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Math Help - The square root of i

  1. #1
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    The square root of i

    Find the square root of i where i is the square root of (-1).

    Show your working.

    Thanks guys
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  2. #2
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    Hello, anthmoo!

    I assume you're not allowed to use DeMoivre's Theorem.
    . . It would be too easy!


    Find the square root of i where i = √(-1)
    Let: .z .= .a + bi, where a and b are real numbers and z = i.

    Then we have: .(a + bi) .= .i

    Expand: .a + 2abi + bi .= .i

    We have: .(a - b) + 2abi .= .i


    Equate real and imaginary components:

    . . a - b .= .0 .[1]
    - - - .2ab . = .1 .[2]


    [1] gives us: .b = a . . b = a


    Solve [2] for b: .b = 1/(2a)

    Substitute into [1]: .a - (1/2a) .= .0 . . a - 1/(4a) .= .0

    Multiply by 4a: .4a^4 - 1 .= .0 . . a^4 .= .1/4

    . . . . - . - . - . ._ . . . . . . . - . . ._
    Hence: .a = √2/2 .and .b = √2/2

    . . . . . . . . . _ . . . . . _
    Therefore: .√i .= .(√2/2)(1 + i)

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  3. #3
    Senior Member ecMathGeek's Avatar
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  4. #4
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    Quote Originally Posted by anthmoo View Post
    Find the square root of i where i is the square root of (-1).

    Show your working.

    Thanks guys
    That question is not defined
    (Not joking, I am serious).

    But what you should have asked was.
    Find, x such that,
    x^2=i
    (Basically what you are trying to ask).
    Which is provided by Soroban.
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    That question is not defined
    (Not joking, I am serious).

    But what you should have asked was.
    Find, x such that,
    x^2=i
    (Basically what you are trying to ask).
    Which is provided by Soroban.
    I would like to interject. I may be mistaken, but the question is defined. (I wish I had my pre-calculus book with me to back this up) Here goes:

    Let z = (a + bi)
    Where (a + bi) = r(cosx + i*sinx)
    It can be proven that z^n for any real value of n is:
    z^n = (a + bi)^n = [r(cosx + i*sinx)]^n = r^n(cosx + i*sinx)^n = r^n(cos(nx) + i*sin(nx))

    I don't remember the name of this theorem, but I'll try to modivate it:
    Let n = 2
    z^2 = (a+bi)^2 = [r(cosx + i*sinx)]^2
    r^2(cosx + i*sinx)^2 = r^2[(cosx)^2 + 2i(sinx)cosx + i^2*(sinx)^2]
    r^2{[(cosx)^2 - (sinx)^2] + i*(2sinx*cosx)}
    Note: cos2x = (cosx)^2 - (sinx)^2 ... sin2x = 2(sinx)cosx
    r^2(cos2x + i*sin2x)

    Let n = 3
    z^3 = z^2*z = r^2(cos2x + i*sin2x)*r(cosx + i*sinx) = r^3(cos2x*cosx + i*cosx*sin2x + i*cos2x*sinx + i^2*sin2x*sinx)
    r^3[(cos2x*cosx - sin2x*sinx) + i*(cosx*sin2x + cos2x*sinx)]
    Note: cos(2x+x) = cos2x*cosx - sin2x*sinx ... sin(2x+x) = sin2x*cosx + cos2x*sinx
    r^3(cos3x + i*sin3x)

    Let n = 4
    ...
    r^4(cos4x + i*sin4x)

    Let n = n
    ...
    r^n(cos(nx) + i*sin(nx))

    Let n = 1/2, a = 0, b = 1
    z = (0 + i*1) = i
    r = |z| = 1
    tan(x) = 1/0 --> x = pi/2

    z^n = (i)^n = r^n(cos(nx) + i*sin(nx))
    z^(1/2) = (i)^(1/2) = r^(1/2)(cos(x/2) + i*sin(x/2)) = 1^(1/2)(cos(pi/4) + i*sin(pi/4)) = 1*(sqrt(2)/2 + i*sqrt(2)/2)
    (i)^(1/2) = sqrt(2)/2 + i*sqrt(2)/2
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  6. #6
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    Quote Originally Posted by ecMathGeek View Post
    I would like to interject. I may be mistaken, but the question is defined. (I wish I had my pre-calculus book with me to back this up) Here goes:
    Please do not argue with me I know what I am talking about. This type of math (about numbers) is something I understand very very well.

