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Solve (z+i)^5 + (z-i)^5 = 0

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    (z+i)5 + (z-i)5 = 0

    I've tried everything I can think of with this question, and I just keep getting that z = 0.

    Can someone help me get started with the proper method?
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    Have you tried expanding out the brackets and getting a polynomial in z?
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    (Original post by ForGreatJustice)
    Have you tried expanding out the brackets and getting a polynomial in z?
    I really wouldn't recommend this method.
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    (Original post by 99wattr89)
    (z+i)5 + (z-i)5 = 0

    I've tried everything I can think of with this question, and I just keep getting that z = 0.

    Can someone help me get started with the proper method?
    Rewrite as (z+i)^5 = -(z-i)^5.

    By considering the modulus of each side, show that z must be real.

    Once you know z is real, suppose \arg(z+i) = \theta. By drawing a diagram, work out what \arg(z-i) must be.

    Finally compare arguments of both sides to find what arg(z+i) must be and hence find z.
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    (Original post by DFranklin)
    Rewrite as (z+i)^5 = -(z-i)^5.

    By considering the modulus of each side, show that z must be real.

    Once you know z is real, suppose \arg(z+i) = \theta. By drawing a diagram, work out what \arg(z-i) must be.

    Finally compare arguments of both sides to find what arg(z+i) must be and hence find z.
    This is very similar to a previous question I asked about, but I couldn't work out what I was trying to prove about z. How did you know that z would be real?

    Also when taking the moduli, should I get |z+i| = -|z-i| ? I can't tell if there should be a minus or not.

    If there is a minus, then I've proven z is real.

    From there I get that arg(z-i) = 2Pi - Theta, and from that I get that Theta = -2Pi -Theta, so that means Theta = Pi

    So that means that (z+i) = reiPi(1+2n), so tan[Pi(1+2n)] = 1/(z+i), so z = cot[Pi(1+2n)] - i

    I think all of that is correct, but I'm struggling to get the desired answer for z. It's meant to come out as z = cot[Pi(1+2n)/10], n = 0, 1, 2, 3, 4.
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    In this case, I knew z would be real because I've helped people do this question about 6 times on here. But it's not hard to show.

    There should not be a minus sign. You have |z+i| = |z-i|. Hint for proving z is real: For two complex numbers a and b, |a-b| is the same as the distance between a and b.

    I get that arg(z-i) = 2Pi - Theta
    Depends on how you define arg, but it's more likely you want arg(z-i) = -theta. But I don't know why you think this implies "Theta = -2Pi -Theta"; it doesn't.

    Obviously your follow on working from this point is incorrect.
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    (Original post by DFranklin)
    I really wouldn't recommend this method.
    Just wondering, why not?

    Looking at the coefficients, you know all the odd powers of i cancel, so you're just left with

    2z^5 -20z^3 + 10z = 2z(z^4 - 10z^2 + 5) which is a quadratic in z^2 and really easy to solve?
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    (Original post by CHY872)
    Just wondering, why not?

    Looking at the coefficients, you know all the odd powers of i cancel, so you're just left with

    2z^5 -20z^3 + 10z = 2z(z^4 - 10z^2 + 5) which is a quadratic in z^2 and really easy to solve?
    Basically because the general case (z+i)^n + (z-i)^n = 0 is a fairly standard problem (that I was fairly sure the OP had asked before).
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    (Original post by DFranklin)
    In this case, I knew z would be real because I've helped people do this question about 6 times on here. But it's not hard to show.

    There should not be a minus sign. You have |z+i| = |z-i|. Hint for proving z is real: For two complex numbers a and b, |a-b| is the same as the distance between a and b.

    Depends on how you define arg, but it's more likely you want arg(z-i) = -theta. But I don't know why you think this implies "Theta = -2Pi -Theta"; it doesn't.

    Obviously your follow on working from this point is incorrect.
    So z is real because the condition |z+i| = |z-i| describes the real axis (is it OK to just state that?), since the real axis is the set of points equidistant from i and -i.


    So, I have that arg(z+i) = Theta = -arg(z-i)

    Also, (z+i)5/(z-i)5 = -1 means that arg[(z+i)5/(z-i)5] = Pi (but strictly speaking it's Pi(1+2n), right? Since there are going to be multiple solutions.)

    So, that means that (z+i)/(z-i) = -1, right? Since the 5th root of -1 is -1.

    So then arg(a/b) = arg(a) - arg(b) means that 2 Theta = Pi so Theta = 45 degrees, so z = 1.

