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`

Answer

To solve the problem, we need to find the expression \( \alpha^n + \beta^n \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x + 4 = 0 \) for \( n = 3k \) where \( k \in \mathbb{N} \).

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

We can use the quadratic formula to find the roots of the equation:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = 4 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 4 = 4 - 16 = -12 \]

Now substituting back into the quadratic formula:
\[ x = \frac{2 \pm \sqrt{-12}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} \]

Thus, the roots are:
\[ \alpha = 1 + i\sqrt{3}, \quad \beta = 1 - i\sqrt{3} \]

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

We can express \( \alpha \) and \( \beta \) in polar form. The modulus \( r \) is given by:
\[ r = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

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

Thus, we can write:
\[ \alpha = 2 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right) = 2 e^{i\frac{\pi}{3}} \]
\[ \beta = 2 \left( \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) \right) = 2 e^{-i\frac{\pi}{3}} \]

### Step 3: Calculate \( \alpha^n + \beta^n \)

Using the polar forms:
\[ \alpha^n = (2 e^{i\frac{\pi}{3}})^n = 2^n e^{i\frac{n\pi}{3}} \]
\[ \beta^n = (2 e^{-i\frac{\pi}{3}})^n = 2^n e^{-i\frac{n\pi}{3}} \]

Now, adding these:
\[ \alpha^n + \beta^n = 2^n \left( e^{i\frac{n\pi}{3}} + e^{-i\frac{n\pi}{3}} \right) = 2^n \cdot 2 \cos\left(\frac{n\pi}{3}\right) = 2^{n+1} \cos\left(\frac{n\pi}{3}\right) \]

### Step 4: Substitute \( n = 3k \)

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

Since \( \cos(k\pi) = (-1)^k \):
\[ \alpha^{3k} + \beta^{3k} = 2^{3k + 1} (-1)^k \]

### Final Answer

Thus, the final result is:
\[ \alpha^{3k} + \beta^{3k} = 2^{3k + 1} (-1)^k \]

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