Q.Show that for any polynomial equation $P(x) = 0$, with real coefficients, imaginary roots occur in conjugate pairs.

Take the conjugate of the equation P(z)=0 and use the fact that conjugation commutes with sums/products and fixes real numbers, to show P(z̄)=0 too.

1. Let $P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ where every coefficient $a_0,a_1,\dots,a_n$ is real.
2. Suppose $z=\alpha+i\beta$ (with $\beta\neq0$) is a root of $P(x)=0$, i.e. $P(z)=a_0z^n+a_1z^{n-1}+\cdots+a_n=0$.
3. Take the complex conjugate of both sides: $\overline{a_0z^n+a_1z^{n-1}+\cdots+a_n}=\bar 0=0$.
4. Conjugation distributes over sums: $\overline{a_0z^n}+\overline{a_1z^{n-1}}+\cdots+\overline{a_n}=0$.
5. Conjugation distributes over products, and since each $a_i$ is real ($\overline{a_i}=a_i$), $\overline{a_iz^{k}}=\overline{a_i}\,\overline{z}^{\,k}=a_i\bar z^{\,k}$.
6. So the equation becomes $a_0\bar z^{\,n}+a_1\bar z^{\,n-1}+\cdots+a_n=0$, i.e. $P(\bar z)=0$.
7. Hence $\bar z=\alpha-i\beta$ is also a root of $P(x)=0$. …