4.1 The Need for Complex Numbers

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Solved Examples

  1. Divide the exponent by 4: 135=4×33+3135 = 4 \times 33 + 3.
  2. The remainder is 3.
  3. Therefore, i135=i4(33)+3=(i4)33⋅i3=133⋅(−i)=−ii^{135} = i^{4(33) + 3} = (i^4)^{33} \cdot i^3 = 1^{33} \cdot (-i) = -i.

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