# How do we use De Moivre's Thereom to simplify (2-2i)^8?

May 11, 2017

DeMoivre Theorem states that ${\left(a + b i\right)}^{n} = {\left(r \left(\cos \theta + i \sin \theta\right)\right)}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)$
Answer: $4096$

#### Explanation:

DeMoivre's Theorem allows us to compute the powers and roots of complex trigonometric expressions:
${\left(a + b i\right)}^{n} = {\left(r \left(\cos \theta + i \sin \theta\right)\right)}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)$

Original question: Use DeMoivre's Thereom to simplify ${\left(2 - 2 i\right)}^{8}$

${\left(2 - 2 i\right)}^{8}$

We can factor out a $2$ inside the base of the power:
$= {\left(2 \left(1 - i\right)\right)}^{8}$

Now, we convert $1 - i$ to trigonometric form:

Note: $1 - i = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

So:
=(2sqrt(2)(cos(pi/4)+isin(-pi/4))^8

Now we can apply DeMoivre's Theorem:
$= {\left(2 \sqrt{2}\right)}^{8} \left(\cos \left(- 8 \left(\frac{\pi}{4}\right)\right) + i \sin \left(- 8 \left(\frac{\pi}{4}\right)\right)\right)$
which we can simplify:
$= 4096 \left(\cos \left(- 2 \pi\right) - i \sin \left(- 2 \pi\right)\right)$
$= 4096 \left(1 - i \left(0\right)\right)$
$= 4096$

Therefore, our final answer is $4096$.