Question:

My teacher has asked us to "evaluate i^27." I think she means she wants the answer to this. How do I find it?
Can you show me how to find the answer to this problem and problems like it? Thanks for any help you can offer.

Answers:

A-Best: 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  

A: 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  

A: "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  

A: 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  

A: i^2 is i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i*i i*i = -1 i*i *i = -1 * i = -i i*i*i *i = -i *i = 1 And so on. More efficiently look at it as (i*i)*(i*i).... (i*i)*i Since (i*i) is -1, that's -1 * -1 * -1 .... * -1 *i The final i dangles because there's an odd number of is So (i*i)* thirteen times is -1, time the dangling i = -i  

A: 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  

A: 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.  

A: 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  

A: 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.  

A: 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.  

A: 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}