July 22, 2024
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R2 – Roots – Advanced

This post covers the use or De Moivre’s formula as an advanced method of calculating roots in R2. Some examples are provided.

Formula

The roots of any Opposite Value can be calculated using the formula below which is based on De Moivre’s formula. This formula relies on the fact that the root of an Opposite Value, starting at an Opposite Value on the x-axis with the same radius, is the angle between the Opposite value and x-axis Opposite Value divided by the degree.

Root = nR((cos(θ/n + 2kπ^/n)^ + sin(θ/n + 2kπ^/n)i^)

Where θ is the counterclockwise angle from the x-axis and k is all values from 0 to n-1 to reflect that a degree has n solutions. For example there are 2 solutions for square  roots.

Example 1:    2√(1v):

The point located at 1v or (1v, 0) has a radius R = √|1v2| = 1 => √R = 1. Remember the Radius is a counter.

The ATAN2 function is used to determine the angle of the point (1v, 0) in radians. This is the angle θ between (1^, 0) and (1v, 0). This function takes into account the 4 different quadrants in the circle.

ATAN2(1v) = 3.14^ – note an arc on the unit circle of 3.14 radians distance from the point (1^, 0). i.e. π^ radians. The arc is the angle θ between (1^, 0) and (1v, 0).

  • For k= 0, 2√(1v) :
    • 1*((cos(π^/2 + 2*0*π^/2)^ + sin(π^/2 +2*0*π^/2)i^)
      • = cos(π^/2)^+ sin(π^/2)i^
      • = 0 + 1*i^ = i^

  • k= 1:
    • 1*((cos(π^/2 + 2*1*π^/2)^ + sin(π^/2 +2*1*π^/2)i^)
      • = (cos(3π^/2)^+ sin(3π^/2)i^
      • = 0 +1*i^ = iv

Example 2:    3√(1v):

The point located at 1v or (1v, 0) has a radius of R = √|1v2| = 1 => √R = 1.

ATAN2(1v) = 3.14^ = π^

  • k= 0:
    • 1*((cos(π^/3 + 2*0*π^/3)^ + sin(π^/3 +2*0*π/3)i^)
      • = cos(π^/3)^+ sin(π^/3)i^
      • = 0.5^ +0.866^*i^

  • k= 1:
    • 1*(cos(π^/3 + 2*1*π/3)^ + sin(π^/3 +2*1*π/3)i^)
      • = cos(π^)^+ sin(π^)i^
      • = 1v + 0*i^ = 1v

  • k= 2:
    • 1*(cos(π^/3 + 2*2*π/3)^ + sin(π^/3 +2*2*π/3)i^)
      • = cos(5π^/3 )^+ sin(5π^/3 )i^
      • = 0.5^ + 0866v*i^ = 0.5^ + 0.866iv

Example 3:   √(7v   + 24iv):  

The point located at (7v + 24iv) has a radius of R = √(72 + 242) = √(625)   = 25  => √R = 5

ATAN2(7v + 24iv) = 1.855v

  • k= 0:
    • 5(cos(1.855v/2 + 2*0*3.14^/2)^ + sin(1.855v/2 +2*0*3.14^/2)i^)
      • = 5(cos(0.9275v)^ + sin(0.9275v)i^
      • = 5(0.6^ + 0.8iv) = 3^ + 4iv

  • k= 1:
    • 5(cos(1.855v/2 + 2*1*3.14^/2) + sin(1.855v/2 +2*1*3.14^/2)
      • = 5(cos(2.2125^)^+ sin(2.2125^)i^
      • = 5(0.6v +0.8i^) = 3v + 4i^

Example 4:   3√(7v   + 24iv):

The point located at (7v + 24iv) has a radius of R = √(72 + 242) = √(625)   = 25  => 3R = 2.924

ATAN2(7v   + 24iv) = 1.855v

  • k= 0:
    • 2.924(cos(1.855v/3 + 2*0*3.14^/3)^ + sin(1.855v/3 +2*0*3.14^/3)i^)
    • = 2.924(cos(0.618v)^+ sin(0.618v)i^
    • = 2.924(0.815^ +0.579iv) = 2.383^ + 1.694iv

  • k= 1:
    • 2.924(cos(1.855v/3 + 2*1*3.14^/3)^ + sin(1.855v/3 +2*1*3.14^/3)i^)
      • =2.924(cos(1.475^)^+ sin(1.475^)i^
      • =2.924(0.0957^ +0.995i^) = 0.276^ + 2.91i^

  • k= 2:
    • 2.924(cos(1.855v/3 + 2*2*3.14^/3)^ + sin(1.855v/3 +2*2*3.14^/3)i^)
      • = 2.924 (cos(3.57^)^+ sin(3.57^)i^
      • = 2.924 (0.909v +0.416iv) = 2.659v + 1.216iv

Conclusion

Try these examples of advanced R2 roots with the help of our online calculator.

Next: Exponentiation

Previous: Expressions

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