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Math Help - Complex Numbers

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    Question Complex Numbers

    hi... there has been a question which always crosses my mind whenever i see complex numbers...
    we say..
    |x| is defined as squareroot(x^2)
    so,
    why do we not write |i| as i... (squareroot(i^2)=i)? (note that 'i' means iota)
    now u'll say that modulus function has also been defined as distance of the argument(i, in this case) from the origin..., which u say is i unit from the origin.. but why can't be the distance of a point from the origin be 'i'(iota) units ???? whats wrong with it ???

    let me frame an example here..
    u draw a graph with Y-axis scale as 1 unit = 3... then in reference to this graph, do we write |3|=1 (as distance from origin=1 unit) ????? the answer is a no.... then why cant we treat iota in the same way?? here also we are defining 1 unit of y-axis as 'i'....
    Last edited by pranjvas; February 19th 2013 at 09:05 AM.
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    Re: Complex Numbers

    Quote Originally Posted by pranjvas View Post
    |x| is defined as squareroot(x^2) so,
    why do we not write |i| as i... (squareroot(i^2)=i)? (note that 'i' means iota) now u'll say that modulus function has also been defined as distance of the argument(i, in this case) from the origin..., which u say is i unit from the origin.. but why can't be the distance of a point from the origin be 'i'(iota) units ???? whats wrong with it ???
    There are several problems with the above from my point of view.
    First and foremost, i is defined as principal root of x^2+1=0.
    Thus i^2=-1 as such |i|=1.

    Now we have only one new 'number'. The set of complex numbers is defined as \{a+bi:\{a,b\}\subset\mathbb{R}\}.

    In effect they can be modeled as \mathbb{R}^2.
    As such, the distance a+bi is from (0,0) is |a+bi|=\sqrt{a^2+b^2}.

    As for your comment about distance, distance (or metric) is measured as a non-negative real number. If each of z~\&~w is a complex number then the distance between them is |z-w|.
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    Re: Complex Numbers

    Quote Originally Posted by Plato View Post
    There are several problems with the above from my point of view.
    First and foremost, i is defined as principal root of x^2+1=0.
    Thus i^2=-1 as such |i|=1.


    Now we have only one new 'number'. The set of complex numbers is defined as \{a+bi:\{a,b\}\subset\mathbb{R}\}.


    In effect they can be modeled as \mathbb{R}^2.
    As such, the distance a+bi is from (0,0) is |a+bi|=\sqrt{a^2+b^2}.


    As for your comment about distance, distance (or metric) is measured as a non-negative real number. If each of z~\&~w is a complex number then the distance between them is |z-w|.

    @Plato :
    the explanation u wrote may be valid, but it does not convince me.. i m here talking about the basic definition of iota, which, according to my point of view is different, and so, the formula definition for complex nos. according to this definition changes.. and u are explaining using pre-defined formulas according to the pre-defined definition of iota.. so, it is obvious that your proof will be correct, as it will correspond to the pre defined definition of iota.. so can u please explain on the ground level... i mean.. no formulas.. only statements... and formulas defined on the basis of those statements..
    so, if u say distance a+bi from (0,0) is |a+bi|=underroot(a^2 + b^2), then u must consider the case of real nos.. in a graph, when we mark the co-ordinates (x,y), then the distance from the origin is underroot(x^2+y^2) ... so here, i can mark a point on the graph as (a,bi), and so the distance from the origin will be underroot(a^2 - b^2) (as (bi)^2 = -b^2)... we say that the modulus is a positive no., then iota is a +ve no., and -i is a -ve no. so |i|should be = i (which is a positive no.)...


    and another question : while marking points on a graph with real axes, when we write (x,y), do we write it as x+y????? the answer is no... then y do we write (a,bi) as a+bi???
    Last edited by pranjvas; February 19th 2013 at 06:53 AM.
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    Re: Complex Numbers

    ...
    Last edited by pranjvas; February 19th 2013 at 06:54 AM.
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    Re: Complex Numbers

    No bumping, either.

    -Dan
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    Re: Complex Numbers

    Quote Originally Posted by pranjvas View Post
    @Plato :
    the explanation u wrote may be valid, but it does not convince me.. i m here talking about the basic definition of iota, which, according to my point of view is different, and so, the formula definition for complex nos. according to this definition changes.. and u are explaining using pre-defined formulas according to the pre-defined definition of iota.. so, it is obvious that your proof will be correct, as it will correspond to the pre defined definition of iota.. so can u please explain on the ground level... i mean.. no formulas.. only statements... and formulas defined on the basis of those statements..
    so, if u say distance a+bi from (0,0) is |a+bi|=underroot(a^2 + b^2), then u must consider the case of real nos.. in a graph, when we mark the co-ordinates (x,y), then the distance from the origin is underroot(x^2+y^2) ... so here, i can mark a point on the graph as (a,bi), and so the distance from the origin will be underroot(a^2 - b^2) (as (bi)^2 = -b^2)... we say that the modulus is a positive no., then iota is a +ve no., and -i is a -ve no. so |i|should be = i (which is a positive no.)...


    and another question : while marking points on a graph with real axes, when we write (x,y), do we write it as x+y????? the answer is no... then y do we write (a,bi) as a+bi???
    When we make definitions there is very little room to alter them. i^2 = -1 is the definition of i. No other way to work with it.

