Question 28839: Simplify:
First Question:
i^2522
Second Question:
16/2i
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Simplify:
First Question:
i^2522
Second Question:
16/2i
First Question:
i^2522
2522 = 630X4 + 2
i^2522 = (i)^ (630X4 + 2)
=[(i)^(630X4)] X[(i)^2 ] [using formula (a)^(mn+k) = (a)^(mn) X(a)^k]
=[(i^4)^(630)]X(-1) [using (a)^(mn) = [(a)^n]^m ]
=[(1)^630]X(-1)
=1X(-1)
= -1
Note: 2522 on division by 4 gives remainder 2
Why should we divide by 4?
Because i^4= 1 [ i^1= i, i^2=-1,i^3 = (-1)Xi = -i and i^4 = (i^2)^2 = (-1) ^2 = +1]
And (1)^(any integer) = 1
Second Question:
(16/2i) =[(16/2i)]X[(i)/(i)] (multiplying nr and dr by (i) and canceling 2]
=(8Xi)/(iXi)
=8i/(i^2)
=8i/(-1)
= -8i
Note: Why should we multiply nr and dr by (i)?
Because i is complex and i itself is the real- izing factor
that is, i is the factor which when multiplied by the complex number i gives the real number (i^2) = -1
|
|
|
