# Is {(-2)^5}^(1/5) a complex number ?

1. Aug 7, 2013

### phydis

Last edited by a moderator: Aug 7, 2013
2. Aug 7, 2013

### pwsnafu

Notice how the alpha page says
If you select the real valued option you'll get -2.

3. Aug 7, 2013

### lurflurf

You have a typo
E = {(-2)^5}^(1/5)=(-32)^(1/5)
There are five numbers such that x^5=-32
(-32)^(1/5) should be one of them
they are
1.61803398874989+1.17557050458495 i
-0.61803398874989+1.90211303259031 i
-2
-0.61803398874989-1.90211303259031 i
1.61803398874989-1.17557050458495 i

We chose 1.61803398874989+1.17557050458495 i as it is first on the list and the most reasonable choice. Some people inexplicably pick -2.

4. Aug 7, 2013

### D H

Staff Emeritus
There are *five* solutions to x5=(-2)5. One of them is the real number -2. The other four are complex numbers. Wolfram Alpha gives you the principal root by default. You can force it to yield the real-valued root by clicking on "Use the real‐valued root instead".

You can see all five solutions if you instead ask WA to solve x5=(-2)5.

5. Aug 7, 2013

### phydis

got it. Thanks everyone!