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

Elissa89

Member
Oct 19, 2017
52
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
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 $1$?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
378
You are evaluating sine and cosine incorrectly!

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