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Math Help - Complex Solution Set

  1. #1
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    Complex Solution Set

    Find all complex solutions of z^5=(z+1)^5. I guess it's possible to just substitute a+bi, collect terms, and expand, but the solution intimated a quicker way. Anyone willing to crack it receives muchos kudos from me.
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  2. #2
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    Quote Originally Posted by mstrfrdmx View Post
    Find all complex solutions of z^5=(z+1)^5. I guess it's possible to just substitute a+bi, collect terms, and expand, but the solution intimated a quicker way. Anyone willing to crack it receives muchos kudos from me.
    Edit
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  3. #3
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    I can't see the edit though.
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    Quote Originally Posted by mstrfrdmx View Post
    Find all complex solutions of z^5=(z+1)^5.
    There may be a clever trick. I don't see it right off.
    But this may save you some work.
    w^5-z^5=(w-z)(w^4+w^3z+w^2z^2+wz^3+z^4).
    Letting w=z+1 the first factor is 1.
    Then we get a polynomial in z^4.
    There are four complex answers.
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  5. #5
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    Better still just expand it.
    (z+1)^5=z^5+5z^4+10z^3+10z^2+5z+1
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    Yeah, that's what I had done. The solution is \frac{1}{w-1} where w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4. I just have no idea what to do after putting everything into polar form.
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  7. #7
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    Quote Originally Posted by mstrfrdmx View Post
    Yeah, that's what I had done. The solution is \frac{1}{w-1} where w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4. I just have no idea what to do after putting everything into polar form.
    \displaystyle w_k=r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]
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  8. #8
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    Quote Originally Posted by mstrfrdmx View Post
    Yeah, that's what I had done. The solution is \frac{1}{w-1} where w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4. I just have no idea what to do after putting everything into polar form.
    I can help you a bit more.
    \dfrac{1}{w-1}=\dfrac{\overline{w}-1}{|w-1|^2}

    and \overline{e^{u}}=e^{\overline{u}}.

    I still do not know where your answer comes from.
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  9. #9
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    I may just be slow, but I don't really see how the nth roots formula helps in this case.
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  10. #10
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    Quote Originally Posted by mstrfrdmx View Post
    Yeah, that's what I had done. The solution is \frac{1}{w-1} where w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4. I just have no idea what to do after putting everything into polar form.
    I'm not sure if you meant that you got this answer or you just know it's the answer and need to get to it.

    If you want to know how to get to it, this may help:

    \displaystyle{z^5=(1+z)^5 \Leftrightarrow (1+z^{-1})^5=1}

    which means all the solutions are 1+z^{-1}=w_k where the w_k

    are the roots of unity of order 5. Equivalently, z=\frac{1}{w_k - 1}
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  11. #11
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    Quote Originally Posted by Unbeatable0 View Post
    I'm not sure if you meant that you got this answer or you just know it's the answer and need to get to it.

    If you want to know how to get to it, this may help:

    \displaystyle{z^5=(1+z)^5 \Leftrightarrow (1+z^{-1})^5=1}

    which means all the solutions are 1+z^{-1}=w_k where the w_k

    are the roots of unity of order 5. Equivalently, z=\frac{1}{w_k - 1}
    You the coolest, doc. Muchos kudos.
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