Question

Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Answer 1

Answer:

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$, $z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$, $z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

$z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]$

Where angles are measured in radians and k represents an integer between $0$ and $n - 1$.

The magnitude of the complex number is $27$ and the equivalent angular value is $1.817\pi$. The set of cubic roots are, respectively:

k = 0

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$

k = 1

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$

k = 2

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$