Question
Find the three cube roots of negative 1. Write your answers in polar form.
(1 point)
Responses
cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°
cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°
cos 0° + i sin 0°, cos 120° + i sin 120°, cos 240° + i sin 240°
cos 0° + i sin 0°, cos 120° + i sin 120°, cos 240° + i sin 240°
cos 60° + i sin 60°, cos 120° + i sin 120°, cos 180° + i sin 180°
cos 60° + i sin 60°, cos 120° + i sin 120°, cos 180° + i sin 180°
cos 40° + i sin 40°, cos 160° + i sin 160°, cos 280°, i sin 280°
cos 40° + i sin 40°, cos 160° + i sin 160°, cos 280°, i sin 280°
(1 point)
Responses
cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°
cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°
cos 0° + i sin 0°, cos 120° + i sin 120°, cos 240° + i sin 240°
cos 0° + i sin 0°, cos 120° + i sin 120°, cos 240° + i sin 240°
cos 60° + i sin 60°, cos 120° + i sin 120°, cos 180° + i sin 180°
cos 60° + i sin 60°, cos 120° + i sin 120°, cos 180° + i sin 180°
cos 40° + i sin 40°, cos 160° + i sin 160°, cos 280°, i sin 280°
cos 40° + i sin 40°, cos 160° + i sin 160°, cos 280°, i sin 280°
Answers
GPT-4o mini
To find the cube roots of \(-1\), we start by expressing \(-1\) in polar form. The polar form of a complex number \(r (\cos \theta + i \sin \theta)\) for \(-1\) is as follows:
\[
-1 = 1(\cos 180^\circ + i \sin 180^\circ)
\]
Now, to find the cube roots, we need to use the formula for the roots of complex numbers in polar form. The \(n\)-th roots of a complex number can be found using the formula:
\[
\sqrt[n]{r} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right)
\]
where \(k = 0, 1, 2, \ldots, n-1\).
For our case:
- \(n = 3\) (since we are finding cube roots)
- \(r = 1\)
- \(\theta = 180^\circ\)
The cube roots of \(-1\) will be given by:
1. For \(k = 0\):
\[
\cos\left(\frac{180^\circ + 2(0)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(0)\pi}{3}\right) = \cos(60^\circ) + i \sin(60^\circ)
\]
2. For \(k = 1\):
\[
\cos\left(\frac{180^\circ + 2(1)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(1)\pi}{3}\right) = \cos(180^\circ) + i \sin(180^\circ)
\]
3. For \(k = 2\):
\[
\cos\left(\frac{180^\circ + 2(2)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(2)\pi}{3}\right) = \cos(300^\circ) + i \sin(300^\circ)
\]
Thus, the three cube roots of \(-1\) in polar form are:
\[
\cos 60^\circ + i \sin 60^\circ, \quad \cos 180^\circ + i \sin 180^\circ, \quad \cos 300^\circ + i \sin 300^\circ
\]
So, the correct response is:
**cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°**
\[
-1 = 1(\cos 180^\circ + i \sin 180^\circ)
\]
Now, to find the cube roots, we need to use the formula for the roots of complex numbers in polar form. The \(n\)-th roots of a complex number can be found using the formula:
\[
\sqrt[n]{r} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right)
\]
where \(k = 0, 1, 2, \ldots, n-1\).
For our case:
- \(n = 3\) (since we are finding cube roots)
- \(r = 1\)
- \(\theta = 180^\circ\)
The cube roots of \(-1\) will be given by:
1. For \(k = 0\):
\[
\cos\left(\frac{180^\circ + 2(0)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(0)\pi}{3}\right) = \cos(60^\circ) + i \sin(60^\circ)
\]
2. For \(k = 1\):
\[
\cos\left(\frac{180^\circ + 2(1)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(1)\pi}{3}\right) = \cos(180^\circ) + i \sin(180^\circ)
\]
3. For \(k = 2\):
\[
\cos\left(\frac{180^\circ + 2(2)\pi}{3}\right) + i \sin\left(\frac{180^\circ + 2(2)\pi}{3}\right) = \cos(300^\circ) + i \sin(300^\circ)
\]
Thus, the three cube roots of \(-1\) in polar form are:
\[
\cos 60^\circ + i \sin 60^\circ, \quad \cos 180^\circ + i \sin 180^\circ, \quad \cos 300^\circ + i \sin 300^\circ
\]
So, the correct response is:
**cos 60° + i sin 60°, cos 180° + i sin 180°, cos 300° + i sin 300°**