De Moivre's Theorem (1 Viewer)

Lukybear · Dec 9, 2009

General question in this form.

w is a non real cube root of 1.
Show that (1+w)(1+2w)(1+3w)(1+5w) = 21.

What is your process of solving this kind of quesitons? Since there are so many answers, i.e. so many types of subsitutions to make?

addikaye03 · Dec 9, 2009

Lukybear said:
General question in this form.

w is a non real cube root of 1.
Show that (1+w)(1+2w)(1+3w)(1+5w) = 21.

What is your process of solving this kind of quesitons? Since there are so many answers, i.e. so many types of subsitutions to make?

(1+w)(1+2w)(1+3w)(1+5w)

Since 1+w+w^2=0 and w^3=1

=(1+3w+2w^2)(1+8w+15w^2)

=1(1+3w+2w^2)+8w(1+3w+2w^2)+15w^2(1+3w+2w^2)

=1+3w+2w^2+8w+24w^2+16w^3+15w^2+45w^3+30w^4

=1+11w+41w^2+61w^3+30w^4

=(1+w+w^2)+10w+40w^2+61w^3+30w^4

=0+10w+40w^2+61+30w

=40w+40^2+61

=40(1+w+w^2)+21

therefore

=21

I dk of any substitution

waller · Dec 9, 2009

What do you mean substitutions?

I just did it slightly differently to addikaye but still worked out the same.

Lukybear · Dec 9, 2009

I mean, how do you know to times say (1+w) and (1+2w) where say sub w=1 or smthing? or w= -1-w^2??

cyl123 · Dec 9, 2009

w^3=1
==> (w-1)(w^2+w+1)=0
But w is not real so w=/=1 so w^2+w+1=0 and hence w=-w^2-1

Cazic · Dec 10, 2009

Just a different answer.

With the facts:

$\begin{align*} \omega^3 &= 1,\\ \omega^2 &= \overline{\omega} & \text{(just multiply the LHS of the first equation by } \overline{\omega} \text{ to see this)}, \\ \text{and} \\ \omega + \overline{\omega} &= -1 & (\text{since } \textstyle \cos\left(\frac{2\pi}{3}\right) = \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2})\end{align*}$

in mind, we now quickly count how many of each power of $\omega$ is in the expansion of the brackets like so:

Choosing to multiply all four 1's in the brackets gives us: $1$
Choosing to multiply three of the 1's and one of the $\omega$ 's gives us: $\omega + 2\omega + 3\omega + 5\omega = 11\omega$
Choosing to multiply two of the 1's and two of the $\omega$ 's gives us: $\omega \times 2\omega + \omega \times 3\omega + \omega \times 5\omega + 2\omega \times 3\omega + 2\omega \times 5\omega + 3\omega \times 5\omega = 41\omega^2$
Choosing to multiply one of the 1's and three of the $\omega$ 's gives us: $\omega \times 2\omega \times 3\omega + \omega \times 2\omega \times 5\omega + \omega \times 3\omega \times 5\omega + 2\omega \times 3\omega \times 5\omega = 61\omega^3$
Choosing to multiply none of the 1's and four of the $\omega$ 's gives us: $\omega \times 2\omega \times 3\omega \times 5\omega = 30\omega^4$

Thus,
$\begin{align*}(1+\omega)(1+2\omega)(1+3\omega)(1+5\omega) &= 1 + 11\omega + 41\omega^2 + 61\omega^3 + 30\omega^4 \\ &= 1 + 11\omega + 41\overline{\omega} + 61 + 30\omega \\ &= 62 + 41(\omega + \overline{\omega}) \\ &= 62 - 41 \\ &= 21. \end{align*}$

Trebla · Dec 10, 2009

Without expanding the whole thing into a quartic at first and using the fact that w² + w + 1 = 0 and w³ = 1 (which are easy to obtain) we can simplify just by expanding quadratic factors only (though it probably isn't much quicker just less horrible algebra):

(1 + w)(1 + 2w)(1 + 3w)(1 + 5w)
= (2w² + 3w + 1)(15w² + 8w + 1)
= (w² + 2w + w² + w + 1)(14w² + 7w + w² + w + 1)
= (w² + 2w)(14w² + 7w)
= 7w²(w + 2)(2w + 1)
= 7w²(2w² + 5w + 2)
= 7w²(3w + 2[w² + w + 1])
= 21w³
= 21

De Moivre's Theorem (1 Viewer)

Lukybear

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addikaye03

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