My teacher has asked us to "evaluate i^27." I think she means she wants the answer to this. How do I find it?

Question

Can you show me how to find the answer to this problem and problems like it? Thanks for any help you can offer.

Answer 1

i^27
= (i^3)^9
Since (i^3) IS i
therefore it becomes i^9
= -i


i^27=(-i)^9.
This means, multiply i by itself 9 times, like this:
sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) x sqrt(-1) = -i

Answer 2

Take 27 and divide it by 4 since there are 4 imaginary numbers. It divides 27 six times with a remainder of 3. The remainder is the important part.

i=sqrt(-1) This is the answer if the remainder is 1. Since the exponent on i is 1.

i^2=-1 This is the answer if the remainder is 2.
Since the exponent on i is 2.

i^3=-i This is the answer to your problem because the remainder is 3.

i^4=1 This is the answer if the remainder is 0.

It's just the easiest way that I can think of the find out values of i that have large exponents, and it works every time.

Answer 3

i = sqrt(-1)
i^2 = -1
i^3 = -i
i^4 = 1
after this it repeats again for i^5 and beyond.

so to find i^27, just do some modular math and divide the exponent 27 by 4. determine the remainder. and then your answer can be found by using the table i provided. if the remainder is:
1 then use the result for i,
2 then use the result for i^2
3 then use the result for i^3
0 or no remainder, then use the result for i^4

in this case, dividing 27 by 4 gives a remainder of 3, so your answer is -i.

Answer 4

First, you're right--evaluate means solve, or find the answer.

Now i is the square root of -1. So we know that i^1 = i, and i^2 = -1. Then i^3 must be -i. And i^4 = 1, because -1^2 = 1.

Now, i^5 is the same as i^1 * i^4, meaning it's the same as i^1, or i. Powers of i cycle through the same four values over and over again, because i^4 = 1.

So, i^27 = i^23 = i^19 = i^15 = i^11 = i^7 = i^3, which is -i.

i^27 = -i.

Answer 5

Since i^2 = -1 then i^4 = (i^2)^2 = (-1)^2 = 1

So if you can extract some powers of i^4, you're in good shape.

i^27 = (i^4)^6 * i^3
= 1^6 * i^3
= i^3 = i * i^2
= -i

The key here is to divide the exponent by 4 and use the REMAINDER (the leftover).
i = i {remainder 1}
i^2 = -1 {remainder 2}
i^3 = -i {remainder 3}
i^4 = 1 {remainder 0}

Answer 6

You just need to memorize these 4 things

i = i (which is the sqr(-1))
i^2 = -1
i^3 = -i
i^4 = 1

to figure out i^x, divide x by 4 and take the remainder -- then do i to the remainder (or i^4 if there is no remainder)

examples

to do i^6
6/4 = 1 remainder 2
so it equals i^2 = -1

to do i ^15
15/6 = 3 remainder 3
i^3 = -i

so to do i^27
27/4 = 6 remainder 3
i^3 = -i

answer
i^27 = -i

Answer 7

"i" has a certain periodicity when raising it to powers. It goes like this:
i^1=i
i^2=-1
i^3=-i
i^4=1
i^5=i
i^6=-1
...... and so on
Here we see that i repeats its value with a step of 4.
i^27=i^24*i^3
24 can be divided by 4 so i^24 has same value as i^4 which is 1
so i^27=i^3 which we earlier calculated as being -i

Answer 8

Well, note that i^1 = i
i² = -1,
i³ = -1,
i^4 = 1,
so the powers of i repeat every 4th go-round.
So i^27 = i^24* i^3 = (i^4)^6 * i^3 = 1^6^i^3 = -i

Answer 9

basically for any problem like this divide the power by four with a remender then find the value of i to that power
for exapmle i^27
27/4= 6(4) + 3 so your answer is i^3
so -i

Answer 10

i=i
i^2=1
i^3=-i
i^4=-1
i^5=i
i^6=1
i^7=-i
i^8=-1
this pattern continues by every fourth power so i^27=-i