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666 bro

Member

From: Flatland

Registered: 2019-04-26

Posts: 706

Higher algebra

Now I'm reading chapter no. 5:complex numbers from  higher algebra by barnard & child in that i didn't understand de Moivre's theorem and extension what does it states that and how does it works?

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"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Higher algebra

hi 666 bro,

It's a really useful result in complex number theory.  But first, have you met Euler's formula for complex numbers?  If not, have a look here

https://www.mathsisfun.com/algebra/eulers-formula.html

MAJOR EDIT: I've thought about this some more.  Do you know about modulus argument form for complex numbers?  And also the simple way to multiply two numbers using mod/arg form ?  If so, what you ask is easy to prove.  If not I'm happy to teach you.

I'll await your next post.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

666 bro

Member

From: Flatland

Registered: 2019-04-26

Posts: 706

Re: Higher algebra

No , but I had an idea about argument of an complex number.

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"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Higher algebra

hi 666 bro,

Thanks for the reply.  It's always useful for a teacher to know what his student knows already.

Let a + ib be a complex number at point P.  O is the origin and Q the foot of the perpendicular from P to the x axis.  Let angle POQ be alpha. Let OP = r.

alpha is called the 'argument' of the complex number, and r its modulus.

And a second complex number:

Now multiply them together in modulus argument form:

Using the angle sum identities  (https://www.mathsisfun.com/algebra/trig … ities.html

This is a remarkable result and very useful in complex number theory.  The modulus of the product of two numbers is the product of the moduli (not surprising) but also the argument of the product is the sum of the arguments.

So, for example, if one complex number has argument 45 degrees and another has argument 30 degrees, then the argument of their product is 75 degrees.

Now consider points on a unit circle (radius=1) centred on the origin.  As r = 1 all such numbers can be written as

And if we multiply that number by itself we get

just by adding the arguments.  And if we carry on multiplying the number by itself over and over we get

This is De Moivres' theorem.

There is a result in complex number theory that is so important it is known as The Fundamental Theorem of Algebra

It states that any polynomial in powers of x, will have a solution in complex numbers.  If you say it is x = C and divide by (x-C) you get a polynomial with degree one smaller and it must, by the theorem, also have a solution.  And so on.

So a polynomial of degree n, will have n roots although they may not all be different.

We know the square roots of 1 are +1 and -1.  That fits with the theorem because x^2 = 1 has order 2 so there should be 2 roots.

We know that the cube root of 1 is 1, but, by the theorem, it should have two more.  These are known as the complex cube roots of 1.  Using De Moivre, we can easily find them.

Let's say one of them is

for some theta.

As a complex number 1 is 1 + 0i so we can write

So 3theta = 360 so theta = 120.  So a cube root of 1 is

I think I'll stop there for now but here's some homework.

(1) Prove that cube root works by multiplying it by itself twice more to get, hopefully, 1.

(2) See if you can work out the third cube root of 1.

There's more if you want it

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

666 bro

Member

From: Flatland

Registered: 2019-04-26

Posts: 706

Re: Higher algebra

So, would you tell me how could I learn a theorem and understand it intuitively for example:de moivre theorem ?

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"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Higher algebra

I cannot answer this as I've always had difficulty learning formulas.  In exams, I usually re-worked the formulas from scratch.  I think understanding is the key.  If you have truly understood where a theorem comes from then you shouldn't have a problem.  I set some 'homework' for these reasons: (1) so you could practise what you have learnt and (2) so I can see if you have truly understood what I've tried to teach you.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

Website

Re: Higher algebra

3blue1brown has some nice videos on this topic: https://www.youtube.com/watch?v=v0YEaeIClKY and https://www.youtube.com/watch?v=mvmuCPvRoWQ

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.