Question

32(\cos 60^{\circ} + i \sin 60^{\circ})\
next, using demoivres find the 5th root of\
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 16 + 16\sqrt{3}i =\
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 32(\cos 60^{\circ} + i \sin 60^{\circ})\
find r and \theta where \theta is in the third quadrant.\
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ r^5 = 32\
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 5\theta = 60^{\circ}, 420^{\circ}, 780^{\circ}, 1140^{\circ}, 1500^{\circ}\
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ r = ? \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \theta = \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \circ

Explanation:

Step1: Solve for \( r \)

We have the equation \( r^5 = 32 \). To find \( r \), we take the fifth root of both sides. Since \( 32 = 2^5 \), the fifth root of \( 32 \) is \( 2 \). So, \( r = \sqrt[5]{32} = 2 \).

Step2: Solve for \( \theta \) (third quadrant)

We know that \( 5\theta = 60^\circ, 420^\circ, 780^\circ, 1140^\circ, 1500^\circ \). We need to find \( \theta \) such that it is in the third quadrant (where \( 180^\circ < \theta < 270^\circ \)). Let's solve for \( \theta \) from each equation:

• For \( 5\theta = 60^\circ \), \( \theta = 12^\circ \) (not in third quadrant).
• For \( 5\theta = 420^\circ \), \( \theta = \frac{420^\circ}{5} = 84^\circ \) (not in third quadrant).
• For \( 5\theta = 780^\circ \), \( \theta = \frac{780^\circ}{5} = 156^\circ \) (not in third quadrant).
• For \( 5\theta = 1140^\circ \), \( \theta = \frac{1140^\circ}{5} = 228^\circ \) (this is in the third quadrant as \( 180^\circ < 228^\circ < 270^\circ \)).
• For \( 5\theta = 1500^\circ \), \( \theta = \frac{1500^\circ}{5} = 300^\circ \) (not in third quadrant).

So the \( \theta \) in the third quadrant is \( 228^\circ \).

Answer:

\( r = 2 \), \( \theta = 228^\circ \)