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    I was wondering what \sqrt{i} is and so I went this way

    \sqrt{i}=i^{\frac{1}{2}}=i^{ \frac{4}{8}}

    i^{ \frac{4}{8}}=\sqrt[8]{i^4}

    since i^4=1

    \sqrt{i}=\sqrt[8]{1}=\pm 1


    but that leads to a contradiction,

    (\sqrt{i})^2=i=1

    what's my error?
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    (Original post by Plato's Trousers)
    I was wondering what \sqrt{i} is and so I went this way

    \sqrt{i}=i^{\frac{1}{2}}=i^{ \frac{4}{8}}

    i^{ \frac{4}{8}}=\sqrt[8]{i^4}

    since i^4=1

    \sqrt{i}=\sqrt[8]{1}=\pm 1


    but that leads to a contradiction,

    (\sqrt{i})^2=i=1

    what's my error?
    The laws of indices don't hold for all complex numbers with a non zero imaginary part.
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    (Original post by Farhan.Hanif93)
    The laws of indices don't hold for all complex numbers with a non zero imaginary part.
    ah. ok.

    So how can root i be expressed? Is there something nice like there is for i^i ?
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    (Original post by Plato's Trousers)
    I was wondering what \sqrt{i} is and so I went this way

    \sqrt{i}=i^{\frac{1}{2}}=i^{ \frac{4}{8}}

    i^{ \frac{4}{8}}=\sqrt[8]{i^4}

    since i^4=1

    \sqrt{i}=\sqrt[8]{1}=\pm 1


    but that leads to a contradiction,

    (\sqrt{i})^2=i=1

    what's my error?
    geometrically, multiplication of a complex number by i is equivalent to a rotation of that complex number by 90degrees. Therfore the root of i is the transformation that rotates a complex number by 45 degrees. This is equivalent to root(2)/2 (1+i). Its counterpart solution is -root(2)/2 (1+i) I know it is not what you asked, but is a non-standard way of finding the root of i. Draw a triangle if you dont believe me.


    Also you could use:

    i= cos(pi/2+2pi*k) + isin(pi/2+2pi*k)

    using de moivres theorem

    De moivres theorem states that (cos(theta) +isin(theta))^n= cos(ntheta) +isin(ntheta)

    You can try and prove this if you like..

    (i)^(1/2)= cos(pi/4 +2pi*k/2) +isin(pi/4+2pi*k/2)
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    (Original post by Plato's Trousers)
    ah. ok.

    So how can root i be expressed? Is there something nice like there is for i^i ?
    I'm not sure how it turns out as I've never actually done the legwork for it but if you let:
    x+yi=\sqrt{i}
    Then square both sides, equate imaginary and real parts to solve for x and y, you will get some answer.
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    root i is quite nice:

    Spoiler:
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    \frac{1+i}{\sqrt{2}}
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    (Original post by Plato's Trousers)
    ah. ok.

    So how can root i be expressed? Is there something nice like there is for i^i ?
    Another way to do it is (a+bi)^2 = 0 + 1i, expand and equate

    EDIT: Farhan beat me to it, no surprises :p:
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    (Original post by Tobedotty)
    root i is quite nice:

    \frac{1+i}{\sqrt{2}}
    I think that's only one of the answers. the other is -\frac{1-i}{\sqrt{2}}
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    (Original post by Farhan.Hanif93)
    I'm not sure how it turns out as I've never actually done the legwork for it but if you let:
    x+yi=\sqrt{i}
    Then square both sides, equate imaginary and real parts to solve for x and y, you will get some answer.
    This has always been my method. With it you can find higher roots of i too, although for particularly nasty indicies I would switch to the polar form, so

     i^k = \exp(i * \pi * 0.5 * k)

    and then use the usual sin and cos numerical representation.

    Note that with this method you can also evaluate  i^i . Weirdly notice that it's a real number.
    Cool, eh?
    (Not rigorous obviously...)
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    (Original post by StephenNeill)
    This has always been my method. With it you can find higher roots of i too, although for particularly nasty indicies I would switch to the polar form, so

     i^k = \exp(i * \pi * 0.5 * k)

    and then use the usual sin and cos numerical representation.

