Trig-gono-metry-2

In $\mathbb{C}$ , the zeroes of the polynomial $z^{8}-256$ are:

$\begin{aligned} z_{0} &= 2e^{i0)} = 2 \\ z_{1} &= 2e^{i\frac{\pi}{4}} = \sqrt{2} + \sqrt{2}i \\ z_{2} &= 2e^{i\frac{\pi}{2}} = 2i \\ z_{3} &= 2e^{i\frac{3\pi}{4}} = -\sqrt{2} + \sqrt{2}i \\ z_{4} &= 2e^{i\frac{\pi}{2}} = -2 \\ z_{5} &= 2e^{i\frac{5\pi}{4}} = -\sqrt{2} - \sqrt{2}i \\ z_{6} &= 2e^{i\frac{3\pi}{2}} = -2i \\ z_{7} &= 2e^{i\frac{7\pi}{4}} = \sqrt{2} - \sqrt{2}i \end{aligned}$

Given that all eight roots are equally spaced and thus make a regular octagon in the complex plane; the perimeter of this regular octagon is expressed as 8 times the distance between any two adjacent roots (say, $z_{0}$ and $z_{1}$ ). The distance/norm between the two complex points $z_{0}$ and $z_{1}$ is expressed as:

$\begin{aligned} d &= \lVert z_{1} - z_{0} \rVert \\ &= \lVert (\sqrt{2}-2)+\sqrt{2}i \rVert \\ &= \sqrt{8-4\sqrt{2}} = 2\sqrt{2-\sqrt{2}} \end{aligned}$

Thus, the perimeter of the octagon formed by the $z_{i}$ 's, denoted by $P$ , is:

$P = 8d = 16\sqrt{2-\sqrt{2}}$

which is of the form $a\sqrt{b-\sqrt{b}}$ . Therefore:

$a + b = 16 + 2 = 18$

as required.