    As far as graphing is concerned we are NOT graphing on the usual xy axis system. The (real valued) xy coordinates are graphed called "Cartesian" and have all the usual properties you probably know well. Complex numbers, on the other hand, are graphed on the "Argand" plane which shares a number of convenient features of the Cartesian plane, but also have different properties as well. The usual way to display a complex number a + ib = \( a, b \) is defined such that a is on an "x-axis" and b is on the "y-axis."

    Also, to be clear: a + ib \implies \(a, b \), not \(a, ib \). This is where your definitions appear to be going wrong.

    -Dan
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    Re: Complex Numbers

    Quote Originally Posted by pranjvas View Post
    @Plato :
    the explanation u wrote may be valid, but it does not convince me.. i m here talking about the basic definition of iota, which, according to my point of view is different, and so, the formula definition for complex nos. according to this definition changes.. and u are explaining using pre-defined formulas according to the pre-defined definition of iota.. so, it is obvious that your proof will be correct, as it will correspond to the pre defined definition of iota.. so can u please explain on the ground level... i mean.. no formulas.. only statements... and formulas defined on the basis of those statements..
    so, if u say distance a+bi from (0,0) is |a+bi|=underroot(a^2 + b^2), then u must consider the case of real nos.. in a graph, when we mark the co-ordinates (x,y), then the distance from the origin is underroot(x^2+y^2) ... so here, i can mark a point on the graph as (a,bi), and so the distance from the origin will be underroot(a^2 - b^2) (as (bi)^2 = -b^2)... we say that the modulus is a positive no., then iota is a +ve no., and -i is a -ve no. so |i|should be = i (which is a positive no.)...
    I don't follow this. "we say that the modulus is a positive no., then iota is a +ve no." How does that last follow?


    and another question : while marking points on a graph with real axes, when we write (x,y), do we write it as x+y????? the answer is no... then y do we write (a,bi) as a+bi???
    The "complex plane" is a representation of the complex numbers. We don't "write (a, bi) as a+ bi". We say that the point (a, b) represents, in the complex plane, the number a+ bi.
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    Re: Complex Numbers

    The following statement is NOT correct except in the special case of the real numbers (or a subset of real numbers).

    |x| is defined as squareroot(x^2)

    A better definition that is apropriate for both Real and Complex numbers is:

    |x| is defined as squareroot(x times x*)

    where:
    x = a + ib
    x* = a - ib (the complex conjugate of x)

    Now if x = i then x* = -i and:

    |i| = sqraretoot(i times -i) = squareroot (1) = 1
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    Re: Complex Numbers

    Quote Originally Posted by Kiwi_Dave View Post
    The following statement is NOT correct except in the special case of the real numbers (or a subset of real numbers).

    |x| is defined as squareroot(x^2)

    A better definition that is apropriate for both Real and Complex numbers is:

    |x| is defined as squareroot(x times x*)

    where:
    x = a + ib
    x* = a - ib (the complex conjugate of x)

    Now if x = i then x* = -i and:

    |i| = sqraretoot(i times -i) = squareroot (1) = 1

    i just want to ask one thing.... forget about all the bookish knowledge u have about complex nos.... assume that complex nos. have not been discovered yet... and u have some knowledge about real nos. ... nd u just found a new no. : underroot(-1) .. nd u want to define the properties of this no.... what wud u do???? well, for example, i want to define modulus function on underroot(-1)... i'll observe the case with real nos. ... i'll see that modulus of a no. is always +ve, nd, i also know that underroot(x) is also +ve, and in the case of underroot(-1), i can say that.. according to the above properties, underroot(-1) should be +ve, nd we can also see that -ve of this no. would be -(underroot(-1))... so .. stating that |underroot(-1)| = underroot(-1) seems to be pretty legit.... can u tell me where did i go wrong?? y does my definition not match with those formulas in the books??

    the formula used above (|x| = squareroot(x times x*)) is defined in such a way that it is bound to give u the result as |i| = 1.. because it is defined keeping in mind that u get the result as real... but my question is.. y??? y cant it be the way as stated above??
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    Re: Complex Numbers

    Quote Originally Posted by HallsofIvy View Post
    I don't follow this. "we say that the modulus is a positive no., then iota is a +ve no." How does that last follow?