    Next, do not listen to your precalculus book. It is not a pure book on math. It is a basic math book. It will not explain this detail strictly mathematically.

    I recomment to search for a longgg thread about imaginary numbers over here somewhere.

    Thus, if you want to think about taking a square root think of it as I do. As a solution to the equation x^2=i.
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Please do not argue with me I know what I am talking about. This type of math (about numbers) is something I understand very very well.

    Next, do not listen to your precalculus book. It is not a pure book on math. It is a basic math book. It will not explain this detail strictly mathematically.

    I recomment to search for a longgg thread about imaginary numbers over here somewhere.

    Thus, if you want to think about taking a square root think of it as I do. As a solution to the equation x^2=i.
    Wow. I'm sorry I've offended you. Please don't take anything I say offensively because it's not my intention to make you look wrong. I know you're smart. I have no doubt you know a lot about math. But I would like to point out that I'm not necessarily wrong because I don't know as much.

    I can say that what I did was not incorrect. I don't disagree that the problem can be seen as x^2=i but I assure you there is a 'proven' theorem that allows for the roots of complex numbers to be taken. Whether these roots are defined I cannot argue, but they can be gotten by means other than x^n=i.
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  8. #8
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    Quote Originally Posted by ecMathGeek View Post
    Whether these roots are defined I cannot argue, but they can be gotten by means other than x^n=i.
    Square roots are only defined for the non-negative real numbers. The statement sqrt(-4)=2i is wrong as I explained a lot of times already. I do not care what the pre-calculus book says, it is a pre-calculus book, it is intended for high-school students.

    One problem with square roots for the complex numbers is that they are not "well-defined". Basically, there are multiple values. Functions can only give a unique value, not multiple values.

    For example, the Riemann surface of the complex function
    is shown below. Note as you travel along the surface you get to a different point where you started at. That is a problem. Because it produces multiple values. De moiver's theorem is used to find roots as in zero's of the cyclic equation.

    Note: The same type of problem occur with the logarithmic function on the complex numbers.
    Attached Thumbnails Attached Thumbnails The square root of i-picture10.gif  
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  9. #9
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Square roots are only defined for the non-negative real numbers. The statement sqrt(-4)=2i is wrong as I explained a lot of times already. I do not care what the pre-calculus book says, it is a pre-calculus book, it is intended for high-school students.

    One problem with square roots for the complex numbers is that they are not "well-defined". Basically, there are multiple values. Functions can only give a unique value, not multiple values.

    For example, the Riemann surface of the complex function
    is shown below. Note as you travel along the surface you get to a different point where you started at. That is a problem. Because it produces multiple values. De moiver's theorem is used to find roots as in zero's of the cyclic equation.

    Note: The same type of problem occur with the logarithmic function on the complex numbers.
    Thank you for the clarification. When you said earlier that sqrt(i) is not defined, I thought you were implying that there is no sqrt(i), not that there is no definite value to sqrt(i).
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Please do not argue with me I know what I am talking about. This type of math (about numbers) is something I understand very very well.

    Next, do not listen to your precalculus book. It is not a pure book on math. It is a basic math book. It will not explain this detail strictly mathematically.

    I recomment to search for a longgg thread about imaginary numbers over here somewhere.

    Thus, if you want to think about taking a square root think of it as I do. As a solution to the equation x^2=i.
    why do you sound so harsh TPH, lighten up
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