If `alpha,beta` are the roots of the equation `x^2-2x+4=0`, find `alpha^(n)+beta^(n)` for
(a) `n=3k, k in N`
(b) `n ne 3k, k in N `

Answer

To solve the problem, we need to find \( \alpha^n + \beta^n \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x + 4 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \( x^2 - 2x + 4 = 0 \):

- \( a = 1 \)
- \( b = -2 \)
- \( c = 4 \)

Substituting these values into the formula gives:

\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} \]

This simplifies to:

\[ x = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} \]

Thus, we have:

\[ \alpha = 1 + i\sqrt{3}, \quad \beta = 1 - i\sqrt{3} \]

### Step 2: Express \( \alpha \) and \( \beta \) in polar form

Next, we convert \( \alpha \) and \( \beta \) into polar form. The modulus \( r \) and argument \( \theta \) can be calculated as follows:

- Modulus:
\[ r = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

- Argument:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]

Thus, we can express:

\[ \alpha = 2 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right), \quad \beta = 2 \left( \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) \right) \]

### Step 3: Use De Moivre's Theorem

Using De Moivre's Theorem, we can find \( \alpha^n \) and \( \beta^n \):

\[ \alpha^n = 2^n \left( \cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3} \right) \]
\[ \beta^n = 2^n \left( \cos\left(-\frac{n\pi}{3}\right) + i\sin\left(-\frac{n\pi}{3}\right) \right) \]

### Step 4: Calculate \( \alpha^n + \beta^n \)

Now, we can add \( \alpha^n \) and \( \beta^n \):

\[ \alpha^n + \beta^n = 2^n \left( \cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3} \right) + 2^n \left( \cos\left(-\frac{n\pi}{3}\right) + i\sin\left(-\frac{n\pi}{3}\right) \right) \]

Since \( \cos(-x) = \cos(x) \) and \( \sin(-x) = -\sin(x) \):

\[ \alpha^n + \beta^n = 2^n \left( 2\cos\frac{n\pi}{3} \right) = 2^{n+1} \cos\frac{n\pi}{3} \]

### Step 5: Analyze for \( n = 3k \) and \( n \neq 3k \)

(a) **For \( n = 3k \)**:
\[ \alpha^{3k} + \beta^{3k} = 2^{3k + 1} \cos\left(\frac{3k\pi}{3}\right) = 2^{3k + 1} \cos(k\pi) \]
Since \( \cos(k\pi) = (-1)^k \):
\[ \alpha^{3k} + \beta^{3k} = 2^{3k + 1} (-1)^k \]

(b) **For \( n \neq 3k \)**:
\[ \alpha^n + \beta^n = 2^{n + 1} \cos\left(\frac{n\pi}{3}\right) \]

### Final Answers:

- (a) \( \alpha^{3k} + \beta^{3k} = 2^{3k + 1} (-1)^k \)
- (b) \( \alpha^n + \beta^n = 2^{n + 1} \cos\left(\frac{n\pi}{3}\right) \)