If `omega ne 1` is a cube root of unity, then `z=sum_(k=1)^(60)omega^(k) - prod_(k=1)^(30)omega^(k)` is equal to

Options

A. 0
B. `omega`
C. `omega^(2)`
D. -1

Answer

To solve the problem, we need to evaluate the expression:

\[ z = \sum_{k=1}^{60} \omega^k - \prod_{k=1}^{30} \omega^k \]

where \(\omega\) is a cube root of unity and \(\omega \neq 1\). The cube roots of unity are \(1\), \(\omega\), and \(\omega^2\), where \(\omega = e^{2\pi i / 3}\) and satisfies the equation \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

### Step 1: Evaluate the summation \(\sum_{k=1}^{60} \omega^k\)

The sum of a geometric series can be calculated using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Here, \(a = \omega\), \(r = \omega\), and \(n = 60\).

Thus,

\[ \sum_{k=1}^{60} \omega^k = \omega \frac{1 - \omega^{60}}{1 - \omega} \]

Since \(\omega^3 = 1\), we find \(\omega^{60} = (\omega^3)^{20} = 1^{20} = 1\).

So,

\[ \sum_{k=1}^{60} \omega^k = \omega \frac{1 - 1}{1 - \omega} = \omega \cdot 0 = 0 \]

### Step 2: Evaluate the product \(\prod_{k=1}^{30} \omega^k\)

The product of the first \(n\) terms of a geometric series is given by:

\[ P_n = a^n \cdot r^{\frac{n(n-1)}{2}} \]

In our case, \(a = \omega\), \(r = \omega\), and \(n = 30\):

\[ \prod_{k=1}^{30} \omega^k = \omega^{1 + 2 + \ldots + 30} = \omega^{\frac{30 \cdot 31}{2}} = \omega^{465} \]

Now, since \(\omega^3 = 1\), we can reduce \(465\) modulo \(3\):

\[ 465 \mod 3 = 0 \]

Thus,

\[ \prod_{k=1}^{30} \omega^k = \omega^{465} = (\omega^3)^{155} = 1^{155} = 1 \]

### Step 3: Combine the results

Now substituting back into the expression for \(z\):

\[ z = \sum_{k=1}^{60} \omega^k - \prod_{k=1}^{30} \omega^k = 0 - 1 = -1 \]

### Final Result

Thus, the value of \(z\) is:

\[ \boxed{-1} \]