The complex number $z = -4\sqrt{2} + 4\sqrt{2}i$ is given. Show that $Re\left(\sqrt[6]{z}\right)$ is $\cfrac{\sqrt{2\sqrt{2} +4}}{2}.$
The previous part of the question asks to convert the $z$ into Euler's form.
I'm able to do so and get:
$$z = 8 \, e^{i\frac{3\pi}{4}}$$
For $\sqrt[6]{z}$, I use this form to obtain roots. My reasoning is that the real part of $\sqrt[6]{z}$ will be the same for all roots, as they're periodically repeated.
$$\sqrt[6]{z} = \sqrt[6]{8} \,\left(e^{i\frac{3\pi}{4}+ 2k\pi}\right)^{\frac{1}{6}}$$ $$\iff \sqrt[2]{2} \, \left(e^{i\frac{3\pi + 8k\pi}{24}}\right)$$
It follows that:
$$Re(\sqrt[6]{z}) = \sqrt[2]{2} \, cos\left(\cfrac{\pi}{8}\right) \; \text{, for $k = 0$}$$
How should I go about solving this?
You know that\begin{align}\frac{\sqrt2}2&=\cos\left(\frac\pi4\right)\\&=\cos\left(2\frac\pi8\right)\\&=\cos^2\left(\frac\pi8\right)-\sin^2\left(\frac\pi8\right)\\&=2\cos^2\left(\frac\pi8\right)-1\end{align}and therefore$$\cos\left(\frac\pi8\right)=\sqrt{\frac{\sqrt2}4+\frac12}=\frac{\sqrt{2+\sqrt2}}2.$$