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Math Help - Roots of unity.

  1. #1
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    Roots of unity.

    I'm trying to find the value of (-64)^\frac{1}{6}.

    Expressing -64 in the form r(\cos{\theta}+i\sin{\theta}), we have:

    -64 = 64(\cos{\pi}+i\sin{\pi})

    Which gives:

    (-64)^{\frac{1}{6}} = 64^{\frac{1}{6}}(\cos{\pi}+i\sin{\pi})^{\frac{1}{6  }}

    Then using De Moivre's theorem:

    (-64)^{\frac{1}{6}} = 64^{\frac{1}{6}}(\cos{\pi}+i\sin{\pi})^{\frac{1}{6  }} = 2\left[\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\right]<br />

    But that is just one solution. How do you get the rest?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Captcha View Post
    I'm trying to find the value of (-64)^\frac{1}{6}.

    Expressing -64 in the form r(\cos{\theta}+i\sin{\theta}), we have:

    -64 = 64(\cos{\pi}+i\sin{\pi})

    Which gives:

    (-64)^{\frac{1}{6}} = 64^{\frac{1}{6}}(\cos{\pi}+i\sin{\pi})^{\frac{1}{6  }}

    Then using De Moivre's theorem:

    (-64)^{\frac{1}{6}} = 64^{\frac{1}{6}}(\cos{\pi}+i\sin{\pi})^{\frac{1}{6  }} = 2\left[\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\right]<br />

    But that is just one solution. How do you get the rest?
    Just tweak your answer to generate them: 2\left[\cos{\frac{(2k+1)\pi}{6}}+i\sin{\frac{(2k+1)\pi}{6  }}\right] for k=0,1,2,3,4,5.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Just tweak your answer to generate them: 2\left[\cos{\frac{(2k+1)\pi}{6}}+i\sin{\frac{(2k+1)\pi}{6  }}\right] for k=0,1,2,3,4,5.
    Thanks Chris. Here is what I've done:

    k = 0 \Rightarrow 2\left[\cos{\frac{(2(0)+1)\pi}{6}}+i\sin{\frac{(2(0)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(0+1)\pi}{6}}+i\sin{\frac{(0+1)\pi}{6}}  \right] =  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght)

    k = 1: \Rightarrow  2\left[\cos{\frac{(2(1)+1)\pi}{6}}+i\sin{\frac{(2(1)+1)\p  i}{6}}\right] =  2\left[\cos{\frac{(2+1)\pi}{6}}+i\sin{\frac{(2+1)\pi}{6}}  \right]  = 2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\ri  ght)

    k = 2: \Rightarrow  2\left[\cos{\frac{(2(2)+1)\pi}{6}}+i\sin{\frac{(2(2)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(4+1)\pi}{6}}+i\sin{\frac{(4+1)\pi}{6}}  \right] = {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)}

    k = 3: \Rightarrow  2\left[\cos{\frac{(2(3)+1)\pi}{6}}+i\sin{\frac{(2(3)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(6+1)\pi}{6}}+i\sin{\frac{(6+1)\pi}{6}}  \right] = {2\left(\cos{\frac{7\pi}{6}}+i\sin{\frac{7\pi}{6}}  \right)}

    k = 4: \Rightarrow  2\left[\cos{\frac{(2(4)+1)\pi}{6}}+i\sin{\frac{(2(4)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(8+1)\pi}{6}}+i\sin{\frac{(8+1)\pi}{6}}  \right] = {2\left(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}  \right)}

    k = 5: \Rightarrow  2\left[\cos{\frac{(2(5)+1)\pi}{6}}+i\sin{\frac{(2(5)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(10+1)\pi}{6}}+i\sin{\frac{(10+1)\pi}{6  }}\right] = {2\left(\cos{\frac{11\pi}{6}}+i\sin{\frac{11\pi}{6  }}\right)}

    So, the answers are:  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght), {2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\r  ight)}, {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)}, {2\left(\cos{\frac{7\pi}{6}}+i\sin{\frac{7\pi}{6}}  \right)}, {2\left(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}  \right)}, and {2\left(\cos{\frac{11\pi}{6}}+i\sin{\frac{11\pi}{6  }}\right)}.

    But the book's answers are:  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght), {2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\r  ight)}, {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)},  2\left(\cos{-\frac{\pi}{6}}+i\sin{-\frac{\pi}{6}}\right), {2\left(\cos{-\frac{\pi}{2}}+i\sin{-\frac{\pi}{2}}\right)}, and {2\left(\cos{-\frac{5\pi}{6}}+i\sin{-\frac{5\pi}{6}}\right)}.

