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2 answers:
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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;
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°))
Answer:
Reflect over the y-axis
Step-by-step explanation:
When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is transformed into its opposite (its sign is changed).
Notice that B is 5 horizontal units to the right of the y-axis, and B' is 5 horizontal units to the left of the y-axis.
The reflection of the point (x,y) across
the y-axis is the point (-x,y).
