Qualification > Math
Complex numbers explained
astarmathsandphysics:
For cube roots of unity find with m=1,2,3
nid404:
omega is the cube root of 1
I want derivation of how cube root of unity=(-1+/- root3 i)/2
don't know how to use symbols for root
Ghost Of Highbury:
got this from Dr.Math
The general form of the question is called the nth roots of unity.
In essence, the roots lie on the unit circle of the complex plane,
centred at the origin. These roots are precisely the vertices of the
regular n-gon starting with the first vertex at the real number 1 on
the Argand diagram.
Now, let me show you how you can find the cube roots of 1.
Suppose that z is a cube root of 1, i.e. z^3 = 1
It is easily seen (by taking the modulus on both sides) that
|z^3| = |1|
It follows: |z|^3 = 1
Since |z| is real, |z| has to be 1.
Thus, in the polar form, z may be expressed as cos A + i sin A where
A is the principal argument of z (i.e. -pi < A <= pi).
So, we have z^3 = 1 leading to (cos A + i sin A)^3 = 1.
Applying De Moivre's Theorem, we have
cos 3A + i sin 3A = 1
Comparing the real part of the left and right expressions (and,
respectively, the imaginary part),
cos 3A = 1 and sin 3A = 0
It follows that 3A = 0, 2pi, 4pi
Thus, A = 0, 2pi/3, 4pi/3
Hence, the three distinct complex cube roots of 1 are:
cos 0 + i sin 0 = 1
cos 2pi/3 + i sin 2pi/3 = -1/2 + i sqrt(3)/2
cos 4pi/3 + i sin 4pi/3 = -1/2 - i sqrt(3)/2
These three complex numbers represent the vertices of an equilateral
triangle circumscribed by the unit circle centred at 0, with the first
vertex at the real number 1.
I hope that this is clear.
astarmathsandphysics:
Thanks for that. The cache expired and I lost the post where I tried to explain.
nid404:
oooooooohhhhhhhhhhhhhhhh
merci beacoup :) je comprends....c'est facile actuellement.....j'ai maque cette lecture a IIT....joined late na
thanks dudee :)
no problem astar....ur notes were very helpful
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