If the complex numbers is `(1+ri)^(3)=lambda(1+i)`, when `i=sqrt(-1)`, for some real `lambda`, the value of r can be

Options

A. `cos""pi/5`
B. `cosec""(3pi)/2`
C. `cot""pi/12`
D. `tan""pi/12`

Answer

To solve the given problem, we need to find the value of \( r \) in the equation \( (1 + ri)^3 = \lambda(1 + i) \), where \( i = \sqrt{-1} \) and \( \lambda \) is a real number.

### Step-by-Step Solution:

1. **Expand the Left-Hand Side**:
We start with the expression \( (1 + ri)^3 \). We can use the binomial expansion formula:
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Here, \( a = 1 \) and \( b = ri \):
\[ (1 + ri)^3 = 1^3 + 3(1^2)(ri) + 3(1)(ri)^2 + (ri)^3 \]
This simplifies to:
\[ = 1 + 3ri + 3r^2i^2 + r^3i^3 \]
Since \( i^2 = -1 \) and \( i^3 = -i \), we can substitute these values:
\[ = 1 + 3ri - 3r^2 - r^3i \]
Rearranging gives:
\[ = (1 - 3r^2) + (3r - r^3)i \]

2. **Expand the Right-Hand Side**:
The right-hand side is \( \lambda(1 + i) \):
\[ = \lambda + \lambda i \]

3. **Equate Real and Imaginary Parts**:
Now we equate the real and imaginary parts from both sides:
- Real part: \( 1 - 3r^2 = \lambda \)
- Imaginary part: \( 3r - r^3 = \lambda \)

4. **Set the Two Equations Equal**:
Since both expressions equal \( \lambda \), we can set them equal to each other:
\[ 1 - 3r^2 = 3r - r^3 \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ r^3 - 3r^2 - 3r + 1 = 0 \]

6. **Finding Roots**:
We can use the Rational Root Theorem or trial and error to find possible rational roots. Testing \( r = 1 \):
\[ 1^3 - 3(1)^2 - 3(1) + 1 = 1 - 3 - 3 + 1 = -4 \quad \text{(not a root)} \]
Testing \( r = -1 \):
\[ (-1)^3 - 3(-1)^2 - 3(-1) + 1 = -1 - 3 + 3 + 1 = 0 \quad \text{(is a root)} \]

7. **Factor the Polynomial**:
Since \( r = -1 \) is a root, we can factor the polynomial:
\[ r^3 - 3r^2 - 3r + 1 = (r + 1)(r^2 - 4r + 1) \]

8. **Solve the Quadratic**:
Now we solve \( r^2 - 4r + 1 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} \]
Simplifying gives:
\[ r = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Final Values of \( r \):
Thus, the possible values of \( r \) are:
\[ r = -1, \quad r = 2 + \sqrt{3}, \quad r = 2 - \sqrt{3} \]