Understanding the Quadratic Equation and its Roots

We are given the quadratic equation \(x^2 - x + 1 = 0\). The roots of this equation are denoted by \(\alpha\) and \(\beta\). To find the roots, we can use the quadratic formula:

For a quadratic equation \(ax^2 + bx + c = 0\), the roots are given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In our equation, \(a=1\), \(b=-1\), and \(c=1\). Substituting these values into the formula:

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}\)

\(x = \frac{1 \pm \sqrt{1 - 4}}{2}\)

\(x = \frac{1 \pm \sqrt{-3}}{2}\)

\(x = \frac{1 \pm i\sqrt{3}}{2}\)

So the distinct roots are \(\alpha = \frac{1 + i\sqrt{3}}{2}\) and \(\beta = \frac{1 - i\sqrt{3}}{2}\) (or vice versa).

Expressing Roots in Polar Form

These complex roots can be expressed in polar form, \(re^{i\theta}\), where \(r\) is the modulus and \(\theta\) is the argument.

• For \(\alpha = \frac{1}{2} + i\frac{\sqrt{3}}{2}\):
• Modulus \(r = \left|\frac{1}{2} + i\frac{\sqrt{3}}{2}\right| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).
• Argument \(\theta\): \(\cos\theta = \frac{1/2}{1} = \frac{1}{2}\) and \(\sin\theta = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\). This gives \(\theta = \frac{\pi}{3}\).
• So, \(\alpha = 1 \cdot e^{i\pi/3} = e^{i\pi/3}\).

• For \(\beta = \frac{1}{2} - i\frac{\sqrt{3}}{2}\):
• Modulus \(r = \left|\frac{1}{2} - i\frac{\sqrt{3}}{2}\right| = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).
• Argument \(\theta\): \(\cos\theta = \frac{1/2}{1} = \frac{1}{2}\) and \(\sin\theta = \frac{-\sqrt{3}/2}{1} = -\frac{\sqrt{3}}{2}\). This gives \(\theta = -\frac{\pi}{3}\) (or \(\frac{5\pi}{3}\)).
• So, \(\beta = 1 \cdot e^{-i\pi/3} = e^{-i\pi/3}\).

Thus, the roots are \(\alpha = e^{i\pi/3}\) and \(\beta = e^{-i\pi/3}\).

Calculating Powers of the Roots (\(\alpha^{100}\) and \(\beta^{100}\))

We need to calculate \(\alpha^{100}\) and \(\beta^{100}\). Using De Moivre's theorem, \((re^{i\theta})^n = r^n e^{in\theta}\):

• \(\alpha^{100} = (e^{i\pi/3})^{100} = e^{i100\pi/3}\).
• To simplify the exponent, we write \(100\pi/3 = \frac{99\pi + \pi}{3} = 33\pi + \frac{\pi}{3}\).
• So, \(\alpha^{100} = e^{i(33\pi + \pi/3)} = e^{i33\pi} \cdot e^{i\pi/3}\).
• Recall that \(e^{ik\pi} = \cos(k\pi) + i\sin(k\pi) = (-1)^k\). For \(k=33\), \(e^{i33\pi} = (-1)^{33} = -1\).
• Therefore, \(\alpha^{100} = -1 \cdot e^{i\pi/3} = -e^{i\pi/3}\).

• \(\beta^{100} = (e^{-i\pi/3})^{100} = e^{-i100\pi/3}\).
• Using the same logic for the exponent, \(-100\pi/3 = -(33\pi + \pi/3)\).
• So, \(\beta^{100} = e^{-i(33\pi + \pi/3)} = e^{-i33\pi} \cdot e^{-i\pi/3}\).
• For \(k=33\), \(e^{-i33\pi} = \cos(-33\pi) + i\sin(-33\pi) = \cos(33\pi) - i\sin(33\pi) = -1 - 0 = -1\).
• Therefore, \(\beta^{100} = -1 \cdot e^{-i\pi/3} = -e^{-i\pi/3}\).

Calculating the Sum and Difference of Powers

Now we find the sum and difference of \(\alpha^{100}\) and \(\beta^{100}\).

• \(\alpha^{100} + \beta^{100} = (-e^{i\pi/3}) + (-e^{-i\pi/3}) = -(e^{i\pi/3} + e^{-i\pi/3})\).
• Using Euler's formula \(e^{i\theta} + e^{-i\theta} = 2\cos\theta\):
• \(e^{i\pi/3} + e^{-i\pi/3} = 2\cos(\pi/3) = 2(\frac{1}{2}) = 1\).
• So, \(\alpha^{100} + \beta^{100} = -(1) = -1\).

