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What is the value of i^n if the remainder of n/4 is 2?

A. I

B. -i

C. 1

D. -1

User Danation
by
7.6k points

2 Answers

4 votes

Answer:

C and D

Explanation:

i=√-1

n/4=x.5

since 2 is the remainder

n=2x

then i^n=i^2x

when x is even, i^n =1

when x is odd, i^n=-1

User Kim Miller
by
8.8k points
6 votes

Answer: 1

Explanation:

we are to find the value of i^n if the remainder of n/2 =4

First, we need to find what n is.

To find n;

n/4 = 2

we simply multiply bothside of the equation by four

n/2 ×4 = 2 ×4

n = 8

Next is for us to know what i stand for.

i usually show the imaginary part if a complex number.

In Mathematics, i = √(-1)

which implies i² ⁼ √(-1) ×√(-1) = - 1

Now back to our question,

iⁿ = i⁸ (since n=8)

Now ,

i⁸ = i² × i² × i² × i²

= -1 × -1 × -1 ×-1 (remember i²=-1)

=1

Therefore, iⁿ is 1 if the remainder of n/4 is 2

User Mecho
by
7.4k points

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