The polynomial equation of the lowest degree having roots `1 , sqrt(3)i` is

Options

A. `x^3 +x^2 +3x +3=0`
B. `x^3 -x^2 +3x-3=0`
C. `x^3 +x^2 -3x -3=0`
D. `x^2 -(1+sqrt(3)i)x+sqrt(3)i =0`

Answer

To find the polynomial equation of the lowest degree having roots \(1\) and \(\sqrt{3}i\), we can follow these steps:

### Step 1: Identify the roots
The given roots are:
- \( \alpha = 1 \)
- \( \beta = \sqrt{3}i \)

Since the coefficients of the polynomial must be real, the complex roots must come in conjugate pairs. Therefore, the conjugate of \(\beta\) is:
- \( \gamma = -\sqrt{3}i \)

### Step 2: Write the polynomial in factored form
The polynomial can be expressed as:
\[ f(x) = (x - \alpha)(x - \beta)(x - \gamma) \]
Substituting the roots:
\[ f(x) = (x - 1)(x - \sqrt{3}i)(x + \sqrt{3}i) \]

### Step 3: Simplify the product of the complex roots
First, we simplify the product \((x - \sqrt{3}i)(x + \sqrt{3}i)\):
\[ (x - \sqrt{3}i)(x + \sqrt{3}i) = x^2 - (\sqrt{3}i)^2 \]
Calculating \((\sqrt{3}i)^2\):
\[ (\sqrt{3}i)^2 = 3i^2 = 3(-1) = -3 \]
Thus, we have:
\[ (x - \sqrt{3}i)(x + \sqrt{3}i) = x^2 - (-3) = x^2 + 3 \]

### Step 4: Combine with the real root
Now, substitute back into the polynomial:
\[ f(x) = (x - 1)(x^2 + 3) \]

### Step 5: Expand the polynomial
Expanding the product:
\[ f(x) = x(x^2 + 3) - 1(x^2 + 3) \]
\[ = x^3 + 3x - x^2 - 3 \]
Rearranging gives:
\[ f(x) = x^3 - x^2 + 3x - 3 \]

### Final Polynomial
The polynomial equation of the lowest degree having the roots \(1\) and \(\sqrt{3}i\) is:
\[ f(x) = x^3 - x^2 + 3x - 3 \]