Imaginary Numbers 1

Strategy that I came up with:

Starting with the exponent at 5 onwards, if you subtract it by 1 then divide by 4 and get a whole number then it equals i. Otherwise if not a whole number then do the process again, subtract the original exponent by 2 then divide by 4 and see if you get a whole number, if you do then it equals -1 ...... if you have to subtract it by 3 it equals -i ...... if you have to subtract it by 4 then it equals 1.

Using this strategy: the exponent of i^10 is 10 soooo:

(10-1)/4 =/= whole so reject i

(10-2)/4 = 2 which is a whole number therefore i^10 = -1.

Now tell me what i^2015 equals =)

Well, i^2015 is -i 

I take it you are posting this as a challenge question to other students?  I mean, you already know how to do it.?  

I intend to draw even our young member's and guest's attention to this question so I think I better explain a little.

You cannot find the square root of a negative number because nothing times by itself can be negative.  

Right ?

Well sort of right.   There is NO answer in the real number system and many of you have not yet heard of imaginary numbers.  BUT

There is also a complex number system that has imaginary numbers in it.

$The\;\;\sqrt{-1}\;\;\mbox{is the basic unit of all imaginary numbers, it is called }i\\ so\\ i^2=\sqrt{-1}\times \sqrt{-1} = \sqrt{-1*-1}=\sqrt{1}=1\\ i^3=i^2*i=1*i=i$

and so on.   See if you can work out  i^4 

Heureka has given you the answer below but imaginary numbers are fun to play with  

$i=i \quad i^2=-1\quad i^3=-i\quad i^4=1\\i^5=i....$

Strategy that I came up with:

Starting with the exponent at 5 onwards, if you subtract it by 1 then divide by 4 and get a whole number then it equals i. Otherwise if not a whole number then do the process again, subtract the original exponent by 2 then divide by 4 and see if you get a whole number, if you do then it equals -1 ...... if you have to subtract it by 3 it equals -i ...... if you have to subtract it by 4 then it equals 1.

Using this strategy: the exponent of i^10 is 10 soooo:

(10-1)/4 =/= whole so reject i

(10-2)/4 = 2 which is a whole number therefore i^10 = -1.

Now tell me what i^2015 equals =)

If the sqare root of -1 is "i", then what is i^2, 3, and 4. Do you see the pattern? Now, following this knowledge, what is i^10? (you can do it!)

See the imaginary coordinate system. The real axis in x (-1,+1) and the imaginary axis in y ( -i, + 1).

Set a Point on ( 1,0) for $i^0$ so $i^0 = 1$

Move the Point counter clockwise 90 degrees you have the Point (0,i) for $i^1$ so $i^1 = i$

Move the Point again counter clockwise 90 degrees you have the Point (-1,0) for $i^2$ so $i^2 = -1$

Move the Point again counter clockwise 90 degrees you have the Point (0,-i) for $i^3$ so $i^3 = -i$

Move the Point again counter clockwise 90 degrees you have the Point (1,0) for $i^4$ so $i^4 = 1$

etc.

$i^{10}=-1$

Nicely done, you nailed it