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Modulus of the difference of two complex numbers

  1. Apr 30, 2013 #1
    Hi guys,
    I've been trying to help a friend with something that I learnt in class but I'm now finding it hard to solve myself. The problem goes as follows:

    Use geometry to show that |z3-z-3| = 2sin3θ

    For z=cisθ, 0<θ<∏/6

    Now, I chose ∏/12 as my angle and plotted all this on an Argand diagram, but I don't know how to prove that |z3-z-3| = 2sin3θ in such general form. I know that it is right, but I don't know how to get there. Right now I have a isosceles triangle with two lengths known and two angles known (both 3θ). I know the triangle is right-angled but I don't know if I can use this information since that is specific to my chosen angle, right?

    I've spent quite some time on it and I'd really like to understand it. So far I've got xsin3θ=1 where x = |z3-z-3|.

    Please help!
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2


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    Do you understand what "cis" means? This takes about one line from that. You don't need to do anything at all with "Argand diagrams". If z= cis(θ) then [itex]z^3= cis(3\theta)[/itex] and [itex]z^{-3}= cis(-3\theta)[/itex] so that [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)[/itex]. Write that out in terms of the definition of "ciz".

  4. May 1, 2013 #3
    Could you explain this a little more?

    I managed to solve it by drawing a rhombus with |[itex]z^3- z^{-3}[/itex]|being the vertical length along the imaginary axis from the origin to the point [itex]z^3- z^{-3}[/itex] (by definition) then drawing a perpendicular bisector (that is, parallel to the real axis) which created four right-angled triangles with their 90° angles joined to the midpoint between the origin and point [itex]z^3- z^{-3}[/itex]. Using one of these triangles, I determined the length of OM was [itex]sin3θ[/itex] and therefore O to [itex]z^3- z^{-3}[/itex] was twice that since M is the midpoint. I solved it like this since the problem stated it needed to be solved using geometry specifically.

    I'll attach a picture to show you what I mean. It doesn't show in it but, line OA is equal in length to [itex]z^{3}[/itex]

    Attached Files:

  5. May 1, 2013 #4


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    You didn't say "the problem stated it needed to be solved using geometry specifically"!

    My idea was [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ sin(3\theta)- (cos(-3\theta)+ sin(-3\theta)[/itex][itex]= cos(3\theta)+ sin(3\theta)- cos(3\theta)+ sin(3\theta)= 2sin(3\theta)[/itex].
  6. May 2, 2013 #5
    So I assumed I had to do it in a way similar to the way I did.

    This is probably stupid but when I do it that way, I end up with [itex]2isin3θ[/itex] since:
    [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ isin(3\theta)- (cos(-3\theta)+ isin(-3\theta)[/itex][itex]= cos(3\theta)+ isin(3\theta)- cos(3\theta)+ isin(3\theta)= 2isin(3\theta)[/itex]

    My class just recently started our complex number topic and so my knowledge of [itex]cis[/itex] is extremely basic. Should I not be writing [itex]z^3[/itex] as [itex]cos3θ+isin3θ[/itex].

    Thanks for your input.
  7. May 2, 2013 #6
    The question is asking for the modulus so a factor of i does not change the answer.

    Edit: You asked for a geometric reason, so plot any z with the given constraint. z^3 will have the same magnitude (1) and will make an angle 3θ with the real axis if z makes and angle θ. z^-3 will just be the reflection of z^3 about the real axis. The length |z^3-z^-3| is just the length of the line connecting these two points so sin(3θ) will be half of this.
    Last edited: May 2, 2013
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