MHB DeMoivre's Theorem express (sqrt(2)/2 + sqrt(2)/2 i)^8 in a+bi form

[Elissa89]

So the question is:

express (sqrt(2)/2 + sqrt(2)/2 i)^8 in a+bi form

I know r=1 and tangent=pi/4

Using the theorem i get 1(cos (2pi) +i*sin (2pi)) which becomes 1(1*i)=1*i however WebAssign says this is incorrect. I've also tried "0+1i" and just "i"

What am I doing wrong?

 

[I like Serena]

So the question is:

express (sqrt(2)/2 + sqrt(2)/2 i)^8 in a+bi form

I know r=1 and tangent=pi/4

Using the theorem i get 1(cos (2pi) +i*sin (2pi)) which becomes 1(1*i)=1*i however WebAssign says this is incorrect. I've also tried "0+1i" and just "i"

What am I doing wrong?

Isn't $\sin(2\pi)$ equal to $0$ instead of $1$?

 

[HOI]

You are evaluating sine and cosine incorrectly!

$cos(2\pi)+ i sin(2\pi)= 1+ i(0)= 1$.