Is ± Present When Taking 4th Root?

[Destroxia]

Homework Statement

If you take a 4th root of something, is that answer also plus or minus, just like if you are taking a square root?

Ex. $4^{\frac 1 2} = ±2$

So following that logic, would that mean that:

$16^{\frac 1 4} = ±2$ as well?

Meaning, is the ± still present if you take a 4th root?

Homework Equations

The Attempt at a Solution

 

[Mark44]

Homework Statement

If you take a 4th root of something, is that answer also plus or minus, just like if you are taking a square root?

Ex. $4^{\frac 1 2} = ±2$

While 2 and -2 are square roots of 4, the expression $\sqrt{4}$ denotes the principal (i.e., positive) root, or +2. So $4^{1/2}$ would also be +2.

So following that logic, would that mean that:

$16^{\frac 1 4} = ±2$ as well?

The principal fourth root of 16 is 2. There are three other fourth roots: -2, 2i, and -2i.

Meaning, is the ± still present if you take a 4th root?

Homework Equations

The Attempt at a Solution

 

[Destroxia]

Okay, so I took the 4th root of -1 using a CAS, and ended up getting the result of

$r = \frac {\sqrt(2)} {2} +\frac {\sqrt(2)} {2}i$

So is there any other answer besides this, or is this the single and only ?

 

[Mark44]

Okay, so I took the 4th root of -1 using a CAS, and ended up getting the result of

$r = \frac {\sqrt(2)} {2} +\frac {\sqrt(2)} {2}i$

So is there any other answer besides this, or is this the single and only ?

There are four fourth roots, spaced 90° apart ($\pi/2$ radians). I don't know if there is a principal fourth root of a negative number, similar to what I said before about the fourth root of 16.

Usually the question is asked as, "Find all of the fourth roots of <N>."

 

[vela]

Okay, so I took the 4th root of -1 using a CAS, and ended up getting the result of

$r = \frac {\sqrt(2)} {2} +\frac {\sqrt(2)} {2}i$

So is there any other answer besides this, or is this the single and only ?

The fourth roots of -1 satisfy the equation $r^4 = -1$ or $r^4 + 1 = 0$. Since you have a fourth-degree polynomial, there are going to be four solutions.

To find the square roots of 4, you want to solve the equation $z^2 = 4$. If you express both sides in polar form, with $z = re^{i\theta}$, you have
$$r^2 e^{i2\theta} = 4e^{i0},$$ which has solution $r=2$ and $\theta=0$, which corresponds to $z=2$. But you could also write the righthand side as $4e^{i2\pi}$, which gives you a second solution, $r=2$ and $\theta=\pi$, which corresponds to $z=-2$.

You can use the same method to find all the fourth roots.

 

[Destroxia]

While 2 and -2 are square roots of 4, the expression $\sqrt{4}$ denotes the principal (i.e., positive) root, or +2. So $4^{1/2}$ would also be +2.
The principal fourth root of 16 is 2. There are three other fourth roots: -2, 2i, and -2i.

The fourth roots of -1 satisfy the equation $r^4 = -1$ or $r^4 + 1 = 0$. Since you have a fourth-degree polynomial, there are going to be four solutions.

To find the square roots of 4, you want to solve the equation $z^2 = 4$. If you express both sides in polar form, with $z = re^{i\theta}$, you have
$$r^2 e^{i2\theta} = 4e^{i0},$$ which has solution $r=2$ and $\theta=0$, which corresponds to $z=2$. But you could also write the righthand side as $4e^{i2\pi}$, which gives you a second solution, $r=2$ and $\theta=\pi$, which corresponds to $z=-2$.

You can use the same method to find all the fourth roots.

Okay, thank you both, I understand how to do this now and have applied it to my problem, but my problem is differential calculus based, so I will post that in the calculus and above section.