Question

If \( z \) is a non-real root of \( z^7 = 1 \), then the value of \[ 1 + 3z + 5z^2 + 7z^3 + 9z^4 + 11z^5 + 13z^6 \] is:

• A. \( \frac{14}{1 - z} \)
• B. \( \frac{-14}{1 - z} \)
• C. \( \frac{15}{1 - z} \)
• D. \( \frac{-15}{1 - z} \)

Hint:

When dealing with roots of unity, always consider using symmetry and identities like \( 1 + z + z^2 + \ldots + z^{n-1} = 0 \) for \( z^n = 1 \), \( z \neq 1 \). Multiply by \( 1 - z \) to reduce summations involving powers of roots.

Answer 1

The Correct Option is B

Step 1: Given \( z^7 = 1 \), and \( z \) is a non-real root, so \( z \neq 1 \), but \( z^7 = 1 \Rightarrow z \) is a 7th root of unity. The 7th roots of unity satisfy: \[ 1 + z + z^2 + \ldots + z^6 = 0. \] Step 2: Define the given expression as: \[ S = 1 + 3z + 5z^2 + 7z^3 + 9z^4 + 11z^5 + 13z^6. \] Group the terms to factor common powers: \[ S = 1 + 3z + 5z^2 + 7z^3 + 9z^4 + 11z^5 + 13z^6. \] Now reverse the order of coefficients and look for a pattern. Let’s define: \[ f(z) = 1 + 3z + 5z^2 + 7z^3 + 9z^4 + 11z^5 + 13z^6. \] This can be written as: \[ f(z) = \sum_{k=0}^{6} (2k + 1) z^k. \] Step 3: Use the identity: \[ \sum_{k=0}^{n-1} (2k + 1) z^k = \frac{2(1 - nz^n)}{(1 - z)^2} + \frac{1}{1 - z}. \] But that is a bit involved, so here’s a neater approach. Step 4: Multiply \( S \) by \( 1 - z \), then simplify: \[ (1 - z)S = (1 + 3z + 5z^2 + \ldots + 13z^6) - (z + 3z^2 + 5z^3 + \ldots + 13z^7). \] Now subtract term by term: \[ (1 - z)S = 1 + (3z - z) + (5z^2 - 3z^2) + (7z^3 - 5z^3) + \ldots + (13z^6 - 11z^6) - 13z^7. \] \[ = 1 + 2z + 2z^2 + 2z^3 + 2z^4 + 2z^5 + 2z^6 - 13z^7. \] Since \( z^7 = 1 \), we get: \[ (1 - z)S = 1 + 2(z + z^2 + z^3 + z^4 + z^5 + z^6) - 13. \] Step 5: Use the known identity \( 1 + z + z^2 + \ldots + z^6 = 0 \), so: \[ z + z^2 + \ldots + z^6 = -1. \] Thus: \[ (1 - z)S = 1 + 2(-1) - 13 = -14. \Rightarrow S = \frac{-14}{1 - z}. \]  the value is \( \boxed{\frac{-14}{1 - z}} \).