Question

Write the square roots of 49 i in ascending order of the angle.

Answer 1

Answer:

The square roots of 49·i in ascending order are;

1) -7·(cos(45°) + i·sin(45°))

2) 7·(cos(45°) + i·sin(45°))

Step-by-step explanation:

The square root of complex numbers 49·i is found as follows;

x + y·i = r·(cosθ + i·sinθ)

Where;

r = √(x² + y²)

θ = arctan(y/x)

Therefore;

49·i = 0 + 49·i

Therefore, we have;

r = √(0² + 49²) = 49

θ = arctan(49/0) → 90°

Therefore, we have;

49·i = 49·(cos(90°) + i·sin(90°)

By De Moivre's formula, we have;

$r \cdot (cos(\theta) + i \cdot sin(\theta) )^{\dfrac{1}{2}} = \pm \sqrt{r} \cdot \left (cos\left (\dfrac{\theta}{2} \right ) + i \cdot sin\left (\dfrac{\theta}{2} \right ) \right )$

Therefore;

√(49·i) = √(49·(cos(90°) + i·sin(90°)) = ± √49·(cos(90°/2) + i·sin(90°/2))

∴ √(49·i) = ± √49·(cos(90°/2) + i·sin(90°/2)) = ± 7·(cos(45°) + i·sin(45°))

√(49·i) = ± 7·(cos(45°) + i·sin(45°))

The square roots of 49·i in ascending order are;

√(49·i) = - 7·(cos(45°) + i·sin(45°))  and  7·(cos(45°) + i·sin(45°))