cube root of unity
- Thread starter Sonal7
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\(\displaystyle \sqrt[3]{Z}=1 \sim Z^3 = 1\)
The proper way to approach the problem is
\(\displaystyle Z^{1/3} = 1 =1 \cdot e^{i2\pi}\\
Z = \sqrt[3]{1} \cdot e^{i2\pi k/3},~k=0,1,2\\
Z = 1 \cdot e^{0},~1\cdot e^{2\pi/3},~1\cdot e^{i4\pi/3} \\
Z = \left \{1, e^{i2\pi/3},~e^{i4\pi/3}\right\}
\)
The proper way to approach the problem is
\(\displaystyle Z^{1/3} = 1 =1 \cdot e^{i2\pi}\\
Z = \sqrt[3]{1} \cdot e^{i2\pi k/3},~k=0,1,2\\
Z = 1 \cdot e^{0},~1\cdot e^{2\pi/3},~1\cdot e^{i4\pi/3} \\
Z = \left \{1, e^{i2\pi/3},~e^{i4\pi/3}\right\}
\)
Dr.Peterson
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I'm not sure what you are asking about. The solution to the problem itself doesn't include (1,0); but 1 is one of the roots of unity that they suggest you use in solving the problem.
But then, what "graph" are you saying crosses the axis?
Please state what you did to solve the problem, and what step you are confused by.
D
Deleted member 4993
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Assume that the cube-root of 1 is w. Then
w3 = 1
1 - w3 = 0 \(\displaystyle \ \to \ \) 13 - w3 = 0
Then
(1 - w) * ( w2 + w + 1) = 0
(1 - w) * (w -w1) * (w-w2) = 0
where w1,2 = \(\displaystyle \frac{1 \ \pm \ \sqrt{1-4}}{2}\) = \(\displaystyle \frac{1 \ \pm \ i \sqrt{3}}{2}\)
and continue........
HallsofIvy
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Because your title was "Cube root of unity"! That is solving z^3= 1.
And even reading you first post doesn't help! A person has to click on your link to see the actual problem, which is
And what "graph" were you talking about? There is no graph in the problem. The triangle mentioned is not the graph of any function. And where the sides of the triangle cross the x-axis are not solutions to this problem nor is 1 a solution! Yes, the problem does give the "cube roots of unity" but they are the vertices of an equilateral with one vertex at (1, 0) and center at the origin. What I would do is note that the line from the origin to the vertex \(\displaystyle (\sqrt{3}, 1)\) makes angle \(\displaystyle arctan(1/\sqrt{3})= \pi/6\) or 30 degrees. So the vertices are the "cube roots of unity" rotated counterclockwise by 30 degreed. And in an Argand diagram, that rotation is the same as multiplying by the complex number \(\displaystyle \sqrt{3}+ i\).
On this argrand diagram all three vertices are shown. The vertices arent rotated by 30 degrees but by 120 degrees, am i right? I guess what you are saying is the the triangle can be rotated so that one vertex is (1,0). I dont get why we need to use 1? I guess i dont get this at all.
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I thought the rotation is 120 degrees.
One root is 1. Now move 120 degrees counter clockwise to 120 degrees or \(\displaystyle \dfrac{2\pi}{3}\) radians. This is another root.
Rotate another 120 degrees to 240 degrees or \(\displaystyle \dfrac{4\pi}{3}\) radians. This is the last root.
Note that \(\displaystyle \dfrac{4\pi}{3} ~rad = -\dfrac{2\pi}{3} ~rad\)
So you could have rotated clockwise in 120 degree steps as well.
Why this particular set of 120 degree spacings? Well my previous post explains it one way.
But another explanation is that we need 2 things
a) 1 is a a cube root of unity
b) the remaining two roots are complex conjugates of one another.
This particular set of 3 points separated by 120 degrees each is the only set of 3 that meets these two requirements.
I was already aware the spacings are 120 ie 130/3 as there are 3 roots. I just cant understand unity and what that means. I find the explanations above helpful but the easiest way to think of the solutions are in terms of e^(2*pi*i)/n, when you have make the full rotation, then e^(2 pi i) which is 1 (Euler's identity). I think as S Khan said above you assume that the unity is 1. Its then a ratio of roots 1, w, w^2 ( a geometric series). Thank you very much for explaining it so well.
unity is just 1
though you'll often see it written as \(\displaystyle 1 = e^{i 2\pi k},~k \in \mathbb{Z}\)
D
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Weisstein, Eric W. "Unity." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Unity.html