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Thread: Roots of unity.

  1. #1
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    Roots of unity.

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

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

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

    Which gives:

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

    Then using De Moivre's theorem:

    $\displaystyle (-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]
    $

    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 $\displaystyle (-64)^\frac{1}{6}$.

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

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

    Which gives:

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

    Then using De Moivre's theorem:

    $\displaystyle (-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]
    $

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

    k = 0 $\displaystyle \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] $ = $\displaystyle 2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri ght)$

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

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

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

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

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

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

    (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 $\displaystyle \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] $ = $\displaystyle 2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\ri ght)$

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

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

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

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

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

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

    (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 $\displaystyle -\pi < \theta \leq \pi$ by definition of $\displaystyle Arg{z}$...
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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 $\displaystyle a+ib$. Is it because any angle differring $\displaystyle 2\pi$ 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 $\displaystyle a+ib$. Is it because any angle differring $\displaystyle 2\pi$ 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 $\displaystyle -\pi < \theta \leq \pi$.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Recall that $\displaystyle -\pi < \theta \leq \pi$ by definition of $\displaystyle Arg{z}$...
    Got it! $\displaystyle -\pi < \frac{\pi}{6} \leq \pi$, $\displaystyle -\pi \ < \frac{\pi}{2} \leq \pi$, and $\displaystyle -\pi < \frac{5\pi}{6} \leq \pi$, so no problem. But $\displaystyle -\pi \not < \frac{7\pi}{6} \not\leq \pi$, $\displaystyle -\pi \not < \frac{3\pi}{2} \not\leq \pi$, and $\displaystyle -\pi \not < \frac{11\pi}{6} \not\leq \pi$. But $\displaystyle \frac{7\pi}{6}$ has the same value as $\displaystyle -\frac{5\pi}{6}$, and $\displaystyle -\pi < -\frac{5\pi}{6} \leq \pi$, so we take that. $\displaystyle \frac{3\pi}{2}$ has the same value as $\displaystyle -\frac{\pi}{2}$, and $\displaystyle -\pi < -\frac{\pi}{2} \leq \pi$, so we take that. And $\displaystyle \frac{11\pi}{6}$ has the same value as $\displaystyle -\frac{\pi}{6}$, and $\displaystyle -\pi < -\frac{\pi}{6} \leq \pi$, 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 $\displaystyle -\pi$ and $\displaystyle \pi$?
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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 $\displaystyle -\pi$ and $\displaystyle \pi$?
    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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