If `z_(r)(r=0,1,2,…………,6)` be the roots of the equation
`(z+1)^(7)+z^7=0`, then `sum_(r=0)^(6)"Re"(z_(r))=`

Options

A. 0
B. `3//2`
C. `7//2`
D. `-7//2`

Answer

To solve the equation \((z + 1)^7 + z^7 = 0\) and find the sum of the real parts of its roots, we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ (z + 1)^7 + z^7 = 0 \]
This can be rewritten as:
\[ (z + 1)^7 = -z^7 \]

### Step 2: Express in Polar Form
Let \( z = re^{i\theta} \). Then, we can express \( (z + 1) \) in polar form:
\[ (z + 1) = re^{i\theta} + 1 \]
We can find the magnitude and argument of \( (z + 1) \) and \( z \).

### Step 3: Use Binomial Theorem
Using the binomial theorem, we expand \((z + 1)^7\):
\[ (z + 1)^7 = \sum_{k=0}^{7} \binom{7}{k} z^k \cdot 1^{7-k} = z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 \]
Thus, we have:
\[ z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 + z^7 = 0 \]
This simplifies to:
\[ 2z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0 \]

### Step 4: Identify the Roots
Let \( z_r \) (for \( r = 0, 1, 2, \ldots, 6 \)) be the roots of the polynomial. We need to find:
\[ \sum_{r=0}^{6} \text{Re}(z_r) \]

### Step 5: Use Vieta's Formulas
According to Vieta's formulas, the sum of the roots of the polynomial \( a_n z^n + a_{n-1} z^{n-1} + \ldots + a_0 = 0 \) is given by:
\[ -\frac{a_{n-1}}{a_n} \]
In our case, \( a_7 = 2 \) and \( a_6 = 7 \), so:
\[ \text{Sum of roots} = -\frac{7}{2} \]

### Step 6: Calculate the Sum of Real Parts
Since the roots can be complex, we can split the sum of the roots into real and imaginary parts. The sum of the real parts of the roots is equal to the real part of the sum of the roots:
\[ \sum_{r=0}^{6} \text{Re}(z_r) = -\frac{7}{2} \]

### Final Answer
Thus, the final answer is:
\[ \sum_{r=0}^{6} \text{Re}(z_r) = -\frac{7}{2} \]

---