Question 5

Find all cube roots of the complex number. Leave answers in trigonometric form.
\[
-125
\]

Step 1 ：Express the given complex number -125 in trigonometric form. The complex number is -125, which can be written as -125 + 0i.

Step 2 ：The trigonometric form of a complex number is \(r(\cos \theta + i \sin \theta)\), where \(r\) is the magnitude of the complex number and \(\theta\) is the argument of the complex number.

Step 3 ：The magnitude \(r\) is given by \(\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts of the complex number, respectively. In this case, \(a = -125\) and \(b = 0\), so \(r = \sqrt{(-125)^2 + 0^2} = 125\).

Step 4 ：The argument \(\theta\) is given by \(\arctan\frac{b}{a}\) if \(a > 0\), \(\pi + \arctan\frac{b}{a}\) if \(a < 0\) and \(b > 0\), and \(-\pi + \arctan\frac{b}{a}\) if \(a < 0\) and \(b < 0\). In this case, \(a = -125\) and \(b = 0\), so \(\theta = \pi\).

Step 5 ：So, the trigonometric form of -125 is \(125(\cos \pi + i \sin \pi)\).

Step 6 ：Now, we can find the cube roots using De Moivre's theorem, which states that the n-th roots of a complex number \(r(\cos \theta + i \sin \theta)\) are given by \(r^{1/n}(\cos(\frac{\theta + 2k\pi}{n}) + i \sin(\frac{\theta + 2k\pi}{n}))\), where \(k = 0, 1, ..., n - 1\).

Step 7 ：In this case, \(n = 3\), so the cube roots are \(125^{1/3}(\cos(\frac{\pi + 2*0\pi}{3}) + i \sin(\frac{\pi + 2*0\pi}{3})) = 5(\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}))\), \(125^{1/3}(\cos(\frac{\pi + 2*1\pi}{3}) + i \sin(\frac{\pi + 2*1\pi}{3})) = 5(\cos \pi + i \sin \pi)\), and \(125^{1/3}(\cos(\frac{\pi + 2*2\pi}{3}) + i \sin(\frac{\pi + 2*2\pi}{3})) = 5(\cos(\frac{5\pi}{3}) + i \sin(\frac{5\pi}{3}))\).

Step 8 ：\(\boxed{5(\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}))}\), \(\boxed{5(\cos \pi + i \sin \pi)}\), and \(\boxed{5(\cos(\frac{5\pi}{3}) + i \sin(\frac{5\pi}{3}))}\) are the cube roots of -125.