Roots of unity.

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January 6th 2010, 05:08 PM
Captcha

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?
January 6th 2010, 07:54 PM
Chris L T521

Quote:

Originally Posted by Captcha

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 $k=0,1,2,3,4,5$ .
January 8th 2010, 06:45 PM
Captcha

Quote:

Originally Posted by Chris L T521

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). (Thinking)
January 8th 2010, 06:51 PM
Chris L T521

Quote:

Originally Posted by Captcha

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). (Thinking)

The two answers are the same! (See why?)
January 8th 2010, 06:54 PM
Prove It

Recall that $-\pi < \theta \leq \pi$ by definition of $Arg{z}$ ...
January 8th 2010, 07:15 PM
Captcha

Quote:

Originally Posted by Chris L T521

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?
January 8th 2010, 07:28 PM
Prove It

Quote:

Originally Posted by Captcha

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 $-\pi < \theta \leq \pi$ .
January 8th 2010, 07:49 PM
Captcha

Quote:

Originally Posted by Prove It

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. (Cool)
January 8th 2010, 08:32 PM
Captcha

Hey, I've a follow-up question: What guarantees that all the nth roots of unity will lie between $-\pi$ and $\pi$ ?
January 8th 2010, 08:58 PM
Prove It

Quote:

Originally Posted by Captcha

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.
January 9th 2010, 06:05 PM
.::MadMaX::.

Yea you did have verry same problem :) Glad that we solve it!