let z1 z2 z3 z4 are roots of equation z^6+z^4+z^3+z^2+1=0 such that zr^3 +1 is not equal to 0 for r={1,2,3,4}.then (z1)^99 + (z2^)99 +(z3)^99 +(z4)^99 is equal to

To solve the problem involving the roots of the polynomial equation \( z^6 + z^4 + z^3 + z^2 + 1 = 0 \), we first need to analyze the roots and their properties. The equation can be rewritten in a more manageable form to identify the roots effectively.

Rearranging the Polynomial

The given polynomial can be expressed as:

\( z^6 + z^4 + z^3 + z^2 + 1 = 0 \)

We can factor out \( z^2 \) from the first three terms:

\( z^2(z^4 + z^2 + z + 1) + 1 = 0 \)

However, a more insightful approach is to recognize that this polynomial can be related to the roots of unity. Specifically, we can use the fact that the roots of unity are evenly spaced around the unit circle in the complex plane.

Finding the Roots

To find the roots, we can set \( z^5 = 1 \) (the fifth roots of unity) and check for any additional roots. The polynomial can be transformed using the substitution \( w = z^2 \), leading us to:

\( w^3 + w^2 + w + 1 = 0 \)

This cubic equation can be solved using the rational root theorem or synthetic division, but we can also observe that it has roots that correspond to the complex numbers on the unit circle.

Identifying the Relevant Roots

Given that \( z^3 + 1 \neq 0 \) for \( r = 1, 2, 3, 4 \), we can conclude that the roots \( z_1, z_2, z_3, z_4 \) are not equal to the cube roots of \(-1\). Thus, we are left with the roots of the polynomial that do not satisfy this condition.

Calculating \( z_1^{99} + z_2^{99} + z_3^{99} + z_4^{99} \)

Next, we need to compute \( z_1^{99} + z_2^{99} + z_3^{99} + z_4^{99} \). Since the roots are complex numbers, we can utilize the properties of exponents and roots of unity. The roots of the polynomial can be expressed in the form \( z_k = e^{i\theta_k} \), where \( \theta_k \) are the angles corresponding to the roots.

Since \( z^5 = 1 \), we know that \( z^n \) for \( n \) being a multiple of 5 will cycle through the roots. Therefore, we can reduce \( 99 \mod 5 \):

\( 99 \div 5 = 19 \) remainder \( 4 \)

This means \( z_k^{99} = z_k^4 \) for each root \( z_k \). Thus, we need to compute:

\( z_1^4 + z_2^4 + z_3^4 + z_4^4 \)

Final Calculation

Since the roots \( z_1, z_2, z_3, z_4 \) are symmetric and evenly distributed on the unit circle, their powers will also exhibit symmetry. The sum of the fourth powers of the roots can be calculated directly from the polynomial or using the properties of roots of unity.

Ultimately, the sum \( z_1^{99} + z_2^{99} + z_3^{99} + z_4^{99} \) simplifies to:

\( 0 \)

This is because the roots of the polynomial are evenly spaced around the unit circle, and their contributions cancel out due to symmetry.

In summary, the value of \( z_1^{99} + z_2^{99} + z_3^{99} + z_4^{99} \) is \( 0 \). This illustrates the fascinating interplay between polynomial roots and complex numbers, particularly in the context of roots of unity.