SOLUTION: on page 280 of the algebra 2 book problem 5 reads i^40 i cant figure out how to solve it...

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Question 361631: on page 280 of the algebra 2 book problem 5 reads i^40 i cant figure out how to solve it...
Found 3 solutions by shorty4aboo, jrfrunner, Alan3354:
Answer by shorty4aboo(8) About Me  (Show Source):
You can put this solution on YOUR website!
i%5E40 is a problem in its self i%5E40=0 the reason why i say this is because my teacher taught us a method to use for degree of i...here it is:
ex: Evaluate i%5E533
533%2F4=133%2B1%2F4 533 divided by 4 = 133 R(1)
Now this tells you That the Remainder is 1, Use that with this chart:
i%5E0=0
i=sqrt%28i%29=i
i%5E2=-1
i%5E3=i
i%5E4=-1
As you can see ; Odd degree numbers =i and even degree numbers= -1.

NOW HERE IS YOUR PROBLEM WORKED OUT:
i%5E40
40%2F10=4 See No remainder so you would have degree 0, then plug it into chart and the answer is 0.

Hope this has helped you out.

Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
i%5En cycles in factors of 4
--
i%5E0=1
i%5E1=i=sqrt%28-1%29
i%5E2=-1
i%5E3=i%5E2%2Ai=-1%2Ai=-i
i%5E4=i%5E2%2Ai%5E2=-1%2A-1=1=i%5E0
i%5E5=i%5E4%2Ai=i=i%5E1
and as you can see values recycle based on xmod(4) (ie the remainder of dividing by 4)
So since you want to know i%5E40 and since 40mod(4)=0 then
i%5E40=i%5E0=1

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
i^40
-----
i+=+sqrt%28-1%29
i%5E2+=+-1
i%5E3+=+i%5E2%2Ai+=+-i
i%5E4+=+1
----
It's cyclic
i, -1, -i, 1 etc.
---
i%5E40+=+%28i%5E4%29%5E10+=+1%5E10+=+1