Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 months ago

De Moivre's Theorem?

Help with this proof please. :)

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3 Answers

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  • 5 months ago
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    The answer is as follows:

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  • 5 months ago

    n-th root of r(cos θ + i sin θ) is complex number "z" such that

    zⁿ = r(cos θ + i sin θ) 

    z = ⁿ√(r (cos θ + i sin θ)) = ⁿ√r * (cos θ + i sin θ)^(1/n)

    Every complex number has n different n-th roots in the complex plane.

    Also, observe that the absolute value (magnitude) ⁿ√r of all the n-th roots is the n-th root of the absolute value "r" of the original number.

    At this point, we already may notice that n-th roots (there are "n" of them) are all located on a complex circle with radius equal to absolute value ⁿ√r.

    In a complex plane, a complex number on the unit circle is represented as

    (cos φ + i sin φ)

    where φ is the angle with respect to the real axis

    These numbers all have absolute value

    cos^2 φ + sin^2 φ = 1

    which corresponds to radius of unit circle.

    Complex numbers with absolute value (magnitude) "m" are located on a complex circle with radius "m" and are written in the form

    m (cos φ + i sin φ)

    In our case m = ⁿ√r

    Let's return to

    z = ⁿ√r * (cos θ + i sin θ)^(1/n)

    We need to stop here.

    You may already know that, by De Moivre's theorem

    (cos θ + i sin θ)ⁿ = cos (nθ) + i sin(nθ)

    This applies only for integer powers n. For non-integer powers, de Moivre's formula in this basic form gives inconsistent results. For example, if n=1/2

    for θ=0, cos (0/2)+i sin(0/2) = cos (0)+i sin(0) = 1 + i 0 = 1

    for θ=2π, cos (2π/2)+i sin(2π/2) = cos (π)+i sin(π) = -1 + i 0 = -1

    that is, the result is multiple-valued.

    To overcome this limitation so that we can use non-integer power like 1/n, we recall the periodicity of sine and cosine to write

    cos θ = cos (θ+2kπ)

    sin θ = sin (θ+2kπ)

    Where k is an integer

    Our n-th root becomes

    z = ⁿ√r * (cos ((θ+2kπ)/n) + i sin ((θ+2kπ)/n))

    The values of k=0,1,2,…,(n−1) yield distinct values of argument

    For fixed n, the argument

    φ = (θ+2kπ)/n

    have values

    θ/n

    θ/n+2π/n

    θ/n+4π/n

    θ/n+6π/n 

    ......

    θ/n+2(n-1)kπ/n

    Since the angle between any two consecutive complex roots is 2π/n , they are evenly spaced around the circle of radius ⁿ√r 

  • Vaman
    Lv 7
    5 months ago

    You can write

    r cos a+ i r sin a= r exp ( i a +2 m pi). m goes from 1 to infinity.

    (r cos a+ i r sin a)^(1/n)= r ^(1/n) exp ( i a/n +2 m/n pi).

    Draw a circle of radius 1 unit. Select a. Then n the root will be a/n. Now draw radial lines at angle 2m/n pi for different m values adding a/n.

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