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Imaginary Numbers 1

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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!)

Mathhemathh  Nov 11, 2015

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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

Mathhemathh  Nov 11, 2015
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#1
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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

Melody  Nov 11, 2015
edited by Melody  Nov 11, 2015
edited by Melody  Nov 11, 2015
#2
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$$i=i \quad i^2=-1\quad i^3=-i\quad i^4=1\\i^5=i....$$

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 =)

Maximillian  Nov 11, 2015
#3
+18927
+5

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.

$$i^{10}=-1$$

heureka  Nov 11, 2015
#4
+302
+5

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

Mathhemathh  Nov 11, 2015
#5
+42
+5

Nicely done, you nailed it

Maximillian  Nov 11, 2015

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