    (I'm thinking that this has to do with the conjugacy of complex roots).
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Captcha View Post
    Thanks Chris. Here is what I've done:

    k = 0 \Rightarrow 2\left[\cos{\frac{(2(0)+1)\pi}{6}}+i\sin{\frac{(2(0)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(0+1)\pi}{6}}+i\sin{\frac{(0+1)\pi}{6}}  \right] =  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght)

    k = 1: \Rightarrow  2\left[\cos{\frac{(2(1)+1)\pi}{6}}+i\sin{\frac{(2(1)+1)\p  i}{6}}\right] =  2\left[\cos{\frac{(2+1)\pi}{6}}+i\sin{\frac{(2+1)\pi}{6}}  \right]  = 2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\ri  ght)

    k = 2: \Rightarrow  2\left[\cos{\frac{(2(2)+1)\pi}{6}}+i\sin{\frac{(2(2)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(4+1)\pi}{6}}+i\sin{\frac{(4+1)\pi}{6}}  \right] = {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)}

    k = 3: \Rightarrow  2\left[\cos{\frac{(2(3)+1)\pi}{6}}+i\sin{\frac{(2(3)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(6+1)\pi}{6}}+i\sin{\frac{(6+1)\pi}{6}}  \right] = {2\left(\cos{\frac{7\pi}{6}}+i\sin{\frac{7\pi}{6}}  \right)}

    k = 4: \Rightarrow  2\left[\cos{\frac{(2(4)+1)\pi}{6}}+i\sin{\frac{(2(4)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(8+1)\pi}{6}}+i\sin{\frac{(8+1)\pi}{6}}  \right] = {2\left(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}  \right)}

    k = 5: \Rightarrow  2\left[\cos{\frac{(2(5)+1)\pi}{6}}+i\sin{\frac{(2(5)+1)\p  i}{6}}\right] = 2\left[\cos{\frac{(10+1)\pi}{6}}+i\sin{\frac{(10+1)\pi}{6  }}\right] = {2\left(\cos{\frac{11\pi}{6}}+i\sin{\frac{11\pi}{6  }}\right)}

    So, the answers are:  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght), {2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\r  ight)}, {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)}, {2\left(\cos{\frac{7\pi}{6}}+i\sin{\frac{7\pi}{6}}  \right)}, {2\left(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}  \right)}, and {2\left(\cos{\frac{11\pi}{6}}+i\sin{\frac{11\pi}{6  }}\right)}.

    But the book's answers are:  2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri  ght), {2\left(\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}\r  ight)}, {2\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}  \right)},  2\left(\cos{-\frac{\pi}{6}}+i\sin{-\frac{\pi}{6}}\right), {2\left(\cos{-\frac{\pi}{2}}+i\sin{-\frac{\pi}{2}}\right)}, and {2\left(\cos{-\frac{5\pi}{6}}+i\sin{-\frac{5\pi}{6}}\right)}.

    (I'm thinking that this has to do with the conjugacy of complex roots).
    The two answers are the same! (See why?)
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  5. #5
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    Recall that -\pi < \theta \leq \pi by definition of Arg{z}...
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    The two answers are the same! (See why?)
    I've noticed that while changing the answers into the form a+ib. Is it because any angle differring 2\pi than the one we have previously found has exactly the same value?
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  7. #7
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    Quote Originally Posted by Captcha View Post
    I've noticed that while changing the answers into the form a+ib. Is it because any angle differring 2\pi than the one we have previously found has exactly the same value?
    Correct. And as I said, we say the principal argument is defined for -\pi < \theta \leq \pi.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Recall that -\pi < \theta \leq \pi by definition of Arg{z}...
    Got it! -\pi < \frac{\pi}{6}  \leq \pi, -\pi \ < \frac{\pi}{2}  \leq \pi, and -\pi < \frac{5\pi}{6}  \leq \pi, so no problem. But -\pi \not < \frac{7\pi}{6}  \not\leq \pi, -\pi \not < \frac{3\pi}{2}  \not\leq \pi, and -\pi \not < \frac{11\pi}{6}  \not\leq \pi. But \frac{7\pi}{6} has the same value as -\frac{5\pi}{6}, and -\pi < -\frac{5\pi}{6}  \leq \pi, so we take that. \frac{3\pi}{2} has the same value as -\frac{\pi}{2}, and -\pi < -\frac{\pi}{2}  \leq \pi, so we take that. And \frac{11\pi}{6} has the same value as -\frac{\pi}{6}, and -\pi < -\frac{\pi}{6}  \leq \pi, so we take that.

    Thanks a lot, guys.
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  9. #9
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    Hey, I've a follow-up question: What guarantees that all the nth roots of unity will lie between -\pi and \pi?
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  10. #10
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    Quote Originally Posted by Captcha View Post
    Hey, I've a follow-up question: What guarantees that all the nth roots of unity will lie between -\pi and \pi?
    Because all the roots are evenly spaced around a circle.
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  11. #11
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    Yea you did have verry same problem Glad that we solve it!
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