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Thread: Complex Number

  1. #1
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    Complex Number

    Let z_1,z_2,z_3 \in \mathbb{C} which satisfy z_1 + z_2 + z_3 = 0 and  |z_1| = |z_2|=|z_3|=1. Show that z_1^2 + z_2^2 + z_3^2 = 0.

    Any help would be appreciated.
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  2. #2
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    Re: Complex Number

    let $z_k = e^{i \theta_k}$

    consider what the sum being zero implies for the sum of the $\theta_k$'s

    then note that

    $z_k^2 = e^{ i 2\theta_k}$

    and consider what that means given the original sum of the $\theta_k$'s
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  3. #3
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    Re: Complex Number

    Quote Originally Posted by romsek View Post
    let $z_k = e^{i \theta_k}$
    consider what the sum being zero implies for the sum of the $\theta_k$'s
    $z_1+z_2+z_3=\exp(\bf{i}\theta_1)+\exp(\bf{i} \theta_2)+\exp(\bf{i}\theta_3)=0$

    That does imply that $ \cos(\theta_1)+\cos(\theta_2)+\cos(\theta_3)=0~\&~ \sin(\theta_1)+\sin(\theta_2)+\sin(\theta_3)=0$
    because real and imaginary parts are zero.

    But what implies anything about $\theta_1+\theta_2+\theta_3~?$
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  4. #4
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    Re: Complex Number

    Quote Originally Posted by Plato View Post

    But what implies anything about $\theta_1+\theta_2+\theta_3~?$
    I have the same question
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    Re: Complex Number

    Quote Originally Posted by dedust View Post
    I have the same question
    Are you sure that you have given all the details of this question?
    Are there other details about the $z_k$?
    From the given we do know the set $\{z_1,z_2,z_3\}$ is a subset of the unit circle.
    Also we know that $(z_1+z_2+z_3)^2=0$.

    Can you rearrange that, see here.
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  6. #6
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    Re: Complex Number

    Quote Originally Posted by Plato View Post
    Are you sure that you have given all the details of this question?
    Are there other details about the $z_k$?
    From the given we do know the set $\{z_1,z_2,z_3\}$ is a subset of the unit circle.
    Also we know that $(z_1+z_2+z_3)^2=0$.

    Can you rearrange that, see here.
    Yes, that is the complete question.
    I've tried to manipulate $(z_1+z_2+z_3)^2$, but still got nothing
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  7. #7
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    Re: Complex Number

    The idea I had in my head which I failed to properly describe is that as these are all unit vectors they are arranged so their vector sum is 0.

    Plato properly described this as the two sets of equations relating the cosines and sines of the angles to each other.

    Now, given 3 complex numbers we can rotate the unit circle until one of them corresponds with $1+0i$, i.e. $\theta = 0$

    Then we can solve for the other two angles given that the cosines and sines sum to zero respectively.

    For angles $0 \leq \theta < 2\pi$ there is only one solution

    $\theta_1 = 0$

    $\theta_2 = \dfrac{2\pi}{3}$

    $\theta_3 = \dfrac{4\pi}{3}$

    When we square each complex number $\theta$ is doubled and we end up with angles of

    $0, ~\dfrac{4\pi}{3},~\dfrac{8\pi}{3}=\dfrac{2\pi}{3}$

    and we can see we have our original angles just reordered.

    We already know the cosines and sines of these sum to 0 and thus we're done.
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  8. #8
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    Re: Complex Number

    using a,b,c for the three complex numbers

    a+b=-c and \bar{a} + \bar{b} = -\bar{c} (conjugates)

    Multiplying gives

    1+a \bar{b} + \bar{a} b+1=1

    a \bar{b} + \bar{a} b=-1

    Multiply both sides by a b

    a^2+b^2=-a b

    on the other hand

    (a+b)^2=a^2+b^2+2a b=c^2

    Eliminate the a b term

    a^2+b^2+c^2=0
    Last edited by Idea; Mar 17th 2017 at 12:56 AM.
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