I am trying to find the fourth roots of $8\sqrt2(1+i)$.
So then, I was deciding to convert $1+i$ to an $re^{i\theta}$, where $r = \sqrt2$ and $\theta = 45^\circ$ or $\pi/4$.
then; $z = \sqrt2e^{i\pi/4}$, but then we need the fourth roots so then, take $\sqrt2e^{i\pi/4}$ and raise it to the fourth power?
You noticed that $1+i=\sqrt{2}e^{\pi i/4}$. That implies that $$8\sqrt{2}(1+i)=8\cdot\sqrt{2}\cdot\sqrt{2}\cdot e^{\pi i/4}=16e^{\pi i/4}.$$ If $z=re^{i\theta}$, where $r>0$ and $\theta\in[0,2\pi)$, has the property that $z^4=8\sqrt{2}(1+i)$, then we must have that $$z^4=(re^{i\theta})^4=r^4e^{4i\theta}=16e^{\pi i/4}.$$ Because $r$ is a non-negative real number, we know that $r^4=16$ implies that $r=2$.
However, solving for $\theta$ is slightly trickier because angles "wrap around" after $2\pi$. To be precise, $$e^{4i\theta}=e^{\pi i/4}$$ implies that $$4i\theta=\frac{\pi i}{4}+2\pi ik\text{ for some }k\in\mathbb{Z}$$
and therefore, $\theta\in[0,2\pi)$ is one of the four values $\frac{\pi }{16}+k\frac{\pi}{2}$ where $k\in\{0,1,2,3\}$.
$z=8\sqrt 2(1+i)$. In euler form, $z=re^{i\theta}$ where $r=\sqrt {(8\sqrt 2)^2+(8\sqrt 2)^2}=16$ and $\theta = \tan^{-1}(1)=\pi/4$. therefore, $z=16e^{i\pi/4}\implies z^{1/4}=2e^{i\pi/16}.e^{ik\pi/2}$ where number of values of $k$ is the root number($4$ here) and $k=\{1,2,3,4\}$.
Here's an overlooked, fairly simple way to find a complex number, raised to any complex power. Let $a$ be a complex constant and $z \neq 0$. Then $z ^ a = e^{a\log(z)}$. Here $log(z)$ is the complex logarithmic function defined by:
$log(z) = ln(|z|) + i arg(z)$
So in this case, we have $z = 8\sqrt{2}(1+i)$. So $z^{1/4} = e^{1/4\log(z)}$.
$|z| = 16$ and $arg(z) = \pi/4 + 2k\pi.$
Thus $$z^{1/4} = e^{1/4(\ln(16) + i\pi(1/4 + 2k))}$$ $$= e^{\ln2}e^{i\pi/4(1/4 + 2k)}$$ $$= 2e^{i\pi/4(1/4 + 2k)}$$ where $k = 0,1,2,3.$
