Simplify i^47

Problem

i^47

Solution

1. Identify the pattern of the powers of the imaginary unit i The powers of i repeat in a cycle of four: i^1=i i^2=−1 i^3=−i and i^4=1

2. Divide the exponent by 4 to find the remainder.

47÷4=11* remainder *3

3. Rewrite the expression using the property of exponents i^n=i^(4*k+r)=(i^4)^k⋅i^r

i^47=i^(4*(11)+3)

4. Substitute the known value i^4=1 into the expression.

i^47=(i^4)^11⋅i^3

i^47=(1)^11⋅i^3

5. Simplify the result using the value of i^3

i^47=1⋅(−i)

i^47=−i

Final Answer

i^47=−i

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