Two many twos

If we write $x \; = \; \tfrac12\sqrt{2+\sqrt{2+\sqrt{2}}} \hspace{2cm} y \; = \; \tfrac12\sqrt{2- \sqrt{2+\sqrt{2}}}$ then $x^2 + y^2 \; = \; \tfrac14\left[2 + \sqrt{2+\sqrt{2}} + 2 - \sqrt{2+\sqrt{2}}\right] \; = \; 1$ so that $z \; = \; (x + iy)^4$ clearly has modulus $1$ , making the answer $\boxed{100}$ .

By the double angle formulas, used to derive the half angle formulas, the following can be obtained: $\dfrac{1}{2} \sqrt{2+ \sqrt{2 + \sqrt{2}}} = \cos \dfrac{\pi}{16}$ and $\dfrac{i}{2} \sqrt{2- \sqrt{2 + \sqrt{2}}} = \sin \dfrac{\pi}{16}$ By DeMoivre's formula, we obtain

$( \cos \dfrac{\pi}{16} + i \sin \dfrac{\pi}{16})^4 = \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}$

whereof the norm is easily seen to be $1$ , so the final answer is $\boxed{100.00}$ .

DeMoivre's formula may look almost too good to be true, but if you write complex numbers in their polar form and use $e^{i \theta} = \cos \theta + i \sin \theta$ , the additivity of exponents shows that if $a, b \in \mathbb{C}$ , then $\arg ab = \arg a + \arg b$ . In other words, multiplication of complex numbers $a$ and $b$ rotates $1$ by $\arg a + \arg b$ and scales $1$ by $|ab|$ .

$\begin{aligned} z & = \left(\frac 12\sqrt{2+\sqrt{2+\sqrt 2}} + \frac i2 \sqrt{2-\sqrt{2+\sqrt 2}} \right)^4 \\ & = \frac 1{2^4} \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}} \right)^4 \\ & = \frac 1{2^4} \left(\left|\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right|e^{\arg \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right)} \right)^4 \\ & = \frac 1{2^4} \left(\sqrt{2+\sqrt{2+\sqrt 2} + 2-\sqrt{2+\sqrt 2}}e^{\arg \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right)} \right)^4 \\ & = \frac 1{2^4} \left(\sqrt 4 e^{\arg \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right)} \right)^4 \\ & = \frac {2^4}{2^4} e^{4\arg \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right)} \\ & = e^{4\arg \left(\sqrt{2+\sqrt{2+\sqrt 2}} + i\sqrt{2-\sqrt{2+\sqrt 2}}\right)} \end{aligned}$

Therefore $|z| = 1$ and $100|z| = \boxed{100.00}$

If we believe this problem has a nice solution, we may hope to be able to write

$\cos\theta = \frac{1}{2} \sqrt{ 2 + \sqrt{ 2 + \sqrt{2}} },$

for some $\theta \in \left[ 0, 2 \pi \right).$ If that is the case, observe that using the identity we can find the value of the corresponding sine. That'll yield

$\sin\theta = \frac{1}{2} \sqrt{ 2 - \sqrt{ 2 + \sqrt{2}} },$

which is exactly the imaginary part of a number $z^{1/4}.$ In this case, we can write our number as

$z = \left( \cos\theta + i \sin\theta \right)^4 = e^{4i \theta},$

which has modulus $1.$