How do you simplify i^15i15?
1 Answer
Explanation:
Remember that
Thus,
i^4 = (i^2)^2 = (-1)^2 = 1i4=(i2)2=(−1)2=1
Also, remember the power rule
a^m * a^n = a^(m+n)am⋅an=am+n
Thus, you have
i^15 = i^(4 + 4 + 4 + 3) = i^4 * i^4 * i^4 * i^3 = 1 * 1 * 1 * i^3 = i^3 = i^2 * i = -1 * i = -ii15=i4+4+4+3=i4⋅i4⋅i4⋅i3=1⋅1⋅1⋅i3=i3=i2⋅i=−1⋅i=−i
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Also, I'd like to offer you a more general solution for
Try to recognize the pattern:
i = ii=i
i^2 = -1i2=−1
i^3 = i^2 * i = -1 * i = -ii3=i2⋅i=−1⋅i=−i
i^4 = i^3 * i = -i * i = -i^2 = 1i4=i3⋅i=−i⋅i=−i2=1
i^5 = i^4 * i = 1 * i = ii5=i4⋅i=1⋅i=i
i^6 = i^4 * i^2 = -1i6=i4⋅i2=−1
...
So, basically, the power of
Thus, to compute
- if
nn can be divided by44 , theni^n = 1in=1 - if
nn can be divided by22 (but not by44 ), theni^n = -1in=−1 - if
nn is an odd number butn-1n−1 can be divided by44 , theni^n = iin=i - if
nn is an odd number butn+1n+1 can be divided by44 , theni^n = -iin=−i
Described in a more formal way,
i^n = {(1, " " n= 4k),(i, " " n = 4k + 1),(-1, " " n = 4k + 2),(-i, " " n= 4k + 3) :} for
k in NN_0 .