Question

Which simplifications of the powers of i are correct?

Answer 1

The correct simplifications of powers of i are:

a. i^27=−i

d. i^8=1

e. i^5=i

Step-by-step explanation:

We know that

$i^2=-1\\i^3=-i\\i^4=1$

We can use the smaller powers of i to solve the larger powers.

a. i^27=−i

$i^{27}\\={i^{24}}.i^3\\={i^{4*6}}.-i\\=(i^4)^6.-i\ \ \ \ As\ we\ know\ i^4=1\\=(1)^6.-i\\=1*-i\\=-i$

b. i^28=i

$i^{28}\\=i^{4*7}\\=(i^4)^7\\=(1)^7\\=1$

c. i^14=1

$i^{14}\\=i^{12}.i^2\\=(i^{4*3}).i^2\\=> i^2=-1\\=(i^4)^3.-1\\=(1)^3*-1\\=-1$

d. i^8=1

$i^8\\=i^{4*2}\\=(i^4)^2\\=(1)^2\\=1$

e. i^5=i

$i^5\\=i^{2*2*1}\\=i^2*i^2*i\\=(-1)(-1)(i)\\=i$

Hence,

The correct simplifications of powers of i are:

a. i^27=−i

d. i^8=1

e. i^5=i

Keywords: Imaginary units, Iota

Learn more about imaginary units at:

• brainly.com/question/8806598
• brainly.com/question/8805387

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