A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $ R$ be the region outside the hexagon, and let $ S\equal{}\{\frac{1}{z}|z\in R\}$. Then the area of $ S$ has the form $ a\pi\plus{}\sqrt{b}$, where $ a$ and $ b$ are positive integers. Find $ a\plus{}b$.
1. **Understanding the Geometry of the Hexagon:**
   - The hexagon is regular and centered at the origin in the complex plane.
   - The distance between opposite sides is 1 unit.
   - One pair of sides is parallel to the imaginary axis.

2. **Calculating the Side Length of the Hexagon:**
   - The distance between opposite sides of a regular hexagon is twice the height of one of its equilateral triangles.
   - The height \( h \) of an equilateral triangle with side length \( s \) is given by \( h = \frac{\sqrt{3}}{2} s \).
   - Given \( 2h = 1 \), we have:
     \[
     2 \left( \frac{\sqrt{3}}{2} s \right) = 1 \implies s = \frac{1}{\sqrt{3}}
     \]

3. **Region \( R \) and Transformation \( S \):**
   - Region \( R \) is the area outside the hexagon.
   - The transformation \( S = \left\{ \frac{1}{z} \mid z \in R \right\} \) maps the exterior of the hexagon to the interior of another region.

4. **Analyzing the Transformation:**
   - For \( z = re^{i\theta} \) in polar coordinates, \( \frac{1}{z} = \frac{1}{r} e^{-i\theta} \).
   - The transformation maps points outside the hexagon to points inside circles centered at the origin.

5. **Finding the Shape of \( S \):**
   - Consider the region to the right of one side of the hexagon, say \( R_1 \).
   - For \( z \in R_1 \), \( \text{Re}(z) \geq \frac{1}{2} \).
   - In polar coordinates, \( z = re^{i\theta} \) implies \( r \cos \theta \geq \frac{1}{2} \).
   - For \( \frac{1}{z} = \frac{1}{r} e^{-i\theta} \), we have \( \frac{\cos \theta}{r} = \frac{1}{2} \implies 2 \cos \theta = r \).

6. **Converting to Cartesian Coordinates:**
   - Let \( z = a + bi \), then \( r = \sqrt{a^2 + b^2} \) and \( \cos \theta = \frac{a}{\sqrt{a^2 + b^2}} \).
   - Thus, \( 2a = a^2 + b^2 \implies (a-1)^2 + b^2 = 1 \).
   - This represents a circle centered at \( (1,0) \) with radius 1.

7. **Union of Circles:**
   - The region \( S \) is the union of circles centered at \( \text{cis}(\theta) \) for \( \theta = \frac{n\pi}{6} \), \( n = 0, 1, 2, 3, 4, 5 \).

8. **Calculating the Area:**
   - Each circle has radius 1.
   - The area of each circle is \( \pi \).
   - There are 6 such circles, so the total area is \( 6\pi \).

9. **Subtracting Overlapping Areas:**
   - The overlapping areas between adjacent circles need to be subtracted.
   - Each pair of adjacent circles overlaps in a lens-shaped region.
   - The area of each lens is \( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \).

10. **Summing the Areas:**
    - The total area of the union of circles is:
      \[
      6\pi - 6 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = 6\pi - 2\pi + \frac{3\sqrt{3}}{2} = 4\pi + \frac{3\sqrt{3}}{2}
      \]

11. **Final Expression:**
    - The area of \( S \) is \( 4\pi + \frac{3\sqrt{3}}{2} \).
    - Comparing with the form \( a\pi + \sqrt{b} \), we have \( a = 4 \) and \( b = 27 \).

The final answer is \( a + b = 4 + 27 = \boxed{31} \).