Author Topic: Complex numbers explained  (Read 1528 times)

Offline astarmathsandphysics

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Complex numbers explained
« on: October 03, 2009, 12:35:46 am »

nid404

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Re: Complex numbers explained
« Reply #1 on: October 03, 2009, 05:20:26 pm »
Thanks a lot astar :)

Offline astarmathsandphysics

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Re: Complex numbers explained
« Reply #2 on: October 03, 2009, 08:45:46 pm »
I will also do something on sums of combinations of roots.

nid404

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Re: Complex numbers explained
« Reply #3 on: October 04, 2009, 07:41:34 am »
thank you sir...u r sooo helpful

I actually need some help on cube roots of unity and proofs of omega....

Offline astarmathsandphysics

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Re: Complex numbers explained
« Reply #4 on: October 04, 2009, 10:01:09 am »
What is omega?
« Last Edit: October 04, 2009, 10:03:05 am by astarmathsandphysics »

Offline astarmathsandphysics

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Re: Complex numbers explained
« Reply #5 on: October 04, 2009, 10:04:12 am »
For cube roots of unity find e^(\frac {2 \pi mi}{3}) with m=1,2,3

nid404

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Re: Complex numbers explained
« Reply #6 on: October 04, 2009, 10:35:08 am »
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

Offline Ghost Of Highbury

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Re: Complex numbers explained
« Reply #7 on: October 04, 2009, 10:38:26 am »
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.
divine intervention!

Offline astarmathsandphysics

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Re: Complex numbers explained
« Reply #8 on: October 04, 2009, 10:43:09 am »
Thanks for that. The cache expired and I lost the post where I tried to explain.

nid404

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Re: Complex numbers explained
« Reply #9 on: October 04, 2009, 10:45:04 am »
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

Offline Ghost Of Highbury

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Re: Complex numbers explained
« Reply #10 on: October 04, 2009, 10:46:05 am »
D'accord.
divine intervention!