<s>[INST] 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.
 [/INST] $$ z^4 = -8 \Longleftrightarrow $$ 
$$ z^4 = |-8|e^{\arg(-8)i} \Longleftrightarrow $$ 
$$ z^4 = 8e^{\pi i} \Longleftrightarrow $$ 
$$ z^4 = 8e^{\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$ 
$$ z = \left( 8e^{\left( \pi + 2 \pi k \right) i} \right)^{\frac{1}{4}} \Longleftrightarrow $$
$$ z = \sqrt[4]{8}e^{\frac{1}{4}\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$ 
$$ z = \sqrt[4]{8}e^{\left( \frac{\pi}{4} + \frac{\pi k}{2} \right) i} \Longleftrightarrow $$ 
$$ z = \sqrt[4]{8}e^{\frac{\pi(2k+1)}{4} i}  $$ 
With $k \in \mathbb{Z}$ and $k$ goes from $0-3$

So the solutions are:
$$z_0=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 0+1)}{4} i}=\sqrt[4]{8}e^{\frac{\pi}{4}i}$$
$$z_1=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 1+1)}{4} i}=\sqrt[4]{8}e^{\frac{3\pi}{4}i}$$
$$z_2=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 2+1)}{4} i}=\sqrt[4]{8}e^{-\frac{3\pi}{4}i}$$
$$z_3=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 3+1)}{4} i}=\sqrt[4]{8}e^{-\frac{\pi}{4}i}$$
</s>