Simple Question on Roots of Unity

Question

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

Answer 1

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.

Answer 2

$$4+4i=4\sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right)=(\sqrt2)^4\sqrt2e^{i\frac\pi4}$$ (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 $\sqrt2\left(-\frac{1+i}{\sqrt2}\right)=-(1+i)$

Answer 3

$(\sqrt{p^2+q^2})^5=4\sqrt{2}$, 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 $(p,q)=(-1, -1)$.