    Note that with this method you can also evaluate  i^i . Weirdly notice that it's a real number.
    Cool, eh?
    (Not rigorous obviously...)
    Yeah, it's a nice result.
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    I'd like to point out that the value of i^i depends on your choice of logarithm, but regardless of your choice, it is a real number.

    In general, if a and b are complex numbers, there is no natural choice of logarithm, and therefore no natural definition of a^b.
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    (Original post by Zhen Lin)
    I'd like to point out that the value of i^i depends on your choice of logarithm, but regardless of your choice, it is a real number.
    eh?

    Surely the value of ii comes from Euler's relation? it's e-pi/2

    What have logs got to do with it?
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    Consider this: \exp \frac{5}{2} \pi i = i. Hence, i^i = e^{-\frac{5}{2} \pi}. Ah, but wait, e^{-\frac{5}{2} \pi} \ne e^{-\frac{1}{2} \pi}.

    In order to calculate a value for a^b = \exp (b \log a), you must first select a value of \log a to use. There is no natural choice for a general complex number.
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    (Original post by Plato's Trousers)

    \sqrt[8]{1}=\pm 1
    Think about this.
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    (Original post by Zhen Lin)
    Consider this: \exp \frac{5}{2} \pi i = i. Hence, i^i = e^{-\frac{5}{2} \pi}. Ah, but wait, e^{-\frac{5}{2} \pi} \ne e^{-\frac{1}{2} \pi}.

    In order to calculate a value for a^b = \exp (b \log a), you must first select a value of \log a to use. There is no natural choice for a general complex number.

    ok, there are a few things I don't understand here.

    1 I can see that because sin and cos are cyclic functions (their value repeats every 2 pi) that ii can have several values. Generally ab has only one value, so don't we generate a contradiction? Does this mean Euler's relation is wrong? (I am sure it isn't, but why isn't it, if it generates a contradiction?)

    2 surely in a^b = \exp (b \log a), the log must be to the base e (ie a natural log) otherwise it won't reverse the exp? If we were to use log to the base 10 we would need a^b = 10^{b \log a}, no?
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    (Original post by Plato's Trousers)
    1 I can see that because sin and cos are cyclic functions (their value repeats every 2 pi) that ii can have several values. Generally ab has only one value, so don't we generate a contradiction? Does this mean Euler's relation is wrong? (I am sure it isn't, but why isn't it, if it generates a contradiction?)
    Generally, a^b has multiple values.

    2 surely in a^b = \exp (b \log a), the log must be to the base e (ie a natural log) otherwise it won't reverse the exp? If we were to use log to the base 10 we would need a^b = 10^{b \log a}, no?
    Yes, but which log to base e? There is only one "natural" choice for the logarithm on the reals, but extending this the complex numbers is problematic.
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    (Original post by SimonM)
    Generally, a^b has multiple values.



    Yes, but which log to base e? There is only one "natural" choice for the logarithm on the reals, but extending this the complex numbers is problematic.
    how can there be more than one ln ? It's log to the base e. Is there more than one value of e then?

    :confused:
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    (Original post by Plato's Trousers)
    how can there be more than one ln ? It's log to the base e. Is there more than one value of e then?

    :confused:
    So 1 = e^0= e^{2 \pi i} = e^{4 \pi i} =  \cdots, so \log 1 = 0, 2 \pi i, \ldots?
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    De moivres theorem:
    e^ix=cosx+isinx

    you can then set x=pi/2
    so:
    e^ipi/2=i
    so root i = i^1/2 = (e^ipi/2)^1/2 = (e^ipi/4)

    hope this helps

    EDIT: sorry i don't know how to insert all the proper mathematical symbols! bear with my 'pi' and '^'s!
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    (Original post by Plato's Trousers)
    how can there be more than one ln ? It's log to the base e. Is there more than one value of e then?

    :confused:
    The problem is that log is not single valued. logx can be defined on the reals as the (unique) number st e^(logx)=x. If we try to do this with complex numbers, it turns out that it's not unique, because 2n*i*pi+logx satisfies the same equation:
    e^(2npi+logx)=e^2n*i*pi * e^logx=(e^2*i*pi)^n *x=1*x=x
    So log is multivalued, which is why a^b is also.
 
 
 
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