Math Help - DeMoivre’s Theorem

1. DeMoivre’s Theorem

Find and write the answer in standard form.
I got this in the end but it was wrong

Any help would be greatly appreciated

2. Originally Posted by Daniels4691
Find and write the answer in standard form.
I got this in the end but it was wrong

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?)

3. 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