Find number of common roots of the equations `Z^(1901)+Z^(100)+1=0` and `Z_3+2Z^2+2Z+1=0`

Options

A. 2
B. 3
C. 4
D. 6

Answer

To find the number of common roots of the equations \( z^{1901} + z^{100} + 1 = 0 \) and \( z^3 + 2z^2 + 2z + 1 = 0 \), we will follow these steps:

### Step 1: Analyze the second equation
First, we will solve the equation \( z^3 + 2z^2 + 2z + 1 = 0 \).

To factor this equation, we can try to find one root by substituting simple values. Let's check \( z = -1 \):

\[ (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0 \]

Thus, \( z = -1 \) is a root. We can now factor \( z + 1 \) out of the polynomial.

### Step 2: Polynomial long division
We will perform polynomial long division of \( z^3 + 2z^2 + 2z + 1 \) by \( z + 1 \):

1. Divide \( z^3 \) by \( z \) to get \( z^2 \).
2. Multiply \( z^2 \) by \( z + 1 \) to get \( z^3 + z^2 \).
3. Subtract: \( (z^3 + 2z^2 + 2z + 1) - (z^3 + z^2) = z^2 + 2z + 1 \).
4. Now divide \( z^2 \) by \( z \) to get \( z \).
5. Multiply \( z \) by \( z + 1 \) to get \( z^2 + z \).
6. Subtract: \( (z^2 + 2z + 1) - (z^2 + z) = z + 1 \).
7. Finally, divide \( z + 1 \) by \( z + 1 \) to get \( 1 \).

So, we have:

\[ z^3 + 2z^2 + 2z + 1 = (z + 1)(z^2 + z + 1) \]

### Step 3: Find the roots of \( z^2 + z + 1 = 0 \)
To find the roots of \( z^2 + z + 1 = 0 \), we can use the quadratic formula:

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

Let \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \). The roots of the second equation are:

1. \( z = -1 \)
2. \( z = \omega \)
3. \( z = \omega^2 \)

### Step 4: Check common roots in the first equation
Now we need to check if these roots satisfy the first equation \( z^{1901} + z^{100} + 1 = 0 \).

#### Checking \( z = -1 \):
\[ (-1)^{1901} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \quad \text{(not a root)} \]

#### Checking \( z = \omega \):
Using the property \( \omega^3 = 1 \):
\[ \omega^{1901} = \omega^{1901 \mod 3} = \omega^1 = \omega \]
\[ \omega^{100} = \omega^{100 \mod 3} = \omega^1 = \omega \]
So,
\[ \omega + \omega + 1 = 2\omega + 1 = 0 \quad \text{(since \( \omega + \omega^2 + 1 = 0 \))} \]

#### Checking \( z = \omega^2 \):
Similarly,
\[ (\omega^2)^{1901} = \omega^{2} \quad \text{and} \quad (\omega^2)^{100} = \omega^{2} \]
So,
\[ \omega^2 + \omega^2 + 1 = 2\omega^2 + 1 = 0 \quad \text{(since \( \omega + \omega^2 + 1 = 0 \))} \]

### Conclusion
The common roots are \( \omega \) and \( \omega^2 \). Therefore, the number of common roots of the two equations is:

\[ \boxed{2} \]