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Root of a complex number in cartesian

  1. Mar 27, 2005 #1
    Is there any law for finding the root of a complex number in catesian coordinates? without changing to polar,

    I've created 1, i just want to know is it worthy or not, so ...

    everybody who reads the message, please post the ROOT OF A COMPLEX NUMBER IN CARTESIAN COORDINATES LAW and let me see if i made somthing old or new :tongue2:

    Note : not only the square root, i mean the N root,

  2. jcsd
  3. Mar 27, 2005 #2


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    I've never heard of root in cartesian coordinates vs polar,elliptic,spherical,paraboloidal,etc...

    A complex # is

    [tex] z=a+ib=re^{i\varphi} [/tex]

    So what do u mean by cartesian coordinates...?

  4. Mar 27, 2005 #3
    LoL, I mean the a + bi, What's the name of it then? it's cartesian, because it's shown in the XOY Plane as x + iy, any number is like a point,
  5. Mar 27, 2005 #4


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    Okay,have it your way.So what about the algebraic forrm of a complex number...?

  6. Mar 27, 2005 #5


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    If you can find the roots in polar, you can readily convert them to the form a+bi by Euler's formula. What's the prob?
  7. Mar 27, 2005 #6
    LoL, man there is no completely any problem, but i'm asking a CLEAR question, is there a root in ALGEBRIC form? without changing to polar?
  8. Mar 27, 2005 #7


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    The best thing I can think of is to expand (x + iy)^n using Binomial Expansion, equate it to whatever you want to find the nth root of, so:

    (x + iy)^n = a + bi

    you'll then have two equations:
    1) Re((x + iy)^n) = a
    2) Im((x + iy)^n) = b

    and two unknowns (x and y), but the equations won't be linear in x and y. In fact, you'll end up with n solutions in general. So it's a matter of finding the n pairs (x, y) that satisfy these two equations, or if you only need one of the n roots, then just find one pair (x, y) that satisfies both equations.
  9. Mar 27, 2005 #8
    Well, that means there is no mathematical formula for that, right?
    Last edited: Mar 27, 2005
  10. Mar 27, 2005 #9


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    How about u present your ideas,calculations & results and let us decide whether what u'd be posting is
    b)correct,but already discovered.
    c)something really new...

  11. Mar 27, 2005 #10
    Well, I'm gonna show here the square root only, and I hope it'll be choice (c), not (a) (lol), here is it:

    [tex]\sqrt{a+b\imath} = \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}}+\imath \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}[/tex]

    I really hope to take choice (c) because i got very mad to reach this point,

    the N root will give more complicated tailes for it,

    hmmmmmmmmm? how is it? a,b,c?
  12. Mar 27, 2005 #11


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    Combo between a) & c)...

    [tex] \sqrt{a+ib} =\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i\sqrt{\frac{\sqrt{a^{2}+b^{2}}-b}{2}} [/tex]

    I'm sure that,even in the correct version,it is not new...

    As for

    [tex] \sqrt[n]{a+ib} [/tex],it's really simple,once u use Euler's formula...

  13. Mar 27, 2005 #12


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    Is [itex]\sqrt{a + ib}[/itex] even well defined? For real arguments, the radical sign gives the positive root. As far as I know, it doesn't make sense to talk about positive and negative complex numbers in general, so why would [itex]\sqrt{a + ib}[/itex] be what's given there, and not it's additive inverse? For the nth root, which root would the formula pick?

    Also, the formula is indeed nothing new. It's not something you'll see in a text book because it's useless, it's like an intermediate step of a homework problem. It seems that memorizing this formula would be no better than doing the work from scratch. So the formulas for the nth root that you might have are not something that no one knew before, it's just something no one would ever bother writing down, especially given the fact that it's so much easier to compute using polar co-ordinates.

    Moreover, there's certainly something wrong with your formula, since any number in general has 2 roots, and if one of the roots is z, the other is -z. But your formula gives a number that will have positive real and imaginary parts, which would mean that any complex number has one root that has both parts positive, and another (-z) that has both parts negative. So that means that no complex number has roots where, say, the real part is positive and the imaginary part is negative. Clearly, that's absurd because the number (a - bi)², where a and b are real and positive, is a complex number with a root that has positive real part and negative imaginary part.

