Question

For the following question, enter the correct numerical value up to TWO decimal places. (If the numerical value has more than two decimal places, round-off the value to TWO decimal places.) If \(\alpha,\beta \in \mathbb{C}\) are the distinct roots of the equation \(x^2-x+1=0\), then the value of \(\alpha^{101}+\beta^{107}\) is ____

Hint:

For roots of unity:
Write roots in exponential form \(e^{i\theta}\)
Use periodicity: \(e^{i(\theta+2\pi)}=e^{i\theta}\)
Use \(e^{i\theta}+e^{-i\theta}=2\cos\theta\)

Answer 1

Correct Answer: 1

Step 1: Solve the quadratic equation: \[ x^2-x+1=0 \] The discriminant: \[ \Delta = (-1)^2-4(1)(1)=-3 \] Hence, the roots are: \[ \alpha=\frac{1+i\sqrt{3}}{2},\quad \beta=\frac{1-i\sqrt{3}}{2} \]
Step 2: Write the roots in exponential form. \[ \alpha=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=e^{i\pi/3} \] \[ \beta=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}=e^{-i\pi/3} \]
Step 3: Evaluate the powers using periodicity. \[ \alpha^{101}=e^{i\frac{101\pi}{3}}, \qquad \beta^{107}=e^{-i\frac{107\pi}{3}} \] Reduce exponents modulo \(2\pi\) (or powers modulo \(6\)): \[ 101 \equiv 5 \pmod{6}, \qquad 107 \equiv 5 \pmod{6} \] \[ \alpha^{101}=e^{i\frac{5\pi}{3}}, \qquad \beta^{107}=e^{-i\frac{5\pi}{3}} \]
Step 4: Add the two terms: \[ e^{i\frac{5\pi}{3}}+e^{-i\frac{5\pi}{3}} =2\cos\frac{5\pi}{3} \] \[ 2\cos\frac{5\pi}{3}=2\cdot\frac{1}{2}=1 \]
Step 5: Writing the answer up to two decimal places: \[ \boxed{1.00} \]