Question

Which simplifications of the powers of i are correct? There may be more than one correct answer. Select all correct answers.

A) i^6=1
B) i^18=1
C) i^7=-i
D) i^16=1

Answer 1

Option C and D are correct options

Explanation:

The correct options are C and D.

We know that i = √-1

i² = -1

i³ = -i

and i^4 = 1

Lets solve the options one by one:

i^6 =1

Break the power:

i² *i ² * i² = (-1)(-1)(-1)

= -1

Therefore A is wrong

B) i^18 = 1

Lets break the power:

i²* i² *i² *i²*i²*i²*i²*i²*i²

put the value of i^2

= (-1) (-1) (-1) (-1) (-1) (-1) (-1) (-1)(-1)

= -1

Therefore option B is incorrect.

C) i^7 = -i

= i² * i² *i² *i

=(-1) (-1) (-1) * √-1

= - √-1

We know that √-1 = i

So,

- √-1 = -i

Therefore option C is correct.

D) i^16 = 1

= i² * i² * i² * i² * i² * i² *i² *i²

= (-1) (-1) (-1) (-1) (-1) (-1) (-1) (-1)

= 1

Therefore option D is correct.

Thus option C and D are correct option....

Answer 2

Answer: Option C and Option D

Explanation:

Remember that by definition we have to:

`i=√(-1)` and
`i^2=-1`

So for the option A we have to:

`i^6=(√(-1))^2*(√(-1))^4\\\\i^6=-1*(-1)^2\\\\i^6=-1`

Option A is false

So for the option B we have to:

`i^(18)=(i^6)^3`

We know that
`i^6=-1`

So

`i^(18)=(-1)^3`

`i^(18)=-1`

Option B is false

So for the option C we have to:

`i^7=(i)^6*i^1`

`i^7=-i`

Option C is true

Finally for the option D we have to:

`i^(16)=(i^4)^4`

`i^(16)=((-1)^2)^4`

`i^(16)=(1)^4`

`i^(16)=1`

Option D is true