Question

Just want to check my work--

so then (this is follow up from previous q)

i²⁶⁵⁴

is the second number in the pattern because of the remainder, so it's -1?


(and pls check the following questions):
i⁶⁸ = i
i⁰ = -1 [?]
i⁵⁵ = -i
i^-11 = -0.1
i^9 = -1

thanks again :)

Answer 1

hey again

all correct but

i^68 is wrong

i^0 is wrong

i^-11 is wrong

and i^9 is wrong

recheck your work for those ones

(I don't know if you want the answer or just to know if it's wrong so this is all I'm putting)

Answer 2

minor corrections

i^68 = 1

i^0 = 1

i^-11 = i

i^9 = i

Explanation:

hi again!!

okay so this is a common mistake with negative exponents, you only do the 0.1 thing when it's : y ·
`10^-^x`

so, when you are finding a negative exponent, you flip the fraction form and write the exponent on the bottom [divisor] :

(remember, i also can be written as
`(i)/(1)` [like any other integer] )

so if we have
`(i)/(1)^(11)`, we can rewrite it as:
`(1)/(i^(11))`

i¹¹ = -i

(you know how to do this :) )

1 / -i

we can multiply:
`(1)/(-i) *(-i)/(-i)`

(remember, we can multiply anything times x/x, it is equal to 1)

(1)(-i)= -i

(-i)(-i)= -1

-i / -1 = i

so, i^-11 = i

----

and remember, anything to the power of 0 is 1. (I think you were thinking that 0 is the start of the sequence, which makes sense in theory--but it would also be the end, because it sorta divides evenly into 4, it is four places ahead of 4, so it is also the last pattern number theoretically, making it 1)

so, i^0 = 1

---

i^9 = i, we pass through our sequence exactly twice, and end up on the first repeat (i)

---

and i^68 is the same thing, we end up exactly on our last pattern value (because 68 evenly divides by 4), so i^68 = 1 [last of the four is 1]

hope this helps!!