zeros of $p(z)=z^4+2$ I want to find all zeros of $p(z)=z^4+2$ and I'm not sure if I've done everything correctly. Can you correct this if something is wrong?
$$x^4+2=0 \iff x^4=-2=2\cdot(-1)$$
$$\Rightarrow x_k= \sqrt[4]{2}e^{\frac{i(2k+1)\pi}{4}}$$with $k\in \{0,1,2,3\}$ are the zeros of $p$. Is it correct?
Additional question: If I want to determine all zeros of $z^n+a$ with $a\in\mathbb{R}$ and $n\in\mathbb{N}$, are the zeros
$$x_k= \sqrt[n]{a}e^{\frac{i(2k+1)\pi}{n}}$$for $k=0,..,.n-1$ in general?
 A: You can factorize the polynomial $P(Z)=Z^4+2$=$Z^4-(i \sqrt 2)^2$=$(Z^2-i\sqrt 2)$$(Z^2+i \sqrt 2)$ and then you factorize the two factors 
$(Z^2-i\sqrt 2)$=$(Z-m\sqrt [4] 2)$$(Z+m\sqrt [4] 2)$ where m is the complex squart root of $i$, it is equals to $(\sqrt 2/2+i\sqrt 2/2$) or (-$(\sqrt 2/2+i\sqrt 2/2$))
and do the same thing with the second factor
A: It is correct. Now, the expression $e^{\frac{i(2k+1)\pi}{4}}$ can be calculated explicitly, it is $\pm \frac{\sqrt{2}}{2} \pm i\frac{\sqrt{2}}{2}$ which means you can write the zeroes in an explicit Cartesian way. I would expect my students to do this.
A: Suppose $z^n+a=0$ Let $b=a^{\frac{1}{n}}$, then we have $z^n+b^n=0$.
Divide by $b^n$ and let $u=\frac{z}{b}$, then rearrange, we have $u^n=-1$.
We know $-1=e^{\frac{i\pi}{2}}$, so $u^n=e^{\frac{i\pi}{2}}$.
Take the nth root of both sides, $u=e^{\frac{i\pi}{2n}}$.
Keep in mind we have $n$ roots by The Fundamental Theorem of Algebra.
So, $u=e^{\frac{i\pi}{2n}}\times e^{\frac{2\pi ki}{n}}$ for $k\in{0,1,...,n-1}$, where the second term is a Root of Unity for each k. 
Replacing $u$, $$z=be^{\frac{i\pi(1 + 4k)}{2n}}$$ 
