Simplify i^47

Problem

i47

Solution

1. Identify the pattern of the powers of the imaginary unit i The powers of i repeat in a cycle of four: i1=i i2=−1 i3=−i and i4=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 in=i(4*k+r)=(i4)k⋅ir

i47=i(4*(11)+3)

4. Substitute the known value i4=1 into the expression.

i47=(i4)11⋅i3

i47=(1)11⋅i3

5. Simplify the result using the value of i3

i47=1⋅(−i)

i47=−i

Final Answer

i47=−i

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