Simplify i^99

Problem

i99

Solution

1. Identify the pattern of the powers of the imaginary unit i

2. Recall that i1=i i2=−1 i3=−i and i4=1

3. Divide the exponent by 4 to find the remainder, since the values repeat every four powers.

99÷4=24* remainder *3

4. Rewrite the expression using the property of exponents.

i99=i(4*(24)+3)

5. Simplify the expression by substituting the known values.

i99=(i4)24⋅i3

i99=(1)24⋅(−i)

i99=1⋅(−i)

Final Answer

i99=−i

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