Find and write the answer in standard form.
I got this in the end but it was wrong
Please show step by step.
Any help would be greatly appreciated
$\displaystyle 2(\sqrt 3 + i)^7 = 2^8 \left( \frac {\sqrt 3}2 + \frac 12i \right)^7$
$\displaystyle = 2^8 \left( \cos \frac {\pi}6 + i \sin \frac {\pi}6 \right)^7$
$\displaystyle = 2^8 \left( e^{i \pi /6}\right)^7$
$\displaystyle = 2^8 e^{i 7 \pi / 6}$
$\displaystyle = 2^8 \left( \cos \frac {7 \pi}6 + i \sin \frac {7 \pi}6 \right)$
$\displaystyle = 2^8 \left( - \frac {\sqrt 3}2 - \frac 12i \right)$
$\displaystyle = - 2^7 \sqrt 3 - 2^7 i$
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?)
To expand this, you'll have to use binomial theorem
$\displaystyle (x+y)^7= x^7+ 7x^6y+ 21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7$
Let's expand $\displaystyle (\sqrt{3}+i)^7$
$\displaystyle (\sqrt{3})^7+7(\sqrt{3})^6i+21(\sqrt{3})^5i^2+35(\ sqrt{3})^4i^3+35(\sqrt{3})^3i^4+21(\sqrt{3})^2i^5+ 7(\sqrt{3})i^6+i^7$
Simplify
$\displaystyle 27\sqrt{3}+189i-189\sqrt{3}-315i+105\sqrt{3}+63i-7\sqrt{3}-i$
Simplify
$\displaystyle -64\sqrt{3}-64i$
Hence
$\displaystyle 2(\sqrt{3}+i)^7=-128(\sqrt{3}+i)$
Edit:
Jhevon was already here with an alternate method much more elegant that mine