# How do you simplify (2+2i)^6?

Mar 23, 2016

${\left(2 + 2 i\right)}^{6} = - 512 i$

#### Explanation:

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Using polar form

$2 + 2 i = 2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

So:

${\left(2 + 2 i\right)}^{6} = {\left(2 \sqrt{2}\right)}^{6} {\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)}^{6}$

$= {\left({2}^{\frac{3}{2}}\right)}^{6} \left(\cos 6 \left(\frac{\pi}{4}\right) + i \sin 6 \left(\frac{\pi}{4}\right)\right)$

$= {2}^{9} \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

$= {2}^{9} \cdot \left(- i\right)$

$= - 512 i$

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Using rectangular form

${\left(2 + 2 i\right)}^{2} = {\left(2 \left(1 + i\right)\right)}^{2} = 4 \left(1 + i\right) \left(1 + i\right) = 4 \left(1 + 2 i + {i}^{2}\right) = 4 \left(2 i\right) = 8 i$

So:

${\left(2 + 2 i\right)}^{6} = {\left({\left(2 + 2 i\right)}^{2}\right)}^{3} = {\left(8 i\right)}^{3} = {8}^{3} {i}^{3} = - 512 i$