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Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Power of i

4.3.5

Power of i

The Pattern of Powers of ii

The imaginary unit ii is defined by i2=−1i^2 = -1. From this single fact, every higher power of ii can be reduced to one of just four values: 11, ii, −1-1, or −i-i. The pattern repeats in a cycle of length 4.

Start with the smallest powers:

  • i1=ii^1 = i
  • i2=−1i^2 = -1
  • i3=i2⋅i=(−1)⋅i=−ii^3 = i^2 \cdot i = (-1) \cdot i = -i
  • i4=i2⋅i2=(−1)(−1)=1i^4 = i^2 \cdot i^2 = (-1)(-1) = 1

Once you reach i4=1i^4 = 1, the cycle resets. Multiply i4i^4 by ii to get i5i^5, and so on:

  • i5=i4⋅i=1⋅i=ii^5 = i^4 \cdot i = 1 \cdot i = i
  • i6=i4⋅i2=1⋅(−1)=−1i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1
  • i7=i4⋅i3=1⋅(−i)=−ii^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i
  • i8=i4⋅i4=1⋅1=1i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1

The pattern is clear: every fourth power brings you back to 11.

Negative Powers of ii

The same cyclic behaviour extends to negative exponents. Recall that i−1i^{-1} is the multiplicative inverse of ii:

i−1=1ii^{-1} = \frac{1}{i}

To write this in the standard a+bia+bi form, multiply numerator and denominator by ii:

i−1=1i×ii=ii2=i−1=−ii^{-1} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i

Now build the rest:

  • i−2=1i2=1−1=−1i^{-2} = \frac{1}{i^2} = \frac{1}{-1} = -1
  • i−3=1i3=1−ii^{-3} = \frac{1}{i^3} = \frac{1}{-i}

Rationalise i−3i^{-3} by multiplying numerator and denominator by ii:

i−3=1−i×ii=i−i2=i−(−1)=i1=ii^{-3} = \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i

  • i−4=1i4=11=1i^{-4} = \frac{1}{i^4} = \frac{1}{1} = 1

Notice that i−1=−ii^{-1} = -i, i−2=−1i^{-2} = -1, i−3=ii^{-3} = i, i−4=1i^{-4} = 1 — the same four values appear, just in a different order.

Watch out

A common mistake is to think i−1=ii^{-1} = i. It does not. The inverse of ii is −i-i, because i⋅(−i)=−i2=−(−1)=1i \cdot (-i) = -i^2 = -(-1) = 1.

The General Formula for Any Integer kk

The pattern for any integer kk (positive, negative, or zero) is captured by four cases based on the remainder when the exponent is divided by 4.

i4k=1,i4k+1=i,i4k+2=−1,i4k+3=−ii^{4k} = 1, \quad i^{4k+1} = i, \quad i^{4k+2} = -1, \quad i^{4k+3} = -i

Here kk is any integer. The proof follows directly from the cycle:

  • i4k=(i4)k=1k=1i^{4k} = (i^4)^k = 1^k = 1 …