Math Is Fun Forum

cooljackiec

Member

Registered: 2012-12-13

Posts: 186

complex numbers

HELP!!!

Last edited by cooljackiec (2013-02-04 15:01:28)

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I see you have graph paper.
You must be plotting something

Harold

Guest

Re: complex numbers

You can use binomial theorem,or you can use moivre's formula(i may have spelled incorrectly)-

cooljackiec

Member

Registered: 2012-12-13

Posts: 186

Re: complex numbers

whenever i do binomial theorem, i keep getting the same thing, but its wrong

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I see you have graph paper.
You must be plotting something

zetafunc.

Guest

Re: complex numbers

Do you know how to apply De Moivre's theorem here?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: complex numbers

hi cooljackiec

Post the answer you are getting and we can see where you are going wrong.

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

noelevans

Member

Registered: 2012-07-20

Posts: 236

Re: complex numbers

Hi cooljackiec!

Three ways to get the result:
                                                                                 i60°                                 
1)  Convert z = 1+ i√3 into polar (exponential) form  2e       so the 6th power is
                           i60° 6     6  i360°         i0°
                      (2e      )  = 2 e        = 64e    = 64(cos0°+isin0°) = 64( 1 + 0 ) = 64.
        iθ
     re    =  (r,θ) are polar forms for complex numbers.
     The rcisθ = r(cosθ+isinθ) is really a "hybrid" in the sense that it is given in terms of
      r and θ, but when the trig functions are evaluated it gives the rectangular form x+iy.
      So rcisθ is handy for converting complex numbers from polar to rectangular form.

2)  First get z² = 2(-1+i√(3)).  Then square z² to get z^4 = -8(1+i√(3)).  Then multiply
     z^4 by z^2 to get the 64.

3)  Multiply (x - (1+i√3))*(x - (1-i√3)) = x^2-2x+4.  Then divide x^6 by x^2-2x+4 to
     get a remainder of 0*x+64.  Substitute z for x in 0*x+64 to get the result 64 using the
     Extended Remainder Theorem.

The Extended Remainder Theorem:  Given  P(x)/D(x) = Q(x) + R(x)/D(x).  For any x for which
                                                     D(x)=0, P(x)=R(x).

Simple proof:  Multiply by D(x) to obtain P(x) = Q(x)D(x) + R(x).  Then since we are using
only x's that make D(x)=0 this reduces to P(x)=R(x).

Dividing x^6 by x^2-2x+4 yields Q(x)=x^4+2x^3-8x-16 with R(z)=0*z+64  Since z=1+i√3 is
one of the two values that makes D(x)=0, P(z)=R(z)=0*z+64=64.  [Note (1-i√3)^6 = 64 also.]

I find 1) to be easiest and 2) to be the hardest.  The reason I find 3) easier than 2) is that
the division can be done quickly using the Generalization of Synthetic Division (GenSynD)
algorithm.  An article on how to do GenSynD can be found on the site that comes up when
googling "Math: The Original Four Letter Word."

Have a superduper day!

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Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

cooljackiec

Member

Registered: 2012-12-13

Posts: 186

Re: complex numbers

I got -26, I don't remember how, but I used binomial theorem. It is wrong I believe

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I see you have graph paper.
You must be plotting something

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: complex numbers

Hi;

Some good methods have been suggested in previous posts. You can do this in yet one more way, by direct computation,

from Pascals triangle:

put a = 1

separate the odd and even powers and plug in (√3)i for b. Even ones first,

then odd.

Add 64 + 0, answer is 64.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.