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Complex Number, hard to solve

  1. Sep 1, 2004 #1
    The question is :
    The complex number [tex]z[/tex] is given by
    z = 1 + cos (theta) + i*sin (theta)
    where -Pi < theta <= Pi
    Show that for all values of theta, the point representing z in a Argrand Diagram is located on a circle. Find the centre and radius of the circle.

    Note that i understand perhaps some of you don't understand the symbol i used above, hence i will explain it here :
    smaller or equal to : <=
    theta : an angle
    Pi : a usefull constant in circle, i dont know how to define it.
    i = square root of -1
     
  2. jcsd
  3. Sep 1, 2004 #2

    arildno

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    Dearly Missed

    Hint: go into the the reference frame with translated origin at 1+0i.
     
  4. Sep 1, 2004 #3

    matt grime

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    I think we all new what the symbols meant, if not we'd probably not be able to help...

    do you know the general formula for a circle in the complex plane? It's the set S ={ z | |z-w|=r} for some fixed w in the plane and r a positive constant. Do any choices of w and r spring to mind given we know that cos^2+sin^2 is identically 1?

    Of course there are other ways to see the answer since it is translation of the unit circle (centred on the origin).
     
  5. Sep 1, 2004 #4
    Solution

    Let me denote
    : theta as x
    :z* the complex conjugate of z
    :^ denotes power x^2 ==x squared



    then
    Z=1+Cosx+iSinx
    Z*=1+Cosx-iSinx

    ZZ*=2 + 2Cosx^2 //z multiplied by (z*) will give you that

    ZZ*-Z-Z*=
    =2 + 2Cosx^2 - 1 - Cosx - iSinx - 1 - Cosx + iSinx
    =2

    therefore
    ZZ* - Z - Z* + 1=3

    ==>Z(Z*-1) - (Z*-1) = 3
    ==>(Z-1)(Z*-1) = 3
    ==>(Z-1)(Z-1)*=3
    ==>|Z-1|^2=3
    ==>|Z-1|=real
    Which is the general form of a circle in complex plane as matt grime pointed out.

    Hope you found this solution useful :shy:

    cheers
    poolwin2001

    P.S:Where did you get this question ??
     
  6. Sep 1, 2004 #5
    This any help ?

    [itex]
    \mbox{\Huge
    \[
    e^{i\theta } = cos\theta + isin\theta
    \]}
    [/itex]

    Best
     
  7. Sep 2, 2004 #6
    hard to understand

    First of all, thanks to those people who had post reply to my question, your message is meaningful to me, thanks you.
    Well, this question i get it from my work book in complex number exercise. Note that i am not European people, i live in South East Asian, hence i don't think is meaning to you if i told you which book i found this question, ok poolwin2001 ?

    matt grime, i am just begin the Complex Number chapter, hence i really don't know got such Set which define any circle in the Argand Diagram. And to poolwin2001 also, i understand all your working but just don't understand how can u state that once u able to show |Z-1|=real , this implies that the point of complex number Z is located on a circle in Argand Diagram.

    I just begin study this chapter, hence i don't think that the author expect me to use this kind of "theorem" ( which i think is under Further Mathematic ) to show Z is located in a circle, do you have any other easier method ?
    For your information, i just study the operation of Complex Number (plus, minus, multiply and divide), conjugate ,draw Argand Diagram, that is all in my syllabus, and that question i was found it inside my textbook (so i sure the author will expect us use the method i study to solve it, and not using those "theorem" )
     
  8. Sep 2, 2004 #7

    matt grime

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    The idea that these points define a circle doesn't have anything to do with the complex numbers as such.

    You should be thinking of the numbers as points in the complex plane, yes? then | | means distance. so {z| |z-w|=r} means all the points z a distance r from w. This is just geometry in the plane, but with nicer notation.
     
  9. Sep 2, 2004 #8
    what matt grime probably asked you to do was to use the trigonometry identity.

    Let z= x + iy
    =[tex] (1+cos\theta) + isin\theta[/tex]

    [tex]\Rightarrow x = 1+cos\theta[/tex]
    [tex]\Rightarrow y = sin\theta[/tex]

    [tex]sin^2\theta + cos^2\theta = 1[/tex]
    [tex]y^2 + (x-1)^2 = 1[/tex]

    Compare with the circle equation : [tex] (x-a)^2 + (y-b)^2 = r^2[/tex]
    with center coordinate of (a,b) and radius r.
     
  10. Sep 2, 2004 #9
    I think I got the explanation down in the zip file (couldn't upload the image directly). Hope it helps, sure took me a long time to create it.
     

    Attached Files:

  11. Sep 5, 2004 #10
    wow 3 different solutions

    WOW 3 different solutions:bugeye:
    Great !3 different solns each radically different have been given .Awesome!!
    (But mine had to be the longest :frown: :cry: )
     
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