Find the 4 roots of the equation: $z^{4}+4=0$ Find the 4 roots of the equation: $z^{4}+4=0$
Note: I can't use the form: $e^{i\theta}$
My attempt:
Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.
Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$
Then, 
$$w=4\cos\pi+i\sin\pi$$
$$z=p\cos\phi+i\sin\phi$$
This implies: $w=z^4$ iff $4=p^4$ and $4\phi=\pi+2k\pi$ iff $p=4^{1/4}$ and $\phi=\frac{\pi+2k\pi}{4}$ with $k=0,1,2,3$
Then the roots are:
$$z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$
is correct this?
 A: Consider the factorization of $z^4+4$:
$$z^4+4=z^4\color{blue}{+4z^2}+4\color{blue}{-4z^2}=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z-2)$$
Then we get $z^2+2z+2=0$ or $z^2-2z+2=0$, then finish it by using quadratic formula.
A: It is $$(z^2)^2+(2^2)^2=0$$ or $$a^2+b^2=0$$ and this is $$(a-bi)(a+bi)=0$$ where $$a=z^2,b=2^2$$
A: $z^2=\pm 2i$ and then the roots are $\pm1\pm i$. 
A: You solution looks OK; the writing can be improved slightly though. I will follow your argument and rewrite it as follows.
The equation is equivalent to $z^4=-4$. 
Now let $z=r(\cos \theta+i\sin\theta)$ and write 
$$
-4=4(\cos \pi +i\sin \pi)\tag{1}
$$
By the De Moivre's formula, 
$$
z^4=r^4(\cos (4\theta)+i\sin (4\theta))\tag{2}
$$
Comparing (1) and (2), we have
$$
r=\sqrt[4]{4},\quad 4\theta=(2k+1)\pi,\quad k\in\mathbb{Z}.
$$
Thus the four roots are
$$
z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$.

[Added:] The solution can be simplified further:


*

*Observe that $4^{1/4}=2^{2/4}=\sqrt{2}$. 

*The expression $\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}}$ can be found exactly by hand and thus can be used to simply the solution. For instance, 
$$
z_0=\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=1+i.
$$
I will leave the cases $k=1,2,3$ to you.
