Imaginaries don't have signs, however if you put them in an Argand diagram, they are taken as positive.
2006-10-08 02:22:28
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answer #1
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answered by Anonymous
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There is INDEED positive and negative in imaginary numbers. the sqrt(-1), as you have in the question, is +i. Also, the powers of i have a funny quality:
i^1 = i
i^2 = -1
i^3 = i^2 x i = -i
i^4 = i^2 x i^2 = +1
i^5 = i^2 x i^3 = i
and the cycle continues.
Also,
i^(-1) = 1/i = -i = i^3
since i x 1/i = 1 which is equal to i x (-i)
So the cycle goes to negative powers of i as well.
2006-10-08 10:40:32
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answer #2
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answered by willismg1959 2
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I do not believe that the sign function is defined for complex number.
The number is "positive" if you accept the convention that "numbers" in the first quadrant are "positive".
By the way, what is the sign of -5 + â-1?
2006-10-08 11:22:54
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answer #3
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answered by Dr. J. 6
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I think that it may be consider as + since -i is possible as in the solution of x^2+1=0.
2006-10-08 09:26:11
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answer #4
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answered by dart 2
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the sign of sqrt (-1)=i is of cause +
example z=a+ib
example z=c-id you can see its an positive unit
2006-10-08 09:22:48
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answer #5
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answered by Broden 4
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It's equal to i
You have to think of it as positive, however, in the complex number system, you don't have positives or negatives.
2006-10-08 09:16:06
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answer #6
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answered by Melody 3
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i is imaginary number and i^2= -1 but i can have sign -/+ in the end it is just i
2006-10-08 09:47:46
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answer #7
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answered by georgenader_0123 2
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â-1 = i OR -i
You must understand that the original rules of algebraic manipulation still apply.
It is easy to note that
(â-1)^2 would be i.i = -1,
say we take (-i).(-i) = (-1)(-1) .(i^2) = 1 .(-1) =-1 as before.
2006-10-08 10:29:50
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answer #8
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answered by yasiru89 6
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This iz complex number +/- i
tis does not have a sign as this is not a real number.
2006-10-08 09:16:55
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answer #9
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answered by Mein Hoon Na 7
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Oh, i is positive.
If it were negative it would have a minus sign in front of it, like -i.
2006-10-08 09:28:18
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answer #10
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answered by Alan Turing 5
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