Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think it would not be wise to try to expand this, but rather apply de Moivre formula on the complex number in the numerator and denominator, then simplify that complex number within the root, and once again apply moivres formula.
I have tried but then I get the expression:$\sqrt[6]{\frac{\sqrt{2}+\cos{\frac{21 \pi}{4}+i\sin{\frac{21 \pi }{4}}}}{\cos \frac{11\pi}{4}+i\sin\frac{11 \pi}{4}}}$ and don't know what to do with it.
 A: Hint:        $$\cos\left(\dfrac{21}{4}\pi\right)=\cos\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\dfrac{21}{4}\pi\right)=\sin\left(\left(\dfrac{20}{4}+\dfrac{1}{4}\right)\pi\right),\\ \cos\left(\dfrac{11}{4}\pi\right)=\cos\left(\left(\dfrac{10}{4}+\dfrac{1}{4}\right)\pi\right),\ \,\sin\left(\dfrac{11}{4}\pi\right)=\sin\left(\left(\dfrac{10}{4}+\dfrac{1}{4}\right)\pi\right).$$
A: Here's my attempt at a solution using complex exponentials...
$$\begin{align*}
\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}} &= 
\sqrt[6]{\frac{\sqrt{2}+\left(e^{i3\pi/4}\right)^7}{\left(e^{i\pi/4}\right)^{11}}} 
\\
& = \sqrt[6]{\frac{\sqrt{2}+e^{i21\pi/4}}{e^{i11\pi/4}}}
\\
& = \sqrt[6]{\sqrt{2}e^{-i11\pi/4}+e^{i10\pi/4}}
\\
& = \sqrt[6]{\sqrt{2}e^{-i3\pi/4}+e^{i\pi/2}}
\\
& = \sqrt[6]{\sqrt{2}\left(-\frac{\sqrt{2}}{2}\color{red}{-}\frac{\sqrt{2}}{2}i\right)+i}
\\
& = \color{green}{\sqrt[6]{-1}}
\end{align*} $$
Now take the root, again using complex exponentials. Take care, though, as the root is multivalued. You ought to find six distinct points, as this is a sixth root of unity. Let me know if anything looks iffy.
EDIT: I checked the answer with Wolfram. I made a sign error above (red). The green result works out.
A: The binomial quantities being raised to powers are easy to simplify: $$\sqrt[6]{\frac{\sqrt{2}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}$$
Now it's quite easy to continue. $$\sqrt[6]{\frac{\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}=\sqrt[6]{-1}$$