    So your formula is a cross between wrong, useless, not well-defined, and already discovered. Sorry.
  14. Mar 28, 2005 #13
    About wrong it's not wrong, because it can be proofed when squaring the 2 sides, while we can put some constraints which will fix the error you said,

    i know using polar form is easier, i made this formula while i didn't study the complex numbers, i was in the school, i read about them, and i liked them, and told my self why isn't there a clear form of the root, why we should solve 2 equations 2 find the roots (before knowing de'movers theorem), and believe me, i made the first step by creating the root of any constant multiplied by i, and the i started generalizing it one by one, (all this while i NEVER know de'movers), and after that after reading more and more i got disappointed after finding 1 of my 2 theorems available, (de'movers), i found it, but he found it before me, i know this is some kind ridiculous, but that's what happen, and after studying them in the university, i knew the polar form and the algebric form, that's the story,

    anywise, i'm now reading 100000 books more than my study a day, hoping to create somthing new in this world,, hehe,, that theorem was a results of a child of 17 years game with numbers,
    note: demovers = eulers but in our books,
    i'll keep moving, i don't wanna hide i'm now disappointed, and here is the N root while you can give me your idea:

    [tex]\sqrt[n]{a+bi} = \sqrt{\frac{\sqrt[n]{a^2+b^2}+Re(\sqrt[\frac{n}{2}]{a+bi})}{2}} + i\sqrt{\frac{\sqrt[n]{a^2+b^2}-Re(\sqrt[\frac{n}{2}]{a+bi})}{2}}[/tex]

    and thanks for the ideas, please give me your ideas about this also,

  15. Mar 28, 2005 #14
    Your equation certainly is wrong. It only works for positive values of b. It's easy enough to fix; a correct equation would be:

    [tex]\sqrt{a+b\imath} = \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}}+\mathrm{sgn}(b)\imath\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}[/tex]

    Of course, this equation only gives you the root with a positive real part. It certainly isn't a new discovery of any kind; one of the practice questions in my old classical algebra textbook asks for this equation.
  16. Mar 28, 2005 #15


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    Your equation is still wrong. Your new one has an additional problem: what if N is odd? What then is [itex]\sqrt[\frac{n}{2}]{a+bi}[/itex]? I suppose it's not a huge problem, there are ways to work around it. Of course, I'm not sure if your formulas are even close, because it's still evident that they're wrong. With some corrections (like the one master coda suggested), they may be right, but I'm not going to bother to check that. Anyways, the problems that still remains with your solution are that it is:

    a) useless
    b) not well-defined (given that a + bi generally has n nth roots, which root does [itex]\sqrt[n]{a+bi}[/itex] even pick out?)
    c) already discovered (more or less, it's probably like a problem in most textbooks, except no textbook would bother writing it down because of DeMoivre's)
    d) wrong because, as I mentioned and as you seemingly didn't notice at all, it gives a root in the form:

    [tex]\sqrt{X} + i\sqrt{Y}[/tex]

    where X and Y are real numbers. But as you know, the square-root function only returns non-negative values. Therefore, [itex]\sqrt{X}[/itex] and [itex]\sqrt{Y}[/itex] are non-negative values. What your formula amounts to saying is that EVERY complex number has a root that has both real and imaginary parts being non-negative.

    What about the number -2i? It has 2 square roots, namely (i - 1) and (1 - i). Neither of these complex numbers has both real and imaginary parts being non-negative. So your formula cannot possibly work to find the square root(s) of -2i.
  17. Mar 29, 2005 #16
    Hehe, you still don't understand me, i said it's not absolutly right, but it was a game of a chiled, and the wrongs you are talking about can be fixed with some constraints as master coda mentioned

    i thought about having an odd number for n, or not an odd as i can say, the problem occurs when the number n is not one of the 2 powers (2,4,8,16,...), but i said to my self some one who knows more than me can fix it, Plus I mentioned before, when i made this equation i completly didn't know the de'moivers equation

    DAMN THIS CENTURY !!! EVERYTHING IS ALREADY DISCOVERED, WHAT THEY LEFT FOR US !! LOL !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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