Finding the Value of Powers of Complex Cube Roots of Unity

This solution details the steps to calculate the value of the expression $\omega^{10} + \omega^{20} + \omega^{30} + \omega^{40} + \omega^{50}$, where $\omega$ is defined as a complex cube root of unity.

Key Properties of $\omega$

When dealing with a complex cube root of unity, denoted as $\omega$, it's essential to remember its fundamental properties derived from the equation $x^3 = 1$. The complex roots are $1, \omega, \omega^2$. The two most crucial properties are:

• Property 1: The cube of $\omega$ is always 1.

  $$ \omega^3 = 1 $$

• Property 2: The sum of the cube roots of unity is zero.

  $$ 1 + \omega + \omega^2 = 0 $$

From Property 2, we can also infer that $\omega + \omega^2 = -1$. We will use these properties to simplify the given expression.

Simplifying Each Term in the Expression

The expression involves several powers of $\omega$. We can simplify each term by using the property $\omega^3 = 1$. The strategy is to reduce the exponent modulo 3.

• Simplifying $\omega^{10}$:

  Divide the exponent $10$ by $3$: $10 = 3 \times 3 + 1$. The remainder is $1$. Therefore, $\omega^{10} = \omega^{3 \times 3 + 1} = (\omega^3)^3 \cdot \omega^1$. Using $\omega^3 = 1$, we get $(1)^3 \cdot \omega = 1 \cdot \omega = \omega$. $$ \omega^{10} = \omega $$

• Simplifying $\omega^{20}$:

  Divide the exponent $20$ by $3$: $20 = 3 \times 6 + 2$. The remainder is $2$. Therefore, $\omega^{20} = \omega^{3 \times 6 + 2} = (\omega^3)^6 \cdot \omega^2$. Using $\omega^3 = 1$, we get $(1)^6 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$. $$ \omega^{20} = \omega^2 $$

• Simplifying $\omega^{30}$:

  Divide the exponent $30$ by $3$: $30 = 3 \times 10 + 0$. The remainder is $0$. Therefore, $\omega^{30} = \omega^{3 \times 10} = (\omega^3)^{10}$. Using $\omega^3 = 1$, we get $(1)^{10} = 1$. $$ \omega^{30} = 1 $$

• Simplifying $\omega^{40}$:

  Divide the exponent $40$ by $3$: $40 = 3 \times 13 + 1$. The remainder is $1$. Therefore, $\omega^{40} = \omega^{3 \times 13 + 1} = (\omega^3)^{13} \cdot \omega^1$. Using $\omega^3 = 1$, we get $(1)^{13} \cdot \omega = 1 \cdot \omega = \omega$. $$ \omega^{40} = \omega $$

• Simplifying $\omega^{50}$:

  Divide the exponent $50$ by $3$: $50 = 3 \times 16 + 2$. The remainder is $2$. Therefore, $\omega^{50} = \omega^{3 \times 16 + 2} = (\omega^3)^{16} \cdot \omega^2$. Using $\omega^3 = 1$, we get $(1)^{16} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$. $$ \omega^{50} = \omega^2 $$

Putting It All Together

Now, substitute these simplified values back into the original expression:

$$ \omega^{10} + \omega^{20} + \omega^{30} + \omega^{40} + \omega^{50} = \omega + \omega^2 + 1 + \omega + \omega^2 $$

We can rearrange the terms to make the calculation easier:

$$ (\omega + \omega^2) + 1 + (\omega + \omega^2) $$

Alternatively, we can group them using the property $1 + \omega + \omega^2 = 0$:

$$ (1 + \omega + \omega^2) + (\omega + \omega^2) $$

Substitute $1 + \omega + \omega^2 = 0$:

$$ 0 + (\omega + \omega^2) $$

Now use the fact that $\omega + \omega^2 = -1$ (derived from $1 + \omega + \omega^2 = 0$):

$$ 0 + (-1) $$

$$ = -1 $$

Final Result

Therefore, the value of the expression $\omega^{10} + \omega^{20} + \omega^{30} + \omega^{40} + \omega^{50}$ is -1.