Do $z^{3/4}=-1 $ solutions exist? I want to get some perspective on complex exponentiation and the best possible approach to solve them.

$ z^{3/4}=-1  ;z \in \mathbb{C}$

Let $z=x+iy=\rho e^{i \theta} $   where $\rho =|z|$   and tan$\theta=\frac xy$
$w=-1=(1).e^{i(\pi+2k\pi)} $
According to me, there is a simple way to prove there is a solution to this equation. I may very well be wrong, but here I go.
$$ \frac{3\theta}{4}=\pi+2k\pi  $$
$$ \theta =\frac{4\pi}{3}+\frac{8k\pi}{3}$$
Which gives me 3 unique solutions for $\theta$: $\frac{4\pi}{3},\frac{2\pi}{3},0$
I'd then like to check that they verify $|z|=1$, I can disqualify $\theta=0$ in a jiffy. I then check:
$$tan\frac{4\pi}{3}=\frac{y}{x}$$
$$x tan\frac{4\pi}{3}=y$$
So from $|z|=1$:
$$ \sqrt{(x tan\frac{4\pi}{3})^2 +x^2}=1$$
$$x=\frac{+}{-}\frac{1}{ \sqrt{ (tan^2\frac{4\pi}{3} +1)} } $$
So nothing seems out of place to me so far... Wolfram Alpha insists there are no solutions and my calculator doesn't make the cut for complex numbers.

Let $z=r(\cos(\theta)+i\sin(\theta))$, by De-Moivre's theorem, we have: $z^3=r^3(\cos(3\theta)+i\sin(3\theta))$ with $0\leq\theta\leq2\pi$. Now we have to compute the value of: $$\left(z^3\right)^{1/4}=\left(r^3(\cos(3\theta)+i\sin(3\theta)\right)^{1/4}$$
with $0\leq\phi_k\leq2\pi$. Using again De-Moivre' theorem, we have:
$$w^4=z^3\leftrightarrow R^4(cos(4\phi)+i\sin(4\phi))=r^3(\cos(3\theta)+i\sin(3\theta))$$
From here, we deduce that: $$\phi_k=\frac{1}{4}(3\theta+2k\pi), k=0,1,2,3$$ and $R^4=r^3$. Now we can impose $w=-1$ and obtain:
$$(r^3)^{1/4}\left(\cos\left(\frac{3\theta+2k\pi}{4}\right)+i\sin\left(\frac{3\theta+2k\pi}{4}\right)\right)=1(\cos(\pi)+i\sin(\pi))$$
This leads to:
$$3\theta=4\pi+8k_1\pi-2k\pi,k_1=0,1,2$$
The solutions are: $$\theta=\pi/3 \: \vee \theta=5\pi/3 \: \vee \theta=\pi$$
In conclusion, I think that the solutions to this equations are the all $z$ such that $w=z^3=1$ because in that case the equation $w^{1/4}=-1$ has four solutions that in Gauss plane are the vertex of a square. 