• \(\alpha^{100} - \beta^{100} = (-e^{i\pi/3}) - (-e^{-i\pi/3}) = -e^{i\pi/3} + e^{-i\pi/3} = -(e^{i\pi/3} - e^{-i\pi/3})\).
• Using Euler's formula \(e^{i\theta} - e^{-i\theta} = 2i\sin\theta\):
• \(e^{i\pi/3} - e^{-i\pi/3} = 2i\sin(\pi/3) = 2i(\frac{\sqrt{3}}{2}) = i\sqrt{3}\).
• So, \(\alpha^{100} - \beta^{100} = -(i\sqrt{3}) = -i\sqrt{3}\).

Evaluating the Expression and its Absolute Value

We need to find the value of \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\).

First, calculate the fraction:

\(\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}} = \frac{-1}{-i\sqrt{3}} = \frac{1}{i\sqrt{3}}\)

To simplify the complex fraction, multiply the numerator and denominator by \(i\):

\(\frac{1}{i\sqrt{3}} \times \frac{i}{i} = \frac{i}{i^2\sqrt{3}}\)

Since \(i^2 = -1\):

\(\frac{i}{-1 \cdot \sqrt{3}} = \frac{i}{-\sqrt{3}} = -\frac{i}{\sqrt{3}}\)

Now, find the absolute value of this complex number, which is in the form \(x+iy\) where \(x=0\) and \(y = -\frac{1}{\sqrt{3}}\). The absolute value is \(\sqrt{x^2 + y^2}\).

\(\left|-\frac{i}{\sqrt{3}}\right| = \left|0 - \frac{1}{\sqrt{3}}i\right| = \sqrt{0^2 + (-\frac{1}{\sqrt{3}})^2}\)

\(= \sqrt{0 + \frac{1}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\)

The value of the expression is \(\frac{1}{\sqrt{3}}\).

Revision Table: Key Steps and Results

Additional Information on Complex Roots and Powers

The roots of the equation \(x^2 - x + 1 = 0\) are closely related to the roots of unity. Specifically, they are the primitive 6th roots of unity, \(e^{i\pi/3}\) and \(e^{-i\pi/3}\), multiplied by -1 (which doesn't change their argument modulo \(2\pi\)). Alternatively, as shown, they are \(e^{i\pi/3}\) and \(e^{-i\pi/3}\).

• Complex Conjugates: The roots of a quadratic equation with real coefficients are always complex conjugates. Here, \(\beta = \frac{1 - i\sqrt{3}}{2}\) is the conjugate of \(\alpha = \frac{1 + i\sqrt{3}}{2}\). In polar form, \(e^{-i\theta}\) is the conjugate of \(e^{i\theta}\). This relationship \(\beta = \bar{\alpha}\) is useful. Note that \((\bar{\alpha})^{100} = \overline{(\alpha^{100})}\), so \(\beta^{100} = \overline{\alpha^{100}}\).
• Let \(Z = \alpha^{100}\). Then \(\beta^{100} = \bar{Z}\). The expression becomes \(\left|\frac{Z+\bar{Z}}{Z-\bar{Z}}\right|\).
• Recall that for any complex number \(Z = x+iy\), \(Z+\bar{Z} = (x+iy)+(x-iy) = 2x = 2\text{Re}(Z)\) and \(Z-\bar{Z} = (x+iy)-(x-iy) = 2iy = 2i\text{Im}(Z)\).
• So, \(\frac{Z+\bar{Z}}{Z-\bar{Z}} = \frac{2\text{Re}(Z)}{2i\text{Im}(Z)} = \frac{\text{Re}(Z)}{i\text{Im}(Z)} = \frac{\text{Re}(Z)\cdot i}{i^2\text{Im}(Z)} = -\frac{\text{Re}(Z)}{\text{Im}(Z)}i\).
• The absolute value is \(\left|-\frac{\text{Re}(Z)}{\text{Im}(Z)}i\right| = \left|\frac{\text{Re}(Z)}{\text{Im}(Z)}\right|\).
• We found \(\alpha^{100} = -e^{i\pi/3} = -(\cos(\pi/3) + i\sin(\pi/3)) = -(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).
• So \(Z = \alpha^{100}\) has \(\text{Re}(Z) = -\frac{1}{2}\) and \(\text{Im}(Z) = -\frac{\sqrt{3}}{2}\).
• The expression becomes \(\left|\frac{-1/2}{-i\sqrt{3}/2}\right| = \left|\frac{1/2}{i\sqrt{3}/2}\right| = \left|\frac{1}{i\sqrt{3}}\right| = \left|-\frac{i}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}}\). This confirms the result using the conjugate property.