Question

What are the fourth roots of −3+33√i ?

Enter your answer by filling in the boxes. Enter the roots in order of increasing angle measure in simplest form.

Answer 1

Explanation:

Just took the test.

1) 4sqrt6 cis(pie/6)

2) 4sqrt6 cis (2pie/3)

3) 4sqrt6 cis(7pie/6)

4) 4sqrt6 cis (5pie/3)

Answer 2

In order of increasing angle measure, the fourth roots of -3 + 3√3·i are presented as follows;

`\sqrt[4]{6} \cdot \left[cos\left({-(\pi)/(12) } \right) + i \cdot sin\left(-(\pi)/(12) } \right) \right]`

`\sqrt[4]{6} \cdot \left[cos\left({(5 \cdot \pi)/(12) } \right) + i \cdot sin\left((5 \cdot\pi)/(12) } \right) \right]`

`\sqrt[4]{6} \cdot \left[cos\left({(11 \cdot \pi)/(12) } \right) + i \cdot sin\left((11 \cdot\pi)/(12) } \right) \right]`

`\sqrt[4]{6} \cdot \left[cos\left({(17 \cdot \pi)/(12) } \right) + i \cdot sin\left((17 \cdot\pi)/(12) } \right) \right]`

Explanation:

The root of a complex number a + b·i is given as follows;

r = √(a² + b²)

θ = arctan(b/a)

The roots are;

`\sqrt[n]{r}`·[cos((θ + 2·k·π)/n) + i·sin((θ + 2·k·π)/n)]

Where;

k = 0, 1, 2,..., n -2, n - 1

For z = -3 + 3√3·i, we have;

r = √((-3)² + (3·√3)²) = 6

θ = arctan((3·√3)/(-3)) = -π/3 (-60°)

Therefore, we have;

`\sqrt[4]{-3 + 3 \cdot √(3) \cdot i \right)} = \sqrt[4]{6} \cdot \left[cos\left((-60 + 2\cdot k \cdot \pi)/(4) \right) + i \cdot sin\left((-60 + 2\cdot k \cdot \pi)/(4) \right) \right]`

When k = 0, the fourth root is presented as follows;

`\sqrt[4]{6} \cdot \left[cos\left((-(\pi)/(3) + 2\cdot 0 \cdot \pi)/(4) \right) + i \cdot sin\left((-(\pi)/(3) + 2\cdot 0 \cdot \pi)/(4) \right) \right] \\= \sqrt[4]{6} \cdot \left[cos\left({-(\pi)/(12) } \right) + i \cdot sin\left(-(\pi)/(12) } \right) \right]`

When k = 1 the fourth root is presented as follows;

`\sqrt[4]{6} \cdot \left[cos\left((-(\pi)/(3) + 2\cdot 1 \cdot \pi)/(4) \right) + i \cdot sin\left((-(\pi)/(3) + 2\cdot 1 \cdot \pi)/(4) \right) \right] \\= \sqrt[4]{6} \cdot \left[cos\left({(5 \cdot \pi)/(12) } \right) + i \cdot sin\left((5 \cdot\pi)/(12) } \right) \right]`

When k = 2, the fourth root is presented as follows;

`\sqrt[4]{6} \cdot \left[cos\left((-(\pi)/(3) + 2\cdot 2 \cdot \pi)/(4) \right) + i \cdot sin\left((-(\pi)/(3) + 2\cdot 2 \cdot \pi)/(4) \right) \right] \\= \sqrt[4]{6} \cdot \left[cos\left({(11 \cdot \pi)/(12) } \right) + i \cdot sin\left((11 \cdot\pi)/(12) } \right) \right]`

When k = 3, the fourth root is presented as follows;

`\sqrt[4]{6} \cdot \left[cos\left((-(\pi)/(3) + 2\cdot 3 \cdot \pi)/(4) \right) + i \cdot sin\left((-(\pi)/(3) + 2\cdot 3 \cdot \pi)/(4) \right) \right] \\= \sqrt[4]{6} \cdot \left[cos\left({(17 \cdot \pi)/(12) } \right) + i \cdot sin\left((17 \cdot\pi)/(12) } \right) \right]`