jmc

First, we can factor $P(z)=z^8+(4\sqrt{3}+6)z^4-(4\sqrt{3}+7)$ as $$P(z)=(z^4-1)(z^4+4\sqrt{3}+7).$$The solutions to $z^4-1=0$ are 1, $-1,$ $i,$ and $-i$.

If $z^4+4\sqrt{3}+7=0,$ then $$z^4=-4\sqrt{3}-7=(-1)(4\sqrt{3}+7),$$so $z^2=\pm i\sqrt{4\sqrt{3}+7}.$

We try to simplify $\sqrt{4\sqrt{3}+7}.$ Let $\sqrt{4\sqrt{3}+7}=a+b.$ Squaring both sides, we get $$4\sqrt{3}+7=a^2+2ab+b^2.$$Set $a^2+b^2=7$ and $2ab=4\sqrt{3}.$ Then $ab=2\sqrt{3},$ so $a^2{b}^2=12.$ We can then take $a^2=4$ and $b^2=3,$ so $a=2$ and $b=\sqrt{3}.$ Thus, $$\sqrt{4\sqrt{3}+7}=2+\sqrt{3},$$and $$z^2=\pm i(2+\sqrt{3}).$$We now try to find the square roots of $2+\sqrt{3},$ $i,$ and $-i.$

Let $\sqrt{2+\sqrt{3}}=a+b.$ Squaring both sides, we get $$2+\sqrt{3}=a^2+2ab+b^2.$$Set $a^2+b^2=2$ and $2ab=\sqrt{3}.$ Then $a^2{b}^2=\frac{3}{4},$ so by Vieta's formulas, $a^2$ and $b^2$ are the roots of $$t^2-2t+\frac{3}{4}=0.$$This factors as $\left(t-\frac{1}{2}\right)\left(t-\frac{3}{2}\right)=0,$ so $a^2$ and $b^2$ are equal to $\frac{1}{2}$ and $\frac{3}{2}$ in some order, so we can take $a=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ and $b=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}.$ Hence, $$\sqrt{2+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{6}}{2}=\frac{\sqrt{2}}{2}(1+\sqrt{3}).$$Let $(x+yi{)}^2=i,$ where $x$ and $y$ are real numbers. Expanding, we get $x^2+2xyi-y^2=i.$ Setting the real and imaginary parts equal, we get $x^2=y^2$ and $2xy=1.$ Then $4x^2{y}^2=1,$ so $4x^4=1.$ Thus, $x=\pm \frac{1}{\sqrt{2}},$ and the square roots of $i$ are $$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i=\frac{\sqrt{2}}{2}(1+i),-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i=-\frac{\sqrt{2}}{2}(1+i).$$Similarly, we can find that the square roots of $-i$ are $$\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i=\frac{\sqrt{2}}{2}(1-i),-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i=\frac{\sqrt{2}}{2}(-1+i).$$Hence, the solutions to $z^4=-4\sqrt{3}-7$ are $$\frac{1+\sqrt{3}}{2}(1+i),-\frac{1+\sqrt{3}}{2}(1+i),\frac{1+\sqrt{3}}{2}(1-i),\frac{1+\sqrt{3}}{2}(-1+i).$$We plot these, along with 1, $-1,$ $i,$ $-i$ in the complex plane.



The four complex numbers 1, $-1,$ $i,$ $-i$ form a square with side length $\sqrt{2}.$ The distance between $\frac{1+\sqrt{3}}{2}(1+i)$ and 1 is $$\begin{array}{rl}\left|\frac{1+\sqrt{3}}{2}(1+i)-1\right|& =\left|\frac{-1+\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}i\right|\\ & =\sqrt{{\left(\frac{-1+\sqrt{3}}{2}\right)}^2+{\left(\frac{1+\sqrt{3}}{2}\right)}^2}\\ & =\sqrt{\frac{1-2\sqrt{3}+3+1+2\sqrt{3}+3}{4}}\\ & =\sqrt{2}.\end{array}$$Thus, each "outer" root has a distance of $\sqrt{2}$ to its nearest neighbors. So to the form the polygon with the minimum perimeter, we join each outer root to its nearest neighbors, to form an octagon with perimeter $\boxed{8\sqrt{2}}.$