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Homework Help: Some simple doubts

  1. Aug 25, 2012 #1
    I had a few doubts in mind. Please help.

    1) Is -15 the L.C.M. of -5 and -3?
    2) Can all complex numbers be plotted in a number line.
    3) 1^(1/3)= 1, ω, ω^2,
    where ω= -(1/2)+i[3^(1/2)/2]
    Then, if we plot these 3 values along a number line, will we get 3 different points or will it be the same point?
    4) I don't get the use of complex numbers. I mean, they don't even exist, right? Isn't that why we write √-1=i ?
    Last edited: Aug 25, 2012
  2. jcsd
  3. Aug 25, 2012 #2
  4. Aug 25, 2012 #3
    See the real numbers lie on a 1 dimensional line called real axis .
    You can cannot find an imaginary number on this line .
    And imaginary numbers lie again on a 1 dimensional line called imaginary axis .
    These two groups are then married, in the sense the area or the region covered by them is known as argand plane where the COMPLEX NUMBERS LIE .
    So 1 , w , w^2 , lie on a plane rather than line ,
    1 lies on real axis ,
    w lies somewhat between real and imaginary , as it consists a part of real and imaginary .
    w^2 has the same story as w . Just the imaginary part is mirrored ( multiplied by -1)
  5. Aug 25, 2012 #4
    If 11/3= 1, ω, ω2, does it mean that for every real number x, ther is a complex number y in the argand plane such that x=y?
  6. Aug 25, 2012 #5
    the case you are talking about is a line x=y , flowing through argand plane ,
    that is only the complex numbers whose imaginary and real part are equal will satisfy such a thing .
    I guess you are getting confused by the equation of 1 raised to a third ?
    Giving complex ?
    Is it so ?
    Like a function ( raising to a third) giving out complex variable?
  7. Aug 25, 2012 #6
    Can you please give an example.

    That is what I meant to ask. Is it possible?
  8. Aug 25, 2012 #7
    See in the world without imaginary axis , ie the Real numbers world .
    1^(1/3) = f(x) has one solution that is f(x)=1 .
    But when you consider i=(-1)^(1/2) , a new set of numbers are introduced .
    And in these numbers lie w , and w^2 who say " hey I can also fit in that equation as well , 1 your monopoly is over " , and indeed when we broaded our range f(x)=1, w or w^2 .
  9. Aug 25, 2012 #8
    Now coming to x=y
    Ok, I want you is to take a graph paper , sketch two perpendicular axis .
    Name them as real and imaginary (as per the convention)
    now on the legend write
    z=x+iy ( x is real part , lying on x axis and y is imaginary part lying on y axis )
    and you want is x=y
    you get

    Now this result is a variable mutlipled by a constant (1+i)
    ,put values of x like 1,2,-4 , 7,9,-10 and get values of z .
    After getting values of z plot them , you will find them lying on a straight line .
  10. Aug 25, 2012 #9
    I was thinking whether the real number x could be equal to a complex number y=a+ib, b≠0. When I think graphically, it doesn't seem possible; what with x lying on the x-axis and y lying in the x-y plane.

    What I essentially meant to ask was that if 11/3=1, ω, ω2, is 1=ω=ω2?

    In the example you gave, y was the imaginary number. That was not the scenario I was considering.
  11. Aug 25, 2012 #10
    graphically it is possible ,
    The graph which we draw is of real x and the f(x)
    to make sense we add another line called imaginary axis , which makes a 3d space ,
    but since f(x) has only real values , we have to add another axis g(y) for imaginary values which function takes , since it is difficult to comprehend 4d graph . we make the 4th dimension ie g(y) as a color set which shows different color for different values of g(y).
    I guess you in 11 or 12 class , if you are complex analysis is not a part of your syllabus .for more info
  12. Aug 25, 2012 #11
    If it should be possible, shoudn't the 2 points (real and complex) lie on the same point, same plane? Isn't that necessary? Or is it that the graph I know isn't enough to understand this?
  13. Aug 25, 2012 #12
    yea real and imaginary line on same plane , its just that on the real axis imaginary part is zero . and vice-versa , take point 2,1 that is 2+i it lies on a single point and on argant plane , same for 4,0 or 0,6 .
  14. Aug 26, 2012 #13
    What about -15 being the L.C.M. of -3 and -5? Is that correct?
  15. Aug 26, 2012 #14


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    write √-1=i ?It is possible to say that -15 is a multiple of -3 and -5: (-3)(-5)= 15 but then (-1)(-3)(-5) is -15. However, because there is no "least" negative number (-30 is also a "multiple of -3 and -5": -30= (-2)(-3)(-5) and -30< -15), we use the term "least common multiple" only of positive integers.
    No. Not in a way that 'respect' multiplication. That is 'on a number line' immediately gives an inequality: a< b if and only if a is left of b on the number line. And then we want 'if a< b and 0< c, ac< bc'. But there is no way to order the complex numbers so that is true.

    You can't plot them on a number line as I said in (2). They can be plotted on the 'complex plane' and, yes, they are three different points. Having some property in common (that their cube is 1) doesn't mean they are the same number!

