Which simplifications of the powers of i are correct?

there may be more than one correct answer. select all correct answers.

i14=i4⋅i4⋅i4⋅i2=1⋅1⋅1⋅−1=−1
i6=i4⋅i2=1⋅−1=−1
i9=i4⋅i4⋅i1=−1⋅−1⋅−i=−i
i7=i4⋅i3=1⋅i=i

WRONG

CORRECT

WRONG

WRONG

Step-by-step explanation:

Some concept you might want to know:

Adding up exponent is the same thing as multiplying numbers with the same base together. So a⁴ = a²⁺² = (a²)(a²).

Another concept is that imaginary number i=√(-1). Therefore i^(even_numbers) will always give you -1. If you have i^(odd_number), leave it as a variable i and don't change it.

So if you have i^(even_number), change it to -1.

if you have i^(odd_number), keep it as i.

i¹⁴ = i⁴⁺⁴⁺⁴⁺² = (i⁴)(i⁴)(i⁴)(i²) = (-1)(-1)(-1)(-1) = 1

i⁶ = i⁴⁺² = (i⁴)(i²) = (-1)(-1) = 1

i⁹ = i⁴⁺⁴⁺¹ = (i⁴)(i⁴)(i) = (-1)(-1)(i) = i

i⁷ = i⁴⁺³= (i⁴)(i³) = (-1)(i) = -i

answer: the answer is fre sha voca do

step-by-step explanation: