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Find solutions of complex equation

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Give under the exponential form all non-zero solutions of (z3) + (4conjugate(z2)) = 0

    2. Relevant equations

    z=x+iy where i2=-1
    zn=rn(cos(nθ)+isin(nθ))

    3. The attempt at a solution

    First i tried expanding by making z=x+iy so x3+3ix2y+4x2-3xy2-8ixy-iy3-4y2=0, but then I have no idea on how to get the solutions out of this mess.
    Then I tried putting it into exponential and using roots:
    z3=-4conjugate(z2)
    z=(-4)1/3r2/3(cos((2θ+2πk)/3)-sin((2θ+2πk)/3))

    And now I have no idea what to do... :S

    Thank you for your time
     
  2. jcsd
  3. Sep 28, 2011 #2

    dynamicsolo

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    This is more like the thing to do (especially in view of the problem statement), but why did you only apply polar form to one side of the equation? You could write

    [tex]( r \cdot e^{i \theta} )^{3} = -4 \cdot ( r \cdot e^{- i \theta} )^{2} . [/tex]

    [Conjugates are easy to deal with in polar form.]

    And you really don't want to work in Cartesian coordinates in solving equations like this in the complex plane, if you can help it: it's an expressway to Dementia...
     
  4. Sep 28, 2011 #3
    Alright I think I've managed to get to -4=r(cos(5θ)+isin(5θ))... Is that possible? :P

    Since the question asks for solutions, do I have to find values of r and θ that satisfy this equation? How do I go on from there?
     
  5. Sep 28, 2011 #4

    dynamicsolo

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    Does [itex]r = \frac{-4}{e^{i \cdot 5\theta}} [/itex] sound like it could be a polar curve? (Remember that "negative radii" do have a meaning for polar curves.)

    EDIT: I guess I should elaborate a little, since this aspect is new to me too. I would suspect that since r is supposed to be real, the solutions will occur when the imaginary part of the expression becomes zero. So there is a set of solutions ( r, theta ) for this; the arrangement, though, strikes me as akin to plotting a polar curve.

    SECOND REVISIT: If you know what a five-petal rosette looks like, you find something interesting about where the solutions lie...
     
    Last edited: Sep 29, 2011
  6. Sep 29, 2011 #5
    What I don't get is how can I find the values of theta to make the equations equal to r... what is r?? I've always found solutions using square roots but I don't see how this can be relevant here...
     
  7. Sep 29, 2011 #6

    dynamicsolo

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    If we go back to your form of the solution, [itex]r = -4 \cdot e^{-i \cdot 5\theta} = -4 \cdot [ \cos(5\theta) - i \sin(5\theta) ] [/itex], the radius is real when the imaginary part is zero, for [itex]\theta = \frac{k \pi}{5} [/itex] . All of the solutions then have modulus 4 , since [itex]\cos(5\theta) = \pm 1 [/itex] at those angles.

    So, for [itex]\theta = 0 , r = -4 [/itex], which puts it at ( 4, pi ) on a polar diagram (corresponding to z = -4) ; for [itex]\theta = \frac{\pi}{5} , r = 4 [/itex] ; etc. I believe you'll find that there are five distinct solutions.
     
  8. Sep 29, 2011 #7
    I think I get it now... but one last thing: wouldn't there only be two solutions? z=-4 and z=4? Because sin is only equal to 0 at 0,π,2π,etc., so the angle will vary between π and 2π. But 0 and 2π are located at the same place on the complex graph. So when the angle is equal to 2π, isn't it equivalent as it being equal to 0?
     
  9. Sep 29, 2011 #8

    dynamicsolo

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    Sorry for the delay -- I had to return to working with students face-to-face… I also was chasing an elusive minus-sign in this.

    I'm going to return to this equation: [itex]r = -4 \cdot [ \cos(5\theta) - i \sin(5\theta) ] [/itex]. Since we require r to be real, the imaginary part will be zero when [itex]\sin (5\theta) = 0 \Rightarrow 5\theta = k \pi \Rightarrow \theta = \frac{k \pi}{5} .[/itex] So under that condition,

    [tex]r = -4 \cos (k \pi) = -4 \cdot ( -1 )^{k} , [/tex]

    which leads us to the solutions

    [tex]z = 4 \cdot ( -1 )^{k+1} \cdot cis (\frac{k \pi}{5}) . [/tex]

    [Incidentally, z = 4 is not a solution, since that gives 43 = 64 , but [itex] 4 \overline {z^{2}} = 4 \cdot (4^{2}) = 64 [/itex] , so the terms don't cancel in the original equation.]

    We now find from this that

    [tex]z^{3} = 4^{3} \cdot [ ( -1 )^{k+1} ]^{3} \cdot [ \cos (\frac{3k \pi}{5}) + i \cdot \sin (\frac{3k \pi}{5}) ] = 64 \cdot ( -1 )^{3(k+1)} \cdot [ \cos (\frac{3k \pi}{5}) + i \cdot \sin (\frac{3k \pi}{5}) ] = 64 \cdot ( -1 )^{(k+1)} \cdot [ \cos (\frac{3k \pi}{5}) + i \cdot \sin (\frac{3k \pi}{5}) ] ,[/tex]

    while

    [tex]z^{2} = 4^{2} \cdot [ ( -1 )^{k+1} ]^{2} \cdot [ \cos (\frac{2k \pi}{5}) + i \cdot \sin (\frac{2k \pi}{5}) ] = 16 \cdot ( -1 )^{2(k+1)} \cdot [ \cos (\frac{2k \pi}{5}) + i \cdot \sin (\frac{2k \pi}{5}) ] = 16 \cdot 1 \cdot [ \cos (\frac{2k \pi}{5}) + i \cdot \sin (\frac{2k \pi}{5}) ] [/tex]

    [tex]\Rightarrow -4 \cdot \overline{z^{2}} = -64 \cdot [ \cos (\frac{2k \pi}{5}) - i \cdot \sin (\frac{2k \pi}{5}) ] . [/tex]

    We could use trig identities to show that [itex]z^{3} = -4 \cdot \overline{z^{2}}[/itex] are the same, but it might be as quick to just examine the solutions, which you can check out. The distinct values only run from k = 0 to k = 4 , since

    z0 = 4 · (-1)0 + 1 cis (0) = z5 = 4 · (-1)5 + 1 cis (pi) = -4 , and so on ; points beyond k = 4 fall onto already existing points.

    As one example, if we test, say, k = 3 in the results for the two terms, we find

    [tex]z_{3}^{3} = 64 \cdot ( -1 )^{(3+1)} \cdot [ \cos (\frac{9 \pi}{5}) + i \cdot \sin (\frac{9 \pi}{5}) ] = 64 \cdot [ \cos (\frac{9 \pi}{5}) + i \cdot \sin (\frac{9 \pi}{5}) ] [/tex]

    versus

    [tex] -4 \cdot \overline{z_{3}^{2}} = -64 \cdot [ \cos (\frac{6 \pi}{5}) - i \cdot \sin (\frac{6 \pi}{5}) ] = -64 \cdot [ \cos (\frac{4 \pi}{5}) - i \cdot [-\sin (\frac{4 \pi}{5})] ] = -64 \cdot [ \cos (\frac{4 \pi}{5}) + i \cdot \sin (\frac{4 \pi}{5}) ] [/tex]

    [tex] = -64 \cdot -[ \cos (\frac{9 \pi}{5}) + i \cdot \sin (\frac{9 \pi}{5}) ] = 64 \cdot [ \cos (\frac{9 \pi}{5}) + i \cdot \sin (\frac{9 \pi}{5}) ] ,[/tex]

    by applying the symmetry properties of sine and cosine. The other four solutions are easier to check.

    As I mentioned earlier, if you plot the polar curve [itex]r = -4 \cos (5\theta) [/itex], you'll get a five-petal rosette: the solutions lie at the tips of each petal.
     
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