    Unfortunately that's totally wrong. But I can't work out what I'm doing wrong. I just don't know what it is I should do.
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    (Original post by 99wattr89)
    ...
    The explanation for the first part is acceptable, though you may be required for more details.
    As for the other parts, you seem to be skipping an important step and this leads you nowhere.

    After you arrive at the expression

    \displaystyle \left( \frac{z + i}{z - i} \right)^5 = -1

    you could simplify your life a bit by transforming it into

    \displaystyle w^5 = -1, where \displaystyle w = \frac{z + i}{z - i}.

    Spoiler:
    Show

    Now, you have found previously that \displaystyle \arg(w) = \arg\left(\frac{z + i}{z - i}\right) = \arg(z + i) - \arg(z - i) = 2\theta.


    However, your last equation says that \arg(w^5) = \pi (lets say for principal value), and hence 5*(2\theta) = \pi.

    This gives you the first solution, namely \displaystyle \theta = \frac{\pi}{10}.


    Hope this helps.
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    (Original post by jack.hadamard)
    The explanation for the first part is acceptable, though you may be required for more details.
    As for the other parts, you seem to be skipping an important step and this leads you nowhere.

    After you arrive at the expression

    \displaystyle \left( \frac{z + i}{z - i} \right)^5 = -1

    you could simplify your life a bit by transforming it into

    \displaystyle w^5 = -1, where \displaystyle w = \frac{z + i}{z - i}.

    Spoiler:
    Show

    Now, you have found previously that \displaystyle \arg(w) = \frac{\arg(z + i)}{\arg(z - i)} = \arg(z + i) - \arg(z - i) = 2\theta.


    However, your last equation says that \arg(w^5) = \pi (lets say for principal value), and hence 5*(2\theta) = \pi.

    This gives you the first solution, namely \displaystyle \theta = \frac{\pi}{10}.


    Hope this helps.
    That definitely does help, thank you.

    I have Theta = Pi(1+2n)/10
    So arg(z+i) = Pi(1+2n)/10

    The thing is that from there I get that tan Theta = i/z

    Giving z = icot(Theta)

    There shouldn't be an i though, z is real. :confused:
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    (Original post by 99wattr89)
    That definitely does help, thank you.

    I have Theta = Pi(1+2n)/10
    So arg(z+i) = Pi(1+2n)/10

    The thing is that from there I get that tan Theta = i/z

    Giving z = icot(Theta)

    There shouldn't be an i though, z is real. :confused:
    I hope you noticed the hint with w^5 = -1. If you are to find the fifth roots of -1, you do not take the fifth root of both sides first.


    Use your diagram again for the second part.
    To deduce the tangent, you used a triangle with a side of length i?
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    (Original post by jack.hadamard)
    I hope you noticed the hint with w^5 = -1. If you are to find the fifth roots of -1, you do not take the fifth root of both sides first.


    Use your diagram again for the second part.
    To deduce the tangent, you used a triangle with a side of length i?
    I'm sorry, I don't understand.

    I don't know what else to do with w, since I already used it in my last attempt. I also don't understand what you mean when you say that you don't take the 5th root of both sides.

    I did use a triangle with base z and height i.
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    (Original post by 99wattr89)
    I'm sorry, I don't understand.

    I did use a triangle with base z and height i.
    That's fine, I will attempt to explain it otherwise.
    Spoiler:
    Show

    The point I was making with w is nothing more than a notice that you should not take the fifth root of both sides.
    The thing is, if you took the fifth root, you simply say that w = -1 and you are stuck on a completely different equation.


    The basic geometry of the complex plane, in ordinary terms, is no different from simple trigonometry.
    Distances between complex numbers are real, and what you have said in your statement is that |i - 0| = |i| = i which is clearly false.

    Draw the triangle again, the base is z. What is its height again?
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    (Original post by jack.hadamard)
    That's fine, I will attempt to explain it otherwise.
    Spoiler:
    Show

    The point I was making with w is nothing more than a notice that you should not take the fifth root of both sides.
    The thing is, if you took the fifth root, you simply say that w = -1 and you are stuck on a completely different equation.


    The basic geometry of the complex plane, in ordinary terms, is no different from simple trigonometry.
    Distances between complex numbers are real, and what you have said in your statement is that |i - 0| = |i| = i which is clearly false.

    Draw the triangle again, the base is z. What is its height again?
    Ohhh! Of course! Gosh, my face is red!
    I'm really sorry about that, that was terrible of me. Thank you for being magnificently patient, I have the proper answer now.

    [ z = 1/tan(Pi/10 + 2nPi/10) = cot(Pi(1+2n)/10) ]

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Updated: June 11, 2012
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