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Math Help - Solution set using De Moivre's theorem

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    Solution set using De Moivre's theorem

    Let z = \frac{128}{\sqrt2}(1 + i). What are all possible solutions of:

    w = z^{1/7}

    So, after converting to polar form I have:

    z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{  \pi}{28}\right))

    I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by terrorsquid View Post
    Let z = \frac{128}{\sqrt2}(1 + i). What are all possible solutions of: w = z^{1/7}
    z^{1/7} =2(cos\left(\frac{\pi}28}\right)+i~sin\left(\frac{  \pi}{28}\right))

    I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?
    2^{\frac{{13}}{{14}}}(cos\left(\frac{\pi}{28} + \frac{2\pi k}7}\right)+i~sin\left(\frac{\pi}{28}+\frac{2\pi k}{7}\right)) for k=0,1,\cdots,6
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by terrorsquid View Post
    Let z = \frac{128}{\sqrt2}(1 + i). What are all possible solutions of:

    w = z^{1/7}

    So, after converting to polar form I have:

    z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{  \pi}{28}\right))

    I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?
    The crucial step is to write z = 2^7\bigl(\cos(\tfrac\pi4 + 2k\pi) + i\sin(\tfrac\pi4 + 2k\pi)\bigr).
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by Opalg View Post
    The crucial step is to write z = 2^7\bigl(\cos(\tfrac\pi4 + 2k\pi) + i\sin(\tfrac\pi4 + 2k\pi)\bigr).
    Hmm I still seem to be missing a step in my head. I don't understand how the k multiples of pi got inside the sin and cos functions. I can see that

    128(cos\left(\frac{\pi}{4}\right) + i~sin\left(\frac{\pi}{4}\right)) can be written as 2^7(cos\left(\frac{\pi}{4}\right) + i~sin\left(\frac{\pi}{4}\right))

    but from there I thought I just applied the power \frac{1}{7} to the 2^7 and then multiplied \theta by \frac{1}{7} which gives me:

    2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{  \pi}{28}\right))

    What am I missing in between those steps?
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by terrorsquid View Post
    Hmm I still seem to be missing a step in my head. I don't understand how the k multiples of pi got inside the sin and cos functions. ...
    Consider how sine & cosine are defined on the unit circle.

    Adding an integer multiple of 2π to the argument of any trig function gives the same result as the original argument.
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by terrorsquid View Post
    Let z = \frac{128}{\sqrt2}(1 + i). What are all possible solutions of:

    w = z^{1/7}

    So, after converting to polar form I have:

    z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{  \pi}{28}\right))

    I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?
    It helps if your realise that all the solutions will be evenly spaced about a circle. So they all have the same magnitude and are all separated by an angle of \displaystyle \frac{2\pi}{7}. Also note that the principal solutions will be in the domain \displaystyle \theta \in (-\pi, \pi].
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    Re: Solution set using De Moivre's theorem

    Ok, I think I get the process. My understanding is very superficial though. I only learned how to "apply the formula" and didn't cover what I am actually doing by solving one of these which is what I think my problem is.

    Take a similar example:

    Let~z = 8(-1 + i\sqrt3). Find all the solutions for:

    w = z^{1/4}

    I convert to polar form:

    16(-cos\left(\frac{2\pi}{3}\right) + isin\left(\frac{2\pi}{3}\right))

    I notice that 16 = 2^4 so I figure there will be 4 solutions (not sure why).

    I then apply the exponent 1/4 to get:

    2(-cos\left(\frac{\pi}{6}\right) + isin\left(\frac{\pi}{6}\right))

    Then to get my solutions I just generate 4 answers by multiplying the exponent 1/4 by 2k\pi and adding it to \frac{\pi}{6} cycling k from 0 - 3 to get:

    2(-cos\left(\frac{\pi}{6} + \frac{2k\pi}{4}\right) + isin\left(\frac{\pi}{6}+ \frac{2k\pi}{4}\right)) = \frac{\pi}{6}, \frac{2\pi}{3}, \frac{7\pi}{6}, \frac{5\pi}{3}

    I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the 2k\pi increments, and I seem to be able to generate the answers for these types of problems, I just feel awkward not knowing why I'm doing each step if you understand?

    Will these problems always have a coefficient that can be expressed as an integer raised to a power so you can determine how many answers there should be?
    Last edited by terrorsquid; August 23rd 2011 at 09:16 AM.
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    Re: Solution set using De Moivre's theorem

    Quote Originally Posted by terrorsquid View Post
    I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the 2k\pi increments, and I seem to be able to generate the answers for these types of problems, I just feel awkward not knowing why I'm doing each step if you understand?
    Will these problems always have a coefficient that can be expressed as an integer raised to a power so you can determine how many answers there should be?
    There are n nth roots of a complex number z=|z|\mathbf{cis}(\theta)
    Here  \mathbf{cis}(\theta) is shorthand for \cos(\theta)+\mathbf{i}\sin(\theta) where
    \theta=\text{Arg}(z).

    Those n roots are |z|^{1/n}\mathbf{cis}\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right) where k=0,1,\cdots,n-1.

    Those n numbers are all on a circle centered at (0,0) with radius |z|^{1/n}.
    Moreover, those points are equally spaced on arcs subtended by angles of \frac{2\pi}{n}.
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