How do I find poles of: $H(z) = \dfrac{z^3}{z^3+\alpha}$.
I know I must find the z values that do $z^3 = -\alpha$. I know how to do it in Matlab (with "residuez" function) but, how can i solve this simple equation analytically? This is so frustrating I'm finishing my engineering studies and I'm stuck with this. Help please. Thank you.
If $\alpha=0$ then $H(z)=1$ on $\mathbb C-\{0\}$ and has no poles.
If $\alpha\neq0$ then:
$$\left|z\right|^{3}e^{3i\arg z}=\left|z\right|^{3}e^{i\arg z^{3}}=z^{3}=-\alpha=\left|-\alpha\right|e^{i\arg\left(-\alpha\right)}=\left|\alpha\right|e^{i\left(\arg\alpha+\pi\right)}$$
leading to: $$\left|z\right|=\left|\alpha\right|^\frac{1}{3}$$ and: $$\arg z=\frac{1}{3}\arg\alpha+\frac{1}{3}\pi+\frac{2}{3}k\pi$$
true for every $k\in\mathbb Z$.