Complex Numbers Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium
Simplify i23i^{23}.

Solution

  1. 1
    Divide the exponent by 4: 23=4×5+323 = 4 \times 5 + 3, so the remainder is 3.
  2. 2
    Use the cycle: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1.
  3. 3
    Since the remainder is 3, i23=i3=−ii^{23} = i^3 = -i.

Answer

−i-i
The powers of ii repeat every four steps. To simplify ini^n, compute nmod  4n \mod 4 and look up the result in the cycle. The remainder 3 corresponds to i3=−ii^3 = -i.

About Complex Numbers

Numbers of the form a+bia + bi where a,ba, b are real and i=−1i = \sqrt{-1}; they extend the real numbers to solve x2=−1x^2 = -1.

Learn more about Complex Numbers →

More Complex Numbers Examples

Example 1 easy

Simplify [formula], [formula], and [formula].

Example 2 medium

Multiply [formula] and write the result in standard form [formula].

Example 3 easy

Add [formula].

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