Question

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

Answer 1

DeMoivre Theorem states that `(a+bi)^n=(r(costheta+isintheta))^n=r^n(cos(ntheta)+isin(ntheta))`
Answer: `4096`

Explanation:

DeMoivre's Theorem allows us to compute the powers and roots of complex trigonometric expressions:
`(a+bi)^n=(r(costheta+isintheta))^n=r^n(cos(ntheta)+isin(ntheta))`

Original question: Use DeMoivre's Thereom to simplify `(2-2i)^8`

`(2-2i)^8`

We can factor out a `2` inside the base of the power:
`=(2(1-i))^8`

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

Note: `1-i=sqrt(2)(cos(pi/4)+isin(-pi/4))`

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

Now we can apply DeMoivre's Theorem:
`=(2sqrt(2))^8(cos(-8(pi/4))+isin(-8(pi/4)))`
which we can simplify:
`=4096(cos(-2pi)-isin(-2pi))`
`=4096(1-i(0))`
`=4096`

Therefore, our final answer is `4096`.