We had a fun problem for a student activity today:

Let $P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)$. What is the minimum perimeter among all the $8$-sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)$?

Solution: $8\sqrt{2}$.

I managed to find the solutions, change them to points, and use Graphics to plot them.

    sols = Solve[z^8 + (4 Sqrt[3] + 6) z^4 - (4 Sqrt[3] + 7) == 0, z];
    pts = {Re[z], Im[z]} /. sols;
    Graphics[{
      PointSize[Large], Point[pts]}

Which produced this image:

[![enter image description here][1]][1]

I would like to ask: How can I use Mathematica to determine how to find the minimum perimeter among all 8-sided polygonal in the complex plane whose vertices are the vertices indicated in my image? And how to best draw the resulting polygon?

**Update:** Based on Rahul's answer, I was able to do the following. Same start:

    sols = Solve[z^8 + (4 Sqrt[3] + 6) z^4 - (4 Sqrt[3] + 7) == 0, z];
    pts = {Re[z], Im[z]} /. sols;

Then, Rahul's command:

    tour = FindShortestTour[pts] // FullSimplify

Which gave the correct answer for the length of the shortest tour, but it also ordered the points for the shortest tour.

    (* {8 Sqrt[2], {1, 2, 3, 4, 5, 6, 7, 8, 9, 1}} *)

So it looks like they were already in the preferred order. Then I did this:

    pts = pts[[Last[tour]]]

Which sorted my points (which were already sorted, but added point number one at the end). Then:

    Graphics[{
      Line[pts],
      Red, PointSize[Large], Point[pts]
      }]

This produced am image similar to that provided by m_goldberg. Then I worried I wasn't interpreting the sort order properly, so I mixed up the list of points.

    pts = {{1/2 (-1 - Sqrt[3]), -(1/2) Sqrt[4 + 2 Sqrt[3]]}, {1/
        2 (1 + Sqrt[3]), -(1/2) Sqrt[4 + 2 Sqrt[3]]},
      {-1, 0}, {0, -1}, {0, 1}, {1, 0}, {1/2 (-1 - Sqrt[3]), 
       1/2 Sqrt[4 + 2 Sqrt[3]]}, {1/2 (1 + Sqrt[3]), 
       1/2 Sqrt[4 + 2 Sqrt[3]]}}

Then:

    tour = FindShortestTour[pts] // FullSimplify

Provided this order and the same shortest length:

    (* {8 Sqrt[2], {1, 3, 7, 5, 8, 6, 2, 4, 1}} *)

And:

    pts = pts[[Last[tour]]];
    Graphics[{
      Line[pts],
      Red, PointSize[Large], Point[pts]
      }]

Again produced the same result.

[![enter image description here][2]][2]

So I hope I am interpreting the FindShortesTour command correctly.

  [1]: https://i.sstatic.net/kSJgq.png
  [2]: https://i.sstatic.net/oEgEj.png