If `z_(1),z_(2),z_(3) andz_(4)` are the roots of the equation `z^(4)=1,` the value of`sum_(i=1)^(4)z_i^(3)`is

Options

A. 0
B. 1
C. `i,i=sqrt(-1)`
D. `1+i,i=sqrt(-1)`

Answer

To solve the problem, we need to find the value of the sum \( \sum_{i=1}^{4} z_i^3 \) where \( z_1, z_2, z_3, z_4 \) are the roots of the equation \( z^4 = 1 \).

### Step-by-Step Solution:

1. **Identify the Roots of the Equation**:
The equation \( z^4 = 1 \) has four roots, which are the fourth roots of unity. These can be expressed in exponential form:
\[ z_k = e^{i \frac{2k\pi}{4}} \quad \text{for } k = 0, 1, 2, 3 \]
This gives us:
\[ z_1 = e^{i \frac{2 \cdot 0 \pi}{4}} = e^{0} = 1 \]
\[ z_2 = e^{i \frac{2 \cdot 1 \pi}{4}} = e^{i \frac{\pi}{2}} = i \]
\[ z_3 = e^{i \frac{2 \cdot 2 \pi}{4}} = e^{i \pi} = -1 \]
\[ z_4 = e^{i \frac{2 \cdot 3 \pi}{4}} = e^{i \frac{3\pi}{2}} = -i \]

2. **Calculate the Cubes of the Roots**:
Now, we need to find \( z_i^3 \) for each root:
\[ z_1^3 = 1^3 = 1 \]
\[ z_2^3 = (i)^3 = -i \]
\[ z_3^3 = (-1)^3 = -1 \]
\[ z_4^3 = (-i)^3 = i \]

3. **Sum the Cubes**:
Now we sum these values:
\[ \sum_{i=1}^{4} z_i^3 = z_1^3 + z_2^3 + z_3^3 + z_4^3 = 1 + (-i) + (-1) + i \]
Simplifying this:
\[ = 1 - 1 + (-i + i) = 0 \]

4. **Final Result**:
Therefore, the value of \( \sum_{i=1}^{4} z_i^3 \) is:
\[ \boxed{0} \]