\(\cos \left ( \frac {\pi }{6} \right ) + i \, \sin \left ( \frac {\pi }{6} \right )\) is which root of unity?
sixth eighth tenth twelfeth
unit circle
We have seen that every Complex number can be written as \(r \cdot (\cos (\theta ) + i \, \sin (\theta ))\). This is a scalar times a Complex number on the unit circle.
We have also seen that if \(z = r \cdot (\cos (\theta ) + i \, \sin (\theta ))\), then \(z^n = r^n \cdot (\cos (n\theta ) + i \, \sin (n\theta ))\). Raising complex numbers to powers is accomplished by raising the modulus to the power and then multiplying the angle.
\(\blacktriangleright \) Roots of Unity
Roots or unity refer to roots or zeros of the polynomial \(x^n - 1\).
Any such root of unity would be a solution to the equation \(z^n = 1\), which is why they are called roots of unity.
For instance, the square roots of unity are the solutions to \(z^2 = 1\). The two solutions are \(1\) and \(-1\).
For instance, the \(4^{th}\) roots of unity are the solutions to \(z^4 = 1\). The four solutions are \(1\), \(-1\), \(i\), and \(-i\).
\(1\) is always a root of unity for any power. Then, there are \(n-1\) other \(n^{th}\) roots of unity.
If \(z = r \cdot (\cos (\theta ) + i \, \sin (\theta ))\) is going to be a root of unity, then \(z^n = r^n \cdot (\cos (n\theta ) + i \, \sin (n\theta )) = 1\), which means \(r=1\).
\(n^{th}\) roots of unity all look like \(\cos (\theta ) + i \, \sin (\theta )\). They all lie on the unit circle.
In addition, if \(\cos (n\theta ) + i \, \sin (n\theta ) = 1\), then \(n \theta = 2 k \pi \) (a mulitple of \(2 \pi \)), because \(1\) is on the positive real axis.
Therefore, \(\theta = \frac {2 k \pi }{n}\)
\(4^{th}\) roots of \(1\)
We need multiples of \(\frac {2 \pi }{4} = \frac {\pi }{2}\).
- \(\theta = \frac {\pi }{2}\)
- \(\theta = \frac {2 \pi }{2} = \pi \)
- \(\theta = \frac {3 \pi }{2}\)
- \(\theta = \frac {4 \pi }{2} = 2 \pi \)
The fourth roots of unity are:
- \(\cos \left (\frac {\pi }{2}\right ) + i \, \sin \left (\frac {\pi }{2}\right ) = i\)
- \(\cos (\pi ) + i \, \sin (\pi ) = -1\)
- \(\cos \left (\frac {3 \pi }{2}\right ) + i \, \sin \left (\frac {3 \pi }{2}\right ) = -i\)
- \(\cos (2 \pi ) + i \, \sin (2 \pi ) = 1\)
The roots of unity are spread out equidistant along the unit circle.
Cube roots of \(1\)
We need multiples of \(\frac {2 \pi }{3}\).
- \(\theta = \frac {2 \pi }{3}\)
- \(\theta = \frac {4 \pi }{3}\)
- \(\theta = \frac {6 \pi }{3} = 2 \pi \)
The fourth roots of unity are:
- \(\cos \left (\frac {2 \pi }{3}\right ) + i \, \sin \left (\frac {2 \pi }{3}\right ) = \frac {1}{2} + \frac {\sqrt {3}}{2} \, i\)
- \(\cos \left (\frac {4 \pi }{3}\right ) + i \, \sin \left (\frac {4 \pi }{3}\right ) = \frac {1}{2} - \frac {\sqrt {3}}{2} \, i\)
- \(\cos (2 \pi ) + i \, \sin (2 \pi ) = 1\)
Using the quadratic formula, we know that \(\frac {-3 + \sqrt {11} \, i}{2}\) is a root of \(x^2 + 3x + 5\).
We know that \(\frac {-3 + \sqrt {11} \, i}{2}\) can be written in the form \(r (\cos (\theta ) + i \, \sin (\theta ))\).
\(r\) is the modulus of \(\frac {-3 + \sqrt {11} \, i}{2}\).
\(r = \answer {\sqrt {5}}\)
The angle for \(\frac {-3 + \sqrt {11} \, i}{2}\) is in which quadrant?
I II III IV
\(\arctan \left ( -\frac {\sqrt {11}}{3} \right )\) gives the angle, but in quadrant IV. What must be added to \(\arctan \left ( -\frac {\sqrt {11}}{3} \right )\) to get \(\theta \)?
Add \(\answer {\pi }\)
\[ \frac {-3 + \sqrt {11} \, i}{2} = \sqrt {5} \left ( \cos \left ( \arctan \left ( -\frac {\sqrt {11}}{3} \right ) + \pi \right ) + i \, \sin \left ( \arctan \left ( -\frac {\sqrt {11}}{3} \right ) +\pi \right ) \right ) \]
As long as we are here...
In the example above, \(x^2 + 3x + 5\) is a polynomial with real coefficients. Therefore, if \(\frac {-3 + \sqrt {11} \, i}{2}\) is a root, then so is \(\frac {-3 - \sqrt {11} \, i}{2}\).
Therefore, \(x^2 + 3x + 5\) factors as
\[ \left ( x - \frac {-3 - \sqrt {11} \, i}{2} \right ) \left ( x - \frac {-3 + \sqrt {11} \, i}{2} \right ) \]
The constant terms have to be equal, which tells us that
\[ \left ( \frac {-3 - \sqrt {11} \, i}{2} \right ) \left ( \frac {-3 + \sqrt {11} \, i}{2} \right ) = 5 \]
The constant term of the polynomial is the product of the roots, which are conjugates. The constant term is the square of the modulus of either root.
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More Examples of Complex Exponentials