First divide 2008/4 to get 502(to make sure it's a multiple of 4). you're likely aware, but you can just take (i+1)^4, find out it's -4, and (i-1)^4, which is also -4. -4^502-(-4^502)=0.
sacreebluee 3 years ago
I love how these problems only take 30 seconds, but you extend them to 30 minutes.
A few methods:
1. It is easy to see that (1+i)^2=2i
so we have [(1+i)^2]^1004= [2i]^1004=2^2004
2. Multiply sqrt(2)/2 inside parenthesis and its reciprical outside. Understand that the reciprocal outside parenthesis = [2/sqrt(2)]^2008
This gives z={[2/sqrt(2)]^2008}{cis(pi/4)
Funny things: You don't talk about intervals of subtracting 2 pi, instead you draw sin graphs.
De Moivre's=Mobious?
mhisno1 4 years ago
Jeez.... (i meant 1004 above)
ps- I cant type math well... and the darn character limit
I wish I could delete the above comment :( You rock
First divide 2008/4 to get 502(to make sure it's a multiple of 4). you're likely aware, but you can just take (i+1)^4, find out it's -4, and (i-1)^4, which is also -4. -4^502-(-4^502)=0.
sacreebluee 3 years ago
I love how these problems only take 30 seconds, but you extend them to 30 minutes.
A few methods:
1. It is easy to see that (1+i)^2=2i
so we have [(1+i)^2]^1004= [2i]^1004=2^2004
2. Multiply sqrt(2)/2 inside parenthesis and its reciprical outside. Understand that the reciprocal outside parenthesis = [2/sqrt(2)]^2008
This gives z={[2/sqrt(2)]^2008}{cis(pi/4)
Funny things: You don't talk about intervals of subtracting 2 pi, instead you draw sin graphs.
De Moivre's=Mobious?
mhisno1 4 years ago
Jeez.... (i meant 1004 above)
ps- I cant type math well... and the darn character limit
mhisno1 4 years ago
I wish I could delete the above comment :( You rock
mhisno1 4 years ago