Definition of the root of -1 for different roots

[Mr Davis 97]

How does the value of $\displaystyle \sqrt[a]{-1}$ vary as $a$ varies as any real number? When is this value complex and when is it real? For example, we know that when a = 2 it is complex, but when a = 3 it is real. What about when $a = \pi$, for example?

 

[SlowThinker]

You may want to read this Insight:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Mark44]

How does the value of $\displaystyle \sqrt[a]{-1}$ vary as $a$ varies as any real number? When is this value complex and when is it real? For example, we know that when a = 2 it is complex, but when a = 3 it is real. What about when $a = \pi$, for example?

As far as I know, there is no such thing as the "π-th root" of a number. The index of a radical is a positive integer that is 2 or larger. You can however raise a number to an arbitrary power. For example, $2^{1/\pi} = (e^{\ln 2})^{1/\pi} = e^{\frac{\ln 2}{\pi}}$, but see the link that @SlowThinker posted.

 

[Mr Davis 97]

As far as I know, there is no such thing as the "π-th root" of a number. The index of a radical is a positive integer that is 2 or larger. You can however raise a number to an arbitrary power. For example, $2^{1/\pi} = (e^{\ln 2})^{1/\pi} = e^{\frac{\ln 2}{\pi}}$, but see the link that @SlowThinker posted.

if $x^{1/a} = e^{\frac{\ln x}{a}}$, and we know that $(-1)^{1/3} = -1$, does that mean that $(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1$?

 

[Mark44]

if $x^{1/a} = e^{\frac{\ln x}{a}}$, and we know that $(-1)^{1/3} = -1$, does that mean that $(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1$?

Not if ln means the usual natural logarithm function whose domain is positive real numbers.

 

[Mr Davis 97]

So in general when would $(-1)^{1/a}$ be complex and when would it be real?

 

[MAGNIBORO]

if $x^{1/a} = e^{\frac{\ln x}{a}}$, and we know that $(-1)^{1/3} = -1$, does that mean that $(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1$?

$(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}=-1$ this is correct but you don't consider the other results,
$\sqrt4=2$ is correct but not complete, becuase $\sqrt4=\pm 2$
in the same way are other results for $(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}$

look the post https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Mark44]

$(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}=-1$ this is correct but you don't consider the other results,
$\sqrt4=2$ is correct but not complete, becuase $\sqrt4=\pm 2$

No. $\sqrt 4$ is generally accepted to mean the principal square root of 4, a positive number.

in the same way are other results for $(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}$

look the post https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Stephen Tashi]

if $x^{1/a} = e^{\frac{\ln x}{a}}$

That definition only applies when $x$ is positive.

(In the domain of real numbers, there is no standard definition for raising a negative number to an irrational exponent. )