Simplify i^27

Problem

i^27

Solution

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

i^1=i

i^2=−1

i^3=−i

i^4=1

2. Divide the exponent by 4 to find the remainder, since the powers of i repeat every four terms.

27÷4=6* remainder *3

3. Rewrite the expression using the property of exponents a^(m*n+r)=(a^n)^m⋅a^r

i^27=i^(4*(6)+3)

i^27=(i^4)^6⋅i^3

4. Substitute the known values i^4=1 and i^3=−i into the expression.

i^27=(1)^6⋅(−i)

5. Simplify the final result.

i^27=1⋅(−i)

i^27=−i

Final Answer

i^27=−i

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