How do you simplify i^15i15?

1 Answer
Feb 8, 2016

i^15 = -ii15=i

Explanation:

Remember that i^2 = -1i2=1.

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)aman=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=i4i4i4i3=111i3=i3=i2i=1i=i

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Also, I'd like to offer you a more general solution for i^nin, with nn being any positive integer.

Try to recognize the pattern:

i = ii=i
i^2 = -1i2=1
i^3 = i^2 * i = -1 * i = -ii3=i2i=1i=i
i^4 = i^3 * i = -i * i = -i^2 = 1i4=i3i=ii=i2=1
i^5 = i^4 * i = 1 * i = ii5=i4i=1i=i
i^6 = i^4 * i^2 = -1i6=i4i2=1
...

So, basically, the power of ii is always ii, -11, -ii, 11, and then repeat.

Thus, to compute i^nin, there are four possibilites:

  • if nn can be divided by 44, then i^n = 1in=1
  • if nn can be divided by 22 (but not by 44), then i^n = -1in=1
  • if nn is an odd number but n-1n1 can be divided by 44, then i^n = iin=i
  • if nn is an odd number but n+1n+1 can be divided by 44, then i^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.