Think and solve

We first note that $z = 0$ is a solution for all positive $p,q$ .

Next, every non-zero complex number can be written uniquely as $z = r \cdot e^{i\theta}$ where $r>0,0\leq \theta < 2\pi$ . We note that this implies that $||z^a|| = r^a$ when written in this form since $||e^{ix}||=1, \forall x$ . Also, note that $\bar{z} = r\cdot e^{-i\theta}$ .

Thus, if the given equation is correct, then we first have

$||z^p|| = ||\bar{z}^q|| \Rightarrow r^p=r^q$ .

For positive real $r$ , this only has solutions at $r=1$ . Thus our original problem is reduced to

$e^{i\theta p} = e^{-i\theta q} \Rightarrow e^{i\theta (p+q)} = 1$ .

From here, we see that we must have

$\theta \cdot (p+q) = 2\pi n$

for natural $n$ , and restricting to the interval $0\leq \theta < 2\pi$ we have our entire set of solutions:

$\boxed{z=0,e^{2\pi i/(p+q)}, e^{2 \cdot 2\pi i/(p+q)}, \dots, e^{(p+q-1)\cdot 2\pi i/(p+q)} }$