Question

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

Answer 1

Answer:

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

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

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

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

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

Step-by-step 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;

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

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

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

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

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

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

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

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

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