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Roots of infinity...lol jk...unity. Watch

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    The question asks,

    Find the fifth roots of unity and show where they lie in the complex plane.
    I'm not much of a 'drawing' person, so I've decided to go about this through 'algebra'. This is what I've done so far,

    z^5=1

    z=r(cos\theta +isin\theta )

    De Moivre's Theorem tells us that,

    z^5=r^5(cos5\theta +isin5\theta ), so,

    r^5(cos5\theta +isin5\theta )=1

    r^5=1 implies r=1

    Consequently,

    cos5\theta +isin5\theta =1

    Equating real and imaginary parts,

    cos5\theta =1, sin5\theta =0

    Thus \theta =0

    5\theta =2k\pi

    \theta=\frac{2k\pi}{5}

    z=(cos\frac{2k\pi}{5}+isin{2k\pi  }{5})

    I'm shtuck... :cry2:

    Could someone help me please? :puppyeyes:
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    z^5=1\Longleftrightarrow z^5=cos(x+2k\pi)+isin(x+2k\pi)

    \Longleftrightarrow z=(cos(x+2k\pi)+isin(x+2k\pi))^{  \frac{1}{5}}

    then de moivre
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    (Original post by Pheylan)
    z^5=1\Longleftrightarrow z^5=cos(x+2k\pi)+isin(x+2k\pi)

    \Longleftrightarrow z=(cos(x+2k\pi)+isin(x+2k\pi))^{  \frac{1}{5}}

    then de moivre
    I've just ended up with,

    z^5=1

    z^5=cos(x+2k\pi )+isin(x+2k\pi )

    z=(cos(x+2k\pi )+isin(x+2k\pi ))^{\frac{1}{5}}

    z=cos\frac{1}{5}(x+2k\pi)+isin \frac{1}{5}(x+2k\pi)

    z=cos\frac{x+2k\pi}{5}+isin\frac  {x+2k\pi}{5}

    Which is basically what I had before, except that now I have the x to complicate matters further.

    Help? :cry:
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    (Original post by RamocitoMorales)
    except that now I have the x to complicate matters further
    you sort of need the x to make your working correct

    now substitute values of k into your expression to obtain your five solutions

    k\in\mathbb{Z}
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    (Original post by Pheylan)
    you sort of need the x to make your working correct

    now substitute values of k into your expression to obtain your five solutions

    k\in\mathbb{Z}
    Should I just ignore the x when substituting in the different values for k. And how would I know which values to substitute?
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    (Original post by Pheylan)
    z^5=1\Longleftrightarrow z^5=cos(x+2k\pi)+isin(x+2k\pi)

    \Longleftrightarrow z=(cos(x+2k\pi)+isin(x+2k\pi))^{  \frac{1}{5}}

    then de moivre
    Um, no...
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    (Original post by Scipio90)
    Um, no...
    what's wrong with it? haven't done this for a while
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    (Original post by Scipio90)
    Um, no...
    So what do I do then?

    I think I was right up to the point where,

    z=(cos\frac{2k\pi}{5}+isin\frac{  2k\pi}{5}).

    I just don't know which values to insert for k, and how to find the roots of unity... :sad:
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    k is any integer.
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    (Original post by Scipio90)
    k is any integer.
    But if \frac{2k\pi}{5} is the principle argument, then one would assume that it must be between \pi and -\pi. And I also need 5 answers. How would I go about getting these? :hmmmm:
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    (Original post by RamocitoMorales)
    So what do I do then?

    I think I was right up to the point where,

    z=(cos\frac{2k\pi}{5}+isin\frac{  2k\pi}{5}).

    I just don't know which values to insert for k, and how to find the roots of unity... :sad:
    Use k as 1, 4,2,3 ,5
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    (Original post by Carlo08)
    Use k as 1, 4,2,3 ,5
    Any particular reason for this? And why no negative numbers? I would have thought it could be something like -2, -1, 0, 1, 2... :hmmmm2:
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    (Original post by RamocitoMorales)
    Any particular reason for this? And why no negative numbers? I would have thought it could be something like -2, -1, 0, 1, 2... :hmmmm2:
    Yeah, depends on whether you want the values of your argument to be between (-pi,pi] or [0,2pi). The values inbetween either of those ranges are commonly known as the principle arguments. Any value of k is valid but you'll want to add or minus 2pi from them until their in the principle range.
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    (Original post by Clarity Incognito)
    Yeah, depends on whether you want the values of your argument to be between (-pi,pi] or [0,2pi). The values inbetween either of those ranges are commonly known as the principle arguments. Any value of k is valid but you'll want to add or minus 2pi from them until their in the principle range.
    So if I just insert -2, -1, 0, 1, 2 for k, would I get my 5 roots? :confused:
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    (Original post by Pheylan)
    z^5=1\Longleftrightarrow z^5=cos(x+2k\pi)+isin(x+2k\pi)

    \Longleftrightarrow z=(cos(x+2k\pi)+isin(x+2k\pi))^{  \frac{1}{5}}

    then de moivre
     1 \not= cos(x+2k\pi)+isin(x+2k\pi) except for when x=0

     1 = cos(0) +isin(0)

    :teehee:
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    (Original post by RamocitoMorales)
    So if I just insert -2, -1, 0, 1, 2 for k, would I get my 5 roots? :confused:
    Yup, remember to check if the answer is in your principle range (which they are)
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    (Original post by Clarity Incognito)
    Yup, remember to check if the answer is in your principle range (which they are)
    I once stole your avatar and left you for dead.

    Thank you for your compassion.
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    (Original post by RamocitoMorales)
    I once stole your avatar and left you for dead.

    Thank you for your compassion.
    I thought I saw someone steal my avatar! Haha, no worries, at least you changed, enjoy complex numbers!
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    (Original post by Clarity Incognito)
    I thought I saw someone steal my avatar! Haha, no worries, at least you changed, enjoy complex numbers!
    So how would I 'show where they lie on the complex plane'?

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    Sorry, my brain doesn't work in terms of graphs and diagrams. :sad:
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    (Original post by RamocitoMorales)
    So how would I 'show where they lie on the complex plane'?

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    Sorry, my brain doesn't work in terms of graphs and diagrams. :sad:
    They are roots of unity, so their magnitude is 1. You know the argument so you draw the lines on an argand diagram like you would do for polar coordinates i.e. z_1 = (r,theta).
 
 
 
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