Let $ z = \frac{1}{2}(\sqrt{2} + i\sqrt{2}) $. The sum $$ \sum_{k = 0}^{13} \dfrac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} $$
can be written in the form $ a - bi $. Find $ a + b $.
1. Given \( z = \frac{1}{2}(\sqrt{2} + i\sqrt{2}) \), we first simplify \( z \):
   \[
   z = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}
   \]
   This can be written as:
   \[
   z = \frac{1}{\sqrt{2}} (1 + i)
   \]

2. We need to find the sum:
   \[
   \sum_{k=0}^{13} \frac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}}
   \]

3. Notice that \( z^2 = \left( \frac{1}{\sqrt{2}} (1 + i) \right)^2 = \frac{1}{2} (1 + i)^2 = \frac{1}{2} (1 + 2i - 1) = \frac{1}{2} \cdot 2i = i \).

4. We can rewrite the sum as:
   \[
   \sum_{k=0}^{13} \frac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} = \sum_{k=0}^6 \left( \frac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} + \frac{1}{1 + ze^{k \cdot \frac{i\pi}{7}}} \right)
   \]

5. Simplify the terms inside the sum:
   \[
   \frac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} + \frac{1}{1 + ze^{k \cdot \frac{i\pi}{7}}} = \frac{1}{1 - ze^{k \cdot \frac{i\pi}{7}}} + \frac{1}{1 + ze^{k \cdot \frac{i\pi}{7}}}
   \]
   Using the common denominator:
   \[
   = \frac{(1 + ze^{k \cdot \frac{i\pi}{7}}) + (1 - ze^{k \cdot \frac{i\pi}{7}})}{(1 - ze^{k \cdot \frac{i\pi}{7}})(1 + ze^{k \cdot \frac{i\pi}{7}})}
   \]
   \[
   = \frac{2}{1 - (ze^{k \cdot \frac{i\pi}{7}})^2}
   \]

6. Since \( z^2 = i \), we have:
   \[
   (ze^{k \cdot \frac{i\pi}{7}})^2 = z^2 e^{2k \cdot \frac{i\pi}{7}} = i e^{2k \cdot \frac{i\pi}{7}}
   \]
   Therefore:
   \[
   \frac{2}{1 - (ze^{k \cdot \frac{i\pi}{7}})^2} = \frac{2}{1 - i e^{2k \cdot \frac{i\pi}{7}}}
   \]

7. The sum becomes:
   \[
   \sum_{k=0}^6 \frac{2}{1 - i e^{2k \cdot \frac{i\pi}{7}}}
   \]

8. The roots \( 1, e^{\frac{2i\pi}{7}}, e^{\frac{4i\pi}{7}}, \ldots, e^{\frac{12i\pi}{7}} \) are the 7th roots of unity, satisfying \( x^7 - 1 = 0 \).

9. We need to find the sum of the roots of the polynomial:
   \[
   \left( i \left( \frac{2}{x} - 1 \right) \right)^7 - 1 = 0
   \]
   Simplifying:
   \[
   -i \left( \frac{2}{x} - 1 \right)^7 - 1 = 0
   \]
   \[
   -i (2 - x)^7 - x^7 = 0
   \]
   \[
   (-1 + i) x^7 - 14 i x^6 + 84 i x^5 - 280 i x^4 + 560 i x^3 - 672 i x^2 + 448 i x - 128 i = 0
   \]

10. By Vieta's formulas, the sum of the roots of this polynomial is:
    \[
    \frac{14i}{-1 + i}
    \]

11. Simplify the fraction:
    \[
    \frac{14i}{-1 + i} \cdot \frac{-1 - i}{-1 - i} = \frac{14i(-1 - i)}{(-1 + i)(-1 - i)} = \frac{14i(-1 - i)}{1 + 1} = \frac{14i(-1 - i)}{2}
    \]
    \[
    = \frac{14i(-1 - i)}{2} = 7i(-1 - i) = 7(-i - i^2) = 7(-i + 1) = 7 - 7i
    \]

The final answer is \( \boxed{7 - 7i} \).