Question 26058
Imaginary numbers: In math, the lower-case letter i is used to stand for the square root of -1, thus: {{{sqrt(-1) = i}}}

If you multiply i times i or {{{i^2}}} this is the same as multiplying {{{sqrt(-1)sqrt(-1) = (sqrt(-1))^2}}} = -1 so you can see that{{{i^2 = -1}}}

Now if you keep doing this, that is, find {{{i^3}}},{{{i^4}}},{{{i^5}}}, you begin to see a definite pattern:

{{{i = sqrt(-1)}}}
{{{i^2 = -1}}}
{{{i^3 = i(i^2)}}} = {{{i(-1) = -i}}}
{{{i^4 = i(i^3)}}} = {{{i(-i) = -i^2}}} = {{{-(-1) = 1}}}
{{{i^5 = i(i^4)}}} = {{{i(1) = i}}}
{{{i^6 = i(i^5)}}} = {{{i(i) = i^2}}} = {{{-1}}}
{{{i^7 = i(i^6)}}} = {{{i(-1) = -i}}} ...and so on.

Do you see it? Sometimes it's helpful to constuct a table for the values of {{{i^n}}} for different values of n:
n----{{{i^n}}}
--------------
1.....i
2....-1
3....-i
4.....1
5.....i
6....-1
7....-i
8.....1
9.....i
10...-1
11...-i
12....1
and so on
As for the number in front of the i's in your problem, you just multiply by that number. For example:
{{{7i^3 = 7(-i)}}} = {{{-7i}}}
{{{8i^(12) = 8(1)}}} = 8