Finding Roots of Complex Equations

The problem asks us to find which of the given options is a root of equation-II: $z^{1985} + z^{100} + 1 = 0$. We are also given equation-I: $z^3 + 2z^2 + 2z + 1 = 0$. Let's first analyze equation-I, as its roots might be related to the options provided.

Analyzing Equation-I: $z^3 + 2z^2 + 2z + 1 = 0$

This is a cubic equation. We can try to find simple roots by inspection. Let's test $z = -1$:

$(-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2(1) - 2 + 1 = -1 + 2 - 2 + 1 = 0$

Since substituting $z=-1$ makes the equation true, $z=-1$ is a root of equation-I. This means $(z+1)$ is a factor of the polynomial $z^3 + 2z^2 + 2z + 1$.

We can perform polynomial division or synthetic division to find the other factor:

The quotient is $z^2 + z + 1$. So, equation-I can be written as:

$(z+1)(z^2 + z + 1) = 0$

The roots of this equation are $z+1 = 0$ or $z^2 + z + 1 = 0$.

• From $z+1 = 0$, we get $z = -1$.
• The equation $z^2 + z + 1 = 0$ is significant in complex numbers. Its roots are the complex cube roots of unity, denoted by $\omega$ and $\omega^2$. These roots satisfy the property $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.

So, the roots of equation-I are $-1$, $\omega$, and $\omega^2$. Note that the options provided are related to these roots or their negatives.

Checking Roots for Equation-II: $z^{1985} + z^{100} + 1 = 0$

Now we need to check which of the given options satisfies equation-II. The options are $-1$, $-\omega$, $-\omega^2$, and $\omega$. We will substitute each option into equation-II and see if the expression evaluates to zero.

Testing Option 1: $z = -1$

Substitute $z=-1$ into $z^{1985} + z^{100} + 1$:
$(-1)^{1985} + (-1)^{100} + 1$
Since 1985 is an odd number, $(-1)^{1985} = -1$.
Since 100 is an even number, $(-1)^{100} = 1$.
So, the expression becomes $-1 + 1 + 1 = 1$.
Since $1 \neq 0$, $z=-1$ is not a root of equation-II.

Testing Option 2: $z = -\omega$

Substitute $z=-\omega$ into $z^{1985} + z^{100} + 1$:
$(-\omega)^{1985} + (-\omega)^{100} + 1$
$= (-1)^{1985} \omega^{1985} + (-1)^{100} \omega^{100} + 1$
$= -1 \cdot \omega^{1985} + 1 \cdot \omega^{100} + 1$
$= -\omega^{1985} + \omega^{100} + 1$

We use the property $\omega^3 = 1$ to simplify the powers of $\omega$. We find the remainder when the exponent is divided by 3.

• For $\omega^{1985}$: $1985 \div 3$. $1985 = 3 \times 661 + 2$. So, $\omega^{1985} = \omega^2$.
• For $\omega^{100}$: $100 \div 3$. $100 = 3 \times 33 + 1$. So, $\omega^{100} = \omega^1 = \omega$.

Substitute these back into the expression:
$= -(\omega^2) + \omega + 1$
$= 1 + \omega - \omega^2$

Using the property $1 + \omega + \omega^2 = 0$, we know $1 + \omega = -\omega^2$.
So, $1 + \omega - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2$.
Since $\omega^2 \neq 0$, $-2\omega^2 \neq 0$. Thus, $z=-\omega$ is not a root of equation-II.

Testing Option 3: $z = -\omega^2$

Substitute $z=-\omega^2$ into $z^{1985} + z^{100} + 1$:
$(-\omega^2)^{1985} + (-\omega^2)^{100} + 1$
$= (-1)^{1985} (\omega^2)^{1985} + (-1)^{100} (\omega^2)^{100} + 1$
$= -1 \cdot \omega^{3970} + 1 \cdot \omega^{200} + 1$
$= -\omega^{3970} + \omega^{200} + 1$

Simplify powers of $\omega$ using $\omega^3 = 1$:

• For $\omega^{3970}$: $3970 \div 3$. $3970 = 3 \times 1323 + 1$. So, $\omega^{3970} = \omega^1 = \omega$.
• For $\omega^{200}$: $200 \div 3$. $200 = 3 \times 66 + 2$. So, $\omega^{200} = \omega^2$.

Substitute these back into the expression:
$= -(\omega) + \omega^2 + 1$
$= 1 - \omega + \omega^2$

Using $1 + \omega + \omega^2 = 0$, we know $1 + \omega^2 = -\omega$.
So, $1 - \omega + \omega^2 = (-\omega) - \omega = -2\omega$.
Since $\omega \neq 0$, $-2\omega \neq 0$. Thus, $z=-\omega^2$ is not a root of equation-II.

Testing Option 4: $z = \omega$

Substitute $z=\omega$ into $z^{1985} + z^{100} + 1$:
$\omega^{1985} + \omega^{100} + 1$

From our calculations in Option 2:

• $\omega^{1985} = \omega^2$
• $\omega^{100} = \omega$

Substitute these back:
$= \omega^2 + \omega + 1$

Using the fundamental property of cube roots of unity, $1 + \omega + \omega^2 = 0$.
So, $\omega^2 + \omega + 1 = 0$.
Since the expression evaluates to 0, $z=\omega$ is a root of equation-II.

Conclusion

Based on the testing of all options, only $z=\omega$ satisfies equation-II ($z^{1985} + z^{100} + 1 = 0$).

Revision Table: Complex Roots and Equations

Additional Information: Complex Cube Roots of Unity

The complex cube roots of unity are the solutions to the equation $z^3 = 1$. These are $1$, $\omega$, and $\omega^2$. Geometrically, they are points in the complex plane that form vertices of an equilateral triangle inscribed in the unit circle, with one vertex at $(1,0)$.

• The principal cube root is $1$.
• The other two roots are $\omega = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
• The third root is $\omega^2 = e^{i4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. Note that $\omega^2$ is the complex conjugate of $\omega$.

Key properties that are frequently used in problems involving $\omega$:

• $\omega^3 = 1$
• $1 + \omega + \omega^2 = 0$
• $\omega^2 = \frac{1}{\omega}$
• $\omega = \frac{1}{\omega^2}$
• $\omega \cdot \omega^2 = \omega^3 = 1$

These properties are crucial for simplifying expressions involving powers of $\omega$, as demonstrated in solving this problem.