Simple Question on Roots of Unity The question asks:
Find integers $p$ and $q$ such that $(p + qj)^{5} = 4 + 4j$
The question prior to this was:
Find the fifth roots of $4 + 4j$ in the form $re^{j\theta }$, where $r > 0$ and $-\pi < \theta \leq  \pi$. Illustrate these fifth roots on an Argand diagram
I am able to do that but I am unsure which root to use to answer the original question. Help would be much appreciated
 A: You say you found all five roots of the equation, presumably in exponential form. Now write them in trigonometric form and simplify. I'm guessing there will only be one case in which both the $\cos \theta$ and $\sin \theta$ will be integers.
A: $$4+4i=4\sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right)=(\sqrt2)^4\sqrt2e^{i\frac\pi4}$$ (using Euler's Identity)
$$\implies 4+4i=(\sqrt2)^5e^{2ni\pi+\frac{i\pi}4}=(\sqrt2)^5e^{\frac{i(8n+1)\pi}4}$$ where $n$ is any integer
So, $$(4+4i)^\frac15=\sqrt2e^{\frac{(8n+1)i\pi}{20}}=\sqrt2\left(\cos \frac{(8n+1)\pi}{20}+i\sin  \frac{(8n+1)\pi}{20}\right)$$ where $0\le n\le 4$ 
Please find the explanation here regarding the values of $n$
We need $\sqrt2$ at the denominator of $\cos,\sin$ to make $p,q$ integers
Trying with $n=0,1,2,3,4$ we find
for $n=3,\cos \frac{(8n+1)\pi}{20}+i\sin  \frac{(8n+1)\pi}{20}$ becomes $\cos \frac{25\pi}{20}+i\sin  \frac{25\pi}{20}=-\frac{1+i}{\sqrt2}$
$\implies$ one of the vaues of $(4+4i)^\frac15$ is $\sqrt2\left(-\frac{1+i}{\sqrt2}\right)=-(1+i)$
A: $(\sqrt{p^2+q^2})^5=4\sqrt{2}$, so $p^2+q^2=2$, then $(p,q)=(\pm 1, \pm 1)$. It remains to check the argument $ 1\times 5,  3\times 5, -3\times 5, -1\times 5$ which one  $\equiv 1\mod 8$, the answer is $(p,q)=(-1, -1)$.
