De Moivre's Theorem?
Help with this proof please. :)
- micatkieLv 75 months agoFavorite Answer
The answer is as follows:
- SlowfingerLv 65 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
Since the angle between any two consecutive complex roots is 2π/n , they are evenly spaced around the circle of radius ⁿ√r
- VamanLv 75 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.