Solution of $p(z)=0$ with $z\in\mathbb C$ and $a_k\in\mathbb R$ for all $k$

Question

Suppose $p(z)=a_0+...+a_nz^n$ with $a_k\in\mathbb R$ for all $k$.

How can I prove that if $p(z)=0$ then $p(\bar z)=0$? I know it's true, but how can I prove it?

Answer 1

Note that $$p(\bar z) = \sum\limits_{i = 0}^n {{a_i}{{\bar z}^i}} $$now, if ${a_i} \in \mathbb{R}$, then ${a_i} = \overline {{a_i}} $ so that $$p(\bar z) = \sum\limits_{i = 0}^n {{{\bar a}_i}{{\bar z}^i}} $$now, for any pair of complex numbers $a$ and $b$, we have $$\begin{array}{l}\overline {a \times b} = \overline {({x_a} + i{y_a}) \times ({x_b} + i{y_b})} = \overline {({x_a}{x_b} - {y_a}{y_b}) + i({x_a}{y_b} + {x_b}{y_a})} = \\({x_a}{x_b} - {y_a}{y_b}) - i({x_a}{y_b} + {x_b}{y_a}) = \overline {({x_a} + i{y_a})} \times \overline {({x_b} + i{y_b})} = \overline a \times \overline b \end{array}$$so by replacing into our equation, we get $$p(\bar z) = \overline {\sum\limits_{i = 0}^n {{a_i}{z^i}} } = \overline {p(z)} = 0$$Hope it helps ;)