Doppelgangers

Let $z=re^{j\theta}$ . Then, $\bar{z}=re^{-j\theta}$ .

Thus, $\alpha_iz^{\beta_i}$ would be $\alpha_i r^{\beta_i} e^{j\beta_i\theta}$ . Similarly, $\alpha_i\bar{z}^{\beta_i}$ would be $\alpha_i r^{\beta_i} e^{-j\beta_i\theta}$ . Which is the complex conjugate of the former term.

Hence, $p(\bar{z})$ would be the complex conjugate of $p(z)$ .

In other words, if $p(z)=0$ or even a real number, then $p(z)=p(\bar{z})$ .

Therefore, no such case exists with $0=p(z)\neq p(\bar{z})$

p(z)= 0,means 'z' is a solution of this polynomial... If 'z' is a solution to polynomial with real coefficients,then, "z conjugate" is also it's solution (it is a standard result)... :)

If the $\beta_i$ are integers, such a $z$ cannot be found. If the $\beta_i$ are not integers, then $z$ can be found, but we need to make a suitable definition of the functions $z^\beta$ which does not using the principal branch. Cut the complex plane by removing the positive real axis, so that $0 < \mathrm{Arg}(z) < 2\pi$ for all $z$ in the domain, and consider the function $p(w) \; = \; w^{\frac12} + w^{-\frac12} - \sqrt{2}$ If we consider $i=e^{\frac12\pi i}$ , then $p(i) \; = \; \tfrac{1}{\sqrt{2}}(1+i) + \tfrac{1}{\sqrt{2}}(1-i) - \sqrt{2} = 0$ but if we consider $\overline{i}=-i=e^{\frac32\pi i}$ (this form is forced on us the but choice of domain!), so that $(-i)^{\frac12} = e^{\frac34\pi i} = \tfrac{1}{\sqrt{2}}(-1+i)$ , we have $p(-i) \;= \; \tfrac{1}{\sqrt{2}}(-1+i) + \tfrac{1}{\sqrt{2}}(-1-i) - \sqrt{2} \; = \; -2\sqrt{2}$ Since complex powers are multivalued, and the question does not specify the choice of domain to the principal branch, we can solve this problem for rational $\beta_i$ .