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bobbym
2013-02-06 12:32:10

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.

cooljackiec
2013-02-06 11:56:16

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

noelevans
2013-02-06 10:14:21

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!

bob bundy
2013-02-06 04:09:56

hi cooljackiec

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

Bob

zetafunc.
2013-02-06 02:18:36

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

cooljackiec
2013-02-06 02:15:58

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

Harold
2013-02-05 14:47:31

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

cooljackiec
2013-02-05 14:01:08

HELP!!!