    Do you not realize that your last two sentences contradict one another? We couldn't write "[itex]\sqrt{-1}= i[/itex]" if those things did not exist. Please do not think that just because we call a set of numbers "real number" that other types of numbers "don't exist". "Real numbers", "imaginary numbers", and "complex number" are arbitrary labels and do not imply anything about the numbers themselves.
    Last edited by a moderator: Aug 27, 2012
  16. Aug 27, 2012 #15


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    No, that is not possible. By definition, a 'real number' is a complex number with imaginary part 0. If y= a+ ib, b≠ 0, then y is NOT equal to any real number.

    No, and I can't imagine why you would think so. The three numbers, 1, ω, and ω2 happen to have some property in common (that their cube is 1). Why in the world would you think that having one property in common means they are equal? '2' and '4' are both even numbers. Would you conclude from that that 2= 4? [itex]\sqrt{2}[/itex] and [itex]-\sqrt{2}[/itex] both have the property that their square is equal to two. Would you ask "is [itex]\sqrt{2}= -\sqrt{2}[/itex]"?

  17. Aug 28, 2012 #16
    If x=(-1)^(1/2), we can't know the value of x, can we? That is why we write the answer as i, right? So i is no actual number.
  18. Aug 28, 2012 #17


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    i IS a number, but it's not a REAL number, it's an imaginary one.
  19. Aug 28, 2012 #18
    But we can't ever represent i with numbers. I know (-1)^(1/2) is a representation of i. If we are not allowed to use any notations(like root, sigma, integral etc.), what is i? We don't know(or atleast I don't know)! Maybe thats why we call it an imaginary number?
  20. Aug 28, 2012 #19


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    Yes, we can- once again, "i" is a perfectly good number, as real as, say, 0, 1, -1, or [itex]\sqrt{2}[/itex]. The terms "real number" and "imaginary numbers" are just labels- they do NOT mean that such numbers are more "real" or "imaginary", in the "usual" sense of those words, than any other numbers.

    I'm glad you add that "at least I don't know"! i and, in general, complex numbers are given a precise definition: The best way to define the complex numbers is as pairs of real numbers, (x, y), with addition defined as (a, b)+ (c, d)= (a+ b, c+ d) and multiplication defined as (a, b)(c, d)= (ac- bd, ad+ bc). We then identify the real number, "x", with the complex number (x, 0). Notice then that for a real number, x, times the complex number, (a, b), we have (x, 0)(a, b)= (xa- 0b, 0a+ xb)= (xa, xb) which we can write as just x(a, b). We define "i" to be the pair (0, 1). We can then write (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a+ bi.
  21. Aug 28, 2012 #20
    i cannot be represented on real number line .
    what you want is to get i=0.392594434.......... something right?
    Well mathematics is based on axioms and theorems . ( google them)
    You have to begin some where , and you have to define it .Please have a sound understanding of set theory before reading this .
    modern Mathematics defines real numbers as undefined objects, which follow a certain properties .
    a set R of objects called real numbers satisfy these axioms
    I commutative law x+y=y+z
    II associative law
    III distributive law
    IV existence of identity eg x+0=x and x.1=x
    V existence of negatives (for every real number x there is a real number y such that x+y=0
    VI existence of reciprocal xy=1
    VII x and y are in R+ , so are x+y and xy
    VIII for every real x not equal to 0 , either x belongs to R+ or or -xbelongs to R+ , but not both .
    IX 0 doesnt not belong to R+

    These results might be obvious and taken for granted since elementary classes .
    But as a mathematician you need to realize the beauty of axioms .
    They can not be proved and from these theorems are derived which can be proved or falsified .
    Since Real numbers are properly defined now , there should be no ambiguity .
    Talking about number root of -1 or simply i . We need to find a number which when squared gives a negative number , a simple case of
    x.x=-1 , but again we donot get a concrete definition of what is the number .
    We define complex numbers as ,
    If a and b are real numbers , the pair (a,b) is called a complex number , provided that equality , addition and multiplication as follows:
    a)equality (a,b) = (c , d) means a=c and b=d
    b)sum : (a,b) + (c,d) = (a+c,b+d)
    c) product L (a,b)(c,d)=(ac-bd , ad+bc) .
    This definition regards (a,b) as ordered pairs ,
    thus complex number (2,3) is not equal to (3,2) ,
    The components of (a,b) are termed as
    a is real part of complex number (a,b)
    b is imaginary part of complex number (a ,b )
    NOTE: i doesnot appear anywhere is definition. and is introduced as a particular complex number (0,1) which algebraic properties of being i^2=-1 in other sense
    (0,1)x(0,1) =(-1,0) use axiom c .
    As the complex numbers are extension of real numbers ,
    all real numbers can be represented as (a,0)
    and asking about the representation of complex numbers as a+ib .
    Using axiom ,
    (a,b) is a complex number
    which is algebraically a+ib
    Hope this helps .
    Source:Tom M Apostol
  22. Aug 28, 2012 #21
    Ivy, I wrote the samething, lol
    I was busy compiling when you wrote .
    I am sorry
  23. Aug 28, 2012 #22
    I am starting to get it, I think. Thanks for your replies.
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