Example 2.61: Find the cube roots of the complex number 8i. In extraction roots

Answer 1

of complex numbers, the formula is given by:

\( z^{1/n} = \sqrt[n]{|z|} \left[ \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right] \), where n is the index and k = 0, 1, ..., n-1.

Given that the complex number is 8i, we can rewrite it in its polar form as 8i = 8(cos(π/2) + i sin(π/2)).

So, |8i| = 8 and the argument θ = π/2.

Now we can apply the formula for the cube roots (n = 3) of the complex number:

Cube root 1:
\( \sqrt[3]{8} \left[ \cos \left( \frac{\pi/2 + 2(0)\pi}{3} \right) + i \sin \left( \frac{\pi/2 + 2(0)\pi}{3} \right) \right] = 2(\cos(\pi/6) + i \sin(\pi/6)) = 2(cos(30°) + i sin(30°)) = 2(√3/2 + i/2) = √3 + i \)

Cube root 2:
\( \sqrt[3]{8} \left[ \cos \left( \frac{\pi/2 + 2(1)\pi}{3} \right) + i \sin \left( \frac{\pi/2 + 2(1)\pi}{3} \right) \right] = 2(\cos 5\pi/6 + i \sin 5\pi/6) = 2(-√3/2 + i/2) = -√3 + i \)

Cube root 3:
\( \sqrt[3]{8} \left[ \cos \left( \frac{\pi/2 + 2(2)\pi}{3} \right) + i \sin \left( \frac{\pi/2 + 2(2)\pi}{3} \right) \right] = 2(\cos 3\pi/2 + i \sin 3\pi/2) = 2(0 - i) = -2i \)

Therefore, the cube roots of the complex number 8i are √3 + i, -√3 + i, and -2i.