Is the fundamental theorem of algebra valid with polynomial terms like $\bar{z}$ and $\Re (z)$?

If we have a polynomial-equation with complex coefficients (of finite degree), like

$$z^3+5z+22=0$$

where $z\in\mathbb{C}$, then we're guaranteed as many complex solutions as the degree of the polynomial by the fundamental theorem of algebra. However, what if we were to have a polynomial that involved the real and imaginary parts of $z$ in a different way, say

$$z|z|^2+5\Re(z)+22=0$$

are we still guaranteed solutions to this type of equation?

• No: e.g. $z\bar{z}+1=0$ has no solutions. – Chappers May 11 '16 at 18:37

For a quick example just consider $|z|^2 = 1$ and $|z|^2 = -1$. The first has infinitely many solutions the second has no solution.
• Did anybody claim it were one? I did not. But if one expresses $z= x+ i y$ then one can write it as $P_1(x,y) + i P_2(x,y)=0$ where $P_j$ are two variable real polynomials. And the set of $z$ where it vanishes corresponds to the intersection of the sets of (x,y) where $P_1$ vanishes and (x,y) where $P_2$ vanishes, or also the (x,y) where $P_1^2 + P_2^2$, a two variable real polynomial, vanishes.. So it "boils down to is considering the zero set of two variable real polynomials. " – quid May 11 '16 at 23:52