Solve the binomial equation Solve the binomial equation $$z^4 = -8$$ 

Below is the steps i have done

1: I have taken  |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).

2: Since $z=r(cos\alpha+isin\alpha)$ leads me to

$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$ 

Divide by 4 since the z term is raised by four gives $2^{1/4}(cos\pi/8 + k * \pi/2 +sin\pi/8 + k * \pi/2) $

Is this the correct way to solve this problem ? I am asking since i just started with binomic equations and been stuck for some hours with the question.
Another approach that is an efficient way forward is to exploit Euler's Identity and  write 

$$z^4=r^4e^{i4\theta}=-8=8e^{i(2n+1)\pi}$$

for all integer $n$.  Thus, upon inverting we have for $z$

$$z=re^{i\theta}=2^{3/4}e^{i(2n+1)\pi/4}$$

for $n=\pm 1, \pm 3$.  Therefore, the $4$ roots of $z$ are

$$\bbox[5px,border:2px solid #C0A000]{z=2^{3/4}e^{\pm i\pi/4}\,\,\text{and}\,\,2^{3/4}e^{\pm i3\pi/4}}$$

or in rectangular form

$$\bbox[5px,border:2px solid #C0A000]{z=2^{1/4}(1\pm i)\,\,\text{and}\,\,2^{1/4}(-1\pm i)}$$
Notice, $$z^4=-8$$
$$z^4=8i^2$$ $$z=\sqrt[4]{8i^2}$$
$$z=(2)^{3/4}\sqrt{i}$$ $$z=(2)^{3/4}\sqrt{\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}}$$
$$=(2)^{3/4}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{2}\right)+i\sin\left(\frac{(4k+1)\pi}{2}\right)\right)^{1/2}$$
$$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{4}\right)+i\sin\left(\frac{(4k+1)\pi}{4}\right)\right)$$
Setting $k=0$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$$
$$=(2)^{3/4}\left(\frac{1}{\sqrt 2}+i\frac{1}{\sqrt 2}\right)$$
$$z=(2)^{1/4}\left(1+i\right)$$

Setting $k=1$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)\right)$$
$$=(2)^{3/4}\left(\frac{-1}{\sqrt 2}+i\frac{-1}{\sqrt 2}\right)$$
$$z=-(2)^{1/4}\left(1+i\right)$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=(2)^{1/4}\left(1+i\right),\  z=-(2)^{1/4}\left(1+i\right)}}$$