weird! That is one of the reasons why we don’t define the exponential function [MATH]a^x[/MATH] for [MATH]a<0[/MATH]. I’m not really knowledgable about complex numbers and this problem is really confusing, but I have some ideas which the more experienced mathematicians could verify
1) We don’t define the exponential function for negative basis, thus I would mark this expression as undefined (again, this probably can make sense in the set of complex numbers)
2) I think that maybe there is an error hidden in the “right” computation, we know that [MATH]\sqrt{x^2} = \lvert x \rvert[/MATH]. And when you cancelled 2/10 to get 1/5, you basically did [MATH]^5 \sqrt{\sqrt{x^2}}[/MATH] which would be the fifth root of absolute x, aka for x equals -1 that would be 1. I’m not completely sure about this, I hope that someone will correct me if I’m wrong.
3) if you write -1 as i^2, then the answer is i^0.4, I guess that would be the result expressed as a complex number?
 
Non-integral powers are multi-valued functions (in the complex world), and different manipulations like these can result in different choices of values.

There are ten tenth roots of 1, and five fifth roots of -1. One of the former is -1, and one of the latter is -1, but what you did chose the "wrong" one on the left.
 
Why are you confused? 0.2= 2/10= 1/5 so I would expect there to be 5 complex roots.

\(\displaystyle -1= -1+ 0i= cos(\pi)+ i sin(\pi)= e^{i\pi}\)

The five fifth roots are equally spaced around the unit circle in the compex plane. Obviously, \(\displaystyle (-1)^5= -1\) so -1 is a root. \(\displaystyle 1^5= 1\ne -1\) so 1 is NOT a root.

The roots are
\(\displaystyle e^{i\pi/5}= cos(\pi/5)+ i sin(\pi/5)= 0.8090+ 0.58778i\)
\(\displaystyle e^{i(3\pi/5)}= cos(3\pi/5)+ i sin(3\pi/5)= -0.3090+ 0.9510i\)
\(\displaystyle e^{i(5\pi/5)}= e^{i\pi}= cos(\pi)+ i sin(\pi)= -1\)
\(\displaystyle e^{i(7\pi/5}= cos(7\pi/5)+ i sin(7\pi/5)= -0.3090+ 0.5877i\)
\(\displaystyle e^{i(9\pi/5)}= cos(9\pi/5)+ i sin(9\pi/5)= 0.8090- 0.5878i\)
(Each rounded to four decimal places.)

If we went up another "\(\displaystyle 2\pi\)" we would have \(\displaystyle 11\pi/5= 2\pi+ \pi/5\), repeating the first root.
 
[MATH](x^2)^{\frac{1}{10}} \neq x^{\frac{2}{10}} [/MATH]
[MATH](x^2)^{\frac{1}{10}} = (|x|)^{\frac{1}{5}}[/MATH]
When [MATH]x = -1, (|-1|)^{\frac{1}{5}} = 1[/MATH]
[MATH]x^{\frac{2}{10}} = x^{\frac{1}{5}}[/MATH]
When [MATH]x = -1, (-1)^{\frac{1}{5}} = -1[/MATH]
You were playing around with exponents, probably following algebra rules, that's fine. But be careful when you face [MATH]\sqrt{x^2} = |x|[/MATH]
It is a headache for thousands of students. You need a little time to overcome it.

Also, you have to remember that, not all algebra rules can be applied for complex numbers. Otherwise, this statement will be true:

[MATH]\sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1[/MATH]
But [MATH]\sqrt{-1} = i[/MATH], then

[MATH]\sqrt{-1} \cdot \sqrt{-1} = i^2 = -1[/MATH]
But [MATH]1 \neq -1[/MATH]
This is why when you are dealing with complex numbers, or even with exponents, you have to be careful of some rules.
 
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No, no, when it comes to exponents, you can not reduce them!
Ex: (-1)2/10= [(-1)2]1/10 = (1)1/10=1
While (-1)1/5=-1

When I noticed this it was shocking to me because I always thought that you can replace equals with equals!
 
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Kuang said:
Please explain where I was wrong and why and the correct way.

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Just as a quick observation, if we are using complex numbers then [math](1)^{1/10}[/math] has 2 possible real values and 8 possible complex values. Likewise [math](-1)^{1/5}[/math] has 1 possible real value and 4 possible complex values. You would have to compare them all to do this.

-Dan
 
I haven't studied complex numbers in great depth myself. To me it can still seem mind-blowing that the exponent operator can output many values at the same time ?. But, I think I may have just had an epiphany while writing this. The exponent operator can output infinite values if we write \(3^x\) because x can be any value. And you wouldn't think twice if you saw \(3^x=9\). Therefore it shouldn't be so surprising that \(1^{(1/2)} \) could output two values and we sometimes need to consider both. The equals = "operator" only needs one match from LHS to RHS before it's a true statement (more than one match can, of course, be possible). I don't know if this is a good way of thinking? Perhaps one of the more experienced helpers will know!

--

(Probably because I've been involved with programming for years) I often find it helpful to think of ONLY the principal root, or principal value, being returned from the exponent operator. Programs like "Matlab" just return the principal root if you type (-1)^(0.2) So, if you're sticking to principal roots, then it's worth noting that the "power of power" rule:-

[math] \left(a^b\right)^c = a^{(bc)} [/math]
with a,b,c real, can only be applied when one of the following is true:-
  1. a>0
  2. a<0 and c integer (b can be ANY real)
  3. a=0 and b>0 and c>0
You wrote...
[math] \left(-1\right)^{2/10} = \left(\left(-1\right)^2\right)^{\color{red}1/10} [/math]
this doesn't work when using principal roots because c, highlighted red, isn't integer. However, this would be fine...
[math] \left(-1\right)^{2/10} = \left(\left(-1\right)^{1/10}\right)^2 [/math]
 
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