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Math Help - find zeros of (1-z)^6 = (1+z)^6

  1. #1
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    find zeros of (1-z)^6 = (1+z)^6

    Hi all,

    I don't know if this is the right forum.
    I have the following equation and I need to find all the solutions in C.
    \Large (z-1)^6=(z+1)^6

    I'm not supposed to mulitply anything out.
    \Large 1=\ (\frac{1-z}{1+z})^6

    I think this is right. What can I do next?
    Can I simplify this further? Subsitute z for the polar form?
    Thanks
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  2. #2
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    Re: find zeros of (1-z)^6 = (1+z)^6

    This is a trick which is supposed to make you think about exponents.

    -1^{even  exponent}=1 ... always. 6 is obviously an even exponent.

    The reason for this is that:

    1^6=-1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \lor 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1

    The opposite is true with an uneven value in the exponent: Then you don't have a minus for every other minus.

    The general rule: \sqrt{x^2} = \pm x

    So basically you can probably think out the solution from there, I think.

    Last edited by JacobOCD; January 20th 2013 at 10:03 AM.
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    Re: find zeros of (1-z)^6 = (1+z)^6

    LOL, sorry, I misread your post. I don't think my answer is relevant.
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  4. #4
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    Re: find zeros of (1-z)^6 = (1+z)^6

    Quote Originally Posted by JacobOCD View Post
    This is a trick which is supposed to make you think about exponents.
    -1^{even  exponent}=1 ... always. 6 is obviously an even exponent.
    The reason for this is that:
    1^6=-1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \lor 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1

    The general rule: \sqrt{x^2} = \pm x
    There are several mistakes in that reply.

    First The general rule: \sqrt{x^2} \ne \pm x !!!

    This is true: The general rule: \sqrt{x^2} = |x|
    I don't think that it is a trick question at all.


    As to the OP, I don't understand why "I'm not supposed to multiply anything out."

    As you can see here there are five solutions.
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    Re: find zeros of (1-z)^6 = (1+z)^6

    I forgot. What does your notation imply? The same as what I meant? The thing that I tried to imply was that the possible solution to for instance \sqrt{4}= \pm 2, as both 2^2 and (-2)^2 would yield 4. I know that this is somewhat irrelevant, but I would like to know what is "wrong".
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    Re: find zeros of (1-z)^6 = (1+z)^6

    Quote Originally Posted by JacobOCD View Post
    I forgot. What does your notation imply? The same as what I meant? The thing that I tried to imply was that the possible solution to for instance \sqrt{4}= \pm 2, as both 2^2 and (-2)^2 would yield 4. I know that this is somewhat irrelevant, but I would like to know what is "wrong".
    If you are going to try and help, then you need to know the basics.

    \sqrt{4}=2~\&~\sqrt{4}\ne -2.

    If the question were what are the square roots of 16 then the answer would be \pm 4.

    However the symbol \sqrt{16}=4.
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    Re: find zeros of (1-z)^6 = (1+z)^6

    "Square roots of 16" isn't the same as 16 under a square root symbol? But yeah, I do get your point, I would actually define the square roots of 4 and 16 as 2 and 4, but there is always the at least theoretical posibility of the value coming to be on the basis of a n, even quantity of negative numbers multiplied together, giving -x^n=y, y being positive.
    Last edited by JacobOCD; January 20th 2013 at 10:59 AM.
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  8. #8
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    Re: find zeros of (1-z)^6 = (1+z)^6

    No, it's not. "16 under a square root symbol" is 4, by definition of "square root". The "square roots of 16" (note the plural) are -4 and 4.

    director, if you are not allowed to multiply out, then, starting from \left(\frac{1- z}{1+ z}\right)^6= 1, note that the 6 6th roots of 1 are 1, cos(\pi/3)+ i sin(\pi/3)= 1/2+ i\sqrt{3}/2, cos(2\pi/3)+ i sin(2\pi/3)= -1/2+ i\sqrt{3}/2, -1, -1/2- i\sqrt{3}/2, and -1/2+ i\sqrt{3}/2.

    Set \frac{1- z}{1+ z} equal to each of those and solve the resulting linear equation.
    Last edited by HallsofIvy; January 20th 2013 at 01:06 PM.
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