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Homework Help: Basic questions of complex number's geometry?

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    (z' represent conjugate of complex number z,i is iota =sqrt(-1))
    (1)find the locus of z.
    |z|2+4z'=0

    (2)|z1|=1, |z2|=2, then find the value of |z1-z2|2+|z1+z2|2

    (3)z=(k+3)+i[sqrt(3-k2)] for all real k. find locus of z
    2. Relevant equations

    |z|2=z.z'

    3. The attempt at a solution

    (1) |z|2+4z'= |z|2+4z'*z/z
    |z|2{1+4/z}=0
    it is possible iff z= -4. so it has one solution.

    (2) I take z1= cos(a)+i sin(a) and z2=2{cos(b)+i*sin(b)}
    after putting these values in the required equation i got 10.

    (3) let z=x+iy
    x=k+3; y=sqrt(3-k2)
    x-3=k; y2=3-k2
    squaring x and adding it to y2,
    (x-3)2+y2=k2+3-k2
    that's circle.

    unfortunately all of my answers are incorrect.
    I want to know why?
     
  2. jcsd
  3. Nov 18, 2011 #2

    fzero

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    For (1), you are missing one solution. You have two factors and the equation is true when either one is zero. [itex]z=-4[/itex] is the solution where the 2nd factor is zero, what is the solution when the 1st factor vanishes?

    For (2), it would help to expand as

    [itex] | z_1-z_2|^2 + | z_1+z_2|^2 = (z_1-z_2) (\bar{z}_1-\bar{z}_2)+ (z_1+z_2) (\bar{z}_1+\bar{z}_2),[/itex]

    then simplify the resulting expression.

    For (3), you made a mistake to try to compute (x-3)2+y2. (x-3)2 does not appear in |z|2. Instead, it might help to look at the real and imaginary parts independently as functions of k. Looking at some special points like [itex]k=0,\pm 3[/itex] also helps.
     
  4. Nov 19, 2011 #3
    (1) Oh it was y foolishness.

    (2) even after opening the equation you place i got 2(|z1|2+|z2|2) =2(1+22)=10

    (3)I did not understand third answer.

    THANKS FOR REPLY first 2 answer really helps me.
    In third; 3 and-3 lies on the circle but not satisfy the complex number. so i think it is part of circle for all those for which [itex]k<|\frac{1}{\sqrt{3}}|[/itex]
     
  5. Nov 19, 2011 #4

    HallsofIvy

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    For the third one, [itex]z= x+ iy= k+3+ (3- k^2)i[/itex].

    x= k+ 3, [itex]y= \sqrt{3- k^2}[/itex]. Solve the first equation for k, the replace k in the second equation with that.
     
  6. Nov 19, 2011 #5

    ehild

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    And what is the locus of z when 3-k2<0? You need to give the points for all real k values.

    ehild
     
  7. Nov 19, 2011 #6
    see this figure
    this is circle(in real part)
    but why it is different from this figure
     
  8. Nov 20, 2011 #7

    ehild

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    I do not really understand your pictures. In case 3-k2<0 z is real. (The imaginary part is zero.)

    ehild
     
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