# Roots of unity.

• Jan 6th 2010, 06:08 PM
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]
$

But that is just one solution. How do you get the rest?
• Jan 6th 2010, 08:54 PM
Chris L T521
Quote:

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]
$

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 $k=0,1,2,3,4,5$.
• Jan 8th 2010, 07:45 PM
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)
• Jan 8th 2010, 07:51 PM
Chris L T521
Quote:

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?)
• Jan 8th 2010, 07:54 PM
Prove It
Recall that $-\pi < \theta \leq \pi$ by definition of $Arg{z}$...
• Jan 8th 2010, 08:15 PM
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?
• Jan 8th 2010, 08:28 PM
Prove It
Quote:

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 $-\pi < \theta \leq \pi$.
• Jan 8th 2010, 08:49 PM
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)
• Jan 8th 2010, 09:32 PM
Hey, I've a follow-up question: What guarantees that all the nth roots of unity will lie between $-\pi$ and $\pi$?
Hey, I've a follow-up question: What guarantees that all the nth roots of unity will lie between $-\pi$ and $\pi$?