# DeMoivre’s Theorem

• June 15th 2009, 07:01 PM
Daniels4691
DeMoivre’s Theorem
I got this in the end but it was wrong http://classroom.advancedacademics.c...s/image021.gif

Please show step by step.
Any help would be greatly appreciated
• June 15th 2009, 07:41 PM
Jhevon
Quote:

Originally Posted by Daniels4691
I got this in the end but it was wrong http://classroom.advancedacademics.c...s/image021.gif

Please show step by step.
Any help would be greatly appreciated

$2(\sqrt 3 + i)^7 = 2^8 \left( \frac {\sqrt 3}2 + \frac 12i \right)^7$

$= 2^8 \left( \cos \frac {\pi}6 + i \sin \frac {\pi}6 \right)^7$

$= 2^8 \left( e^{i \pi /6}\right)^7$

$= 2^8 e^{i 7 \pi / 6}$

$= 2^8 \left( \cos \frac {7 \pi}6 + i \sin \frac {7 \pi}6 \right)$

$= 2^8 \left( - \frac {\sqrt 3}2 - \frac 12i \right)$

$= - 2^7 \sqrt 3 - 2^7 i$

do you see where we applied the theorem? or rather, where we would have applied it? re-writing in terms of e did away with the need to apply the theorem explicitly (what lines could we skip by applying the theorem?)
• June 15th 2009, 07:50 PM
I-Think
Binomial theorem
To expand this, you'll have to use binomial theorem
$(x+y)^7= x^7+ 7x^6y+ 21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7$

Let's expand $(\sqrt{3}+i)^7$
$(\sqrt{3})^7+7(\sqrt{3})^6i+21(\sqrt{3})^5i^2+35(\ sqrt{3})^4i^3+35(\sqrt{3})^3i^4+21(\sqrt{3})^2i^5+ 7(\sqrt{3})i^6+i^7$

Simplify
$27\sqrt{3}+189i-189\sqrt{3}-315i+105\sqrt{3}+63i-7\sqrt{3}-i$

Simplify
$-64\sqrt{3}-64i$

Hence
$2(\sqrt{3}+i)^7=-128(\sqrt{3}+i)$

Edit:
Jhevon was already here with an alternate method much more elegant that mine