    The "complex plane" is a representation of the complex numbers. We don't "write (a, bi) as a+ bi". We say that the point (a, b) represents, in the complex plane, the number a+ bi.


    just wanted to ask u a question.. maybe a bit off discussion... but... if i say that i have an object whose mass is iota kgs, or i say that an object with length iota meters... does that mean anything???or is it absurd??? what do u think??? if it is absurd, then, keeping mathematical bookish formulas apart... what does the no. "iota" signify according to u?? if i cant use it for assigning quantity, what can i use it for??
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    Re: Complex Numbers

    Quote Originally Posted by pranjvas View Post
    i just want to ask one thing.... forget about all the bookish knowledge u have about complex nos.... assume that complex nos. have not been discovered yet... and u have some knowledge about real nos. ... nd u just found a new no. : underroot(-1) .. nd u want to define the properties of this no....
    First I have a simple question for you: Are you incapable of using proper English constructions? Why do you make yourself look foolish by using "chat-speak" when discussing a serious topic.

    Second, there is no such thing as "underroot(-1)" If you are serious then learn to use symbols. [TEX]\sqrt{-1}[/TEX] gives \sqrt{-1}. You are not too lazy to take the time to type that are you?

    Now to your question. If I were living pre-sixteenth century and saw " \sqrt{-1}", I would say "what fool wrote that?'

    But from the year 1609 on, the idea of complex numbers has become well understood. It has been developed into a consistent theory.

    Quote Originally Posted by pranjvas View Post
    "iota" signify according to u?? if i cant use it for assigning quantity, what can i use it for??

    Do you even understand that "iota", \iota is a Greek letter not to be confused with the mathematical symbol i, which is used to stand for the principal root of x^2+1=0~?

    And that is you answer as to what is is used for: it completes the number system giving us the fundamental theorem of algebra.
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    Re: Complex Numbers

    Quote Originally Posted by Plato View Post
    First I have a simple question for you: Are you incapable of using proper English constructions? Why do you make yourself look foolish by using "chat-speak" when discussing a serious topic.

    Second, there is no such thing as "underroot(-1)" If you are serious then learn to use symbols. [TEX]\sqrt{-1}[/TEX] gives \sqrt{-1}. You are not too lazy to take the time to type that are you?

    Now to your question. If I were living pre-sixteenth century and saw " \sqrt{-1}", I would say "what fool wrote that?'

    But from the year 1609 on, the idea of complex numbers has become well understood. It has been developed into a consistent theory.




    Do you even understand that "iota", \iota is a Greek letter not to be confused with the mathematical symbol i, which is used to stand for the principal root of x^2+1=0~?

    And that is you answer as to what is is used for: it completes the number system giving us the fundamental theorem of algebra.

    I am sorry to waste your precious time, Mr. Plato.... You are an MHF Expert, MHF Helper, who has 560 Thanks and an Award. And I am a newbie, probably a fool who is searching answers of some stupid questions; and I should not waste your time discussing such absurd topics here as I am no expert, no helper and do not posses "badges" like you.. I am sorry that I did not know that English is more important to you than the mathematical problem

    If a fool like me had shown you \sqrt{-1} in the 16th century, you definitely would have said "what fool wrote that?" as I am no MHF Expert, MHF Helper, and do not have awards like you. But if you would have found that, your name would have been there in the text books instead of Argand.. as you are MHF Expert, MHF Helper, have 560 Thanks and an Award and hence you are a great Mathematician. But I am a fool because I was thinking about complex numbers' origin right from scratch and trying to understand basics and questioned YOU (The Great Plato) for such a low level doubt.

    Anyways, "Iota"(Begins with 'I') is the 9th letter of the Greek alphabet, and the lowercase iota(begins with 'i') symbol is sometimes used to write the imaginary unit. Iota. I found this information on wikipedia and in some books on complex numbers. But yeah, I am a fool because I believed wikipedia, some books and what I studied in school that \sqrt{-1} could also be denoted by \iota than believing you, who is MHF Expert, MHF Helper, has 560 Thanks and an Award. Wikipedia and book writers are wrong and you are right.

    While you were making a fool out of me and showing off, most of the questions were still unanswered. But yeah, you have earned the right to piss me off and show off as you are MHF Expert, MHF Helper, have 560 Thanks and an Award, and I am a fool as I do not have any of those things... I am sorry that your EGO was hurt because of my language.
    Last edited by pranjvas; February 20th 2013 at 09:51 AM.
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    Re: Complex Numbers

    Quote Originally Posted by pranjvas View Post
    just wanted to ask u a question.. maybe a bit off discussion... but... if i say that i have an object whose mass is iota kgs, or i say that an object with length iota meters... does that mean anything???or is it absurd??? what do u think??? if it is absurd, then, keeping mathematical bookish formulas apart... what does the no. "iota" signify according to u?? if i cant use it for assigning quantity, what can i use it for??
    There are a number (bad pun!) of uses of imaginary numbers in Physics...It is found in dispersion relations, resistance calculations in electricity, etc. My favorite one by far is that the mass of a tachyon, which is a (so far undetected) particle that travels faster than the speed of light. The mass of a tachyon is imaginary.

    -Dan
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