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14 solutions
This is the best explanation I've seen so far.
Oops, I stopped at 1/i. It all makes sense now!
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Same here, I stopped there as well. Thanks for the solution.
me too, i was like wtf???
Thanks for the responses everyone. I teach math in middle school and I just love telling the kids that we can do real math with numbers that don't exist. ;-)
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aCT ually on the first instance i pressed -i,but when the answer appeared a slip of my finger pressed the wrong answer . NO problems .
Marshall I reached the same answer in a difficult way but yours is better method.
the question is how you can demostrate tha i/i is =1
If 1/i = -i, then 1 = - i(i) = - (i^2) = - (-1) = 1
i^(i)^2=i^(-1)=-1/i=-i *ohh now I got it the problems I was trying to think fast!!
i i 2 = i − 1 = i 1 = − i − 1 = − i i 2 = − i
What about doing: 1/i = 1/√-1 = √1/√-1 = √(1/-1) = √-1 = i
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The radical product rule, a × b = a b is valid as long as a , b ≥ 0 . When complex numbers get involved it does not work the same. So, − 1 1 = − 1 1 .
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Little correction in your statement.. The radical product a × b = a b is valid if either or both a,b ≥ 0 .i mean radical product is valid also if one is negative and other is positive.
When you take underroot of 1 it can be both negative as well as positive so you end up with two possible solutions. However multiplying and dividing by i gives you one perfect answer.
i power i power 2 =1/i=-i power2/i= -i
@Saiful Haque ... plz tell me is 4^-1 is ( -4) or 1/4?
i i 2 = i ( − 1 ) = e 2 i π ( − 1 ) = e 2 − i π = ( cos ( 2 π ) − i sin ( 2 π ) ) = − i
i^i^2=1/i=(i^2 * i^2)/i=i * i^2=i*(-1)=-i
i^i^2 = i^ (-1) = 1/i = i^(4)/i = i^3
Hence i. i^(2) = - i
One should not write i = sqrt(-1). i is by definition the number whose square is -1. If you say i=sqrt(-1) you have the following problem:
Suppose i =sqrt(-1) and consider i^2. i^2 = sqrt(-1) sqrt(-1) =sqrt(-1*-1) =sqrt(1)=1 but on the other hand i^2 =-1 by definition
Sqrt (-1) * Sqrt (-1) = -1
i^-1= i * -i
i^-1= -i
Since i^2 is -1 so we get , i^-1 which on multiplying num and denominator by i, we get -i
Seeing as how squaring a number cancels out a square root, it would simply be i to the -1 power, equaling -i.
i^i^2 = i^(i^2) = i^-1 = i^3 = -i , from the fact i^n = i^(n+4k) where k is an integer. Hence I used k = 1 in this case to show i^-1 = i^(-1+4(1)) = i^3 which we know is -i because i^3 = i * i^2 = i(-1) = -i
Thanks for showing these steps. This explained it perfectly.
i is not square root of -1. The square root can be extracted only from positive numbers. i is that complex number for which i^2 = -1
i is an imaginary number that is the square root of -1.
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check this out: https://en.wikipedia.org/wiki/Imaginary_unit You are kind of right, except you cannot have a square root of a negative number
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The number i is an imaginary number defined in mathematics as i=√-1. It is also defined as i^2=-1. Since in neither case can an answer be real, this is the reason it is called an imaginary number.
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@Brian Thorpe – i is not defined as − 1 , rather it can be defined as i 2 = − 1 or i = ± − 1
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@Tarjan Tamas – The clarification above was given by Brilliant's people...................not my fault
i i 2 = i − 1 = i 1
Then we multiply the top and bottom by i .
i 1 ⋅ i i = − 1 i = − i