The complex conjugate of z, written \overline{z}, is defined by $$\overline{a+i b}=a-i b,$$ where a and b are the imaginary parts of z. It's pretty easy to show that
\begin{eqnarray} \overline{z_1+z_2}&=&\overline{z_1}+\overline{z_2},\\ \overline{z_1-z_2}&=&\overline{z_1}-\overline{z_2},\\ \overline{z_1z_2}&=&\overline{z_1}\overline{z_2},\\ \overline{z_1/z_2}&=&\overline{z_1}/\overline{z_2}. \end{eqnarray}
Here's how. One of the consequences of this is that if a_0, a_1, \dots, a_n are real, then $$\overline{a_n z^n + a_{n-1}z^{n-1}+\dots+a_1 z+a_0} = a_n \overline{z}^n + a_{n-1}\overline{z}^{n-1}+\dots+a_1 \overline{z}+a_0.$$ So if $$a_n z^n + a_{n-1}z^{n-1}+\dots+a_1 z+a_0=0,$$ then $$a_n \overline{z}^n + a_{n-1}\overline{z}^{n-1}+\dots+a_1 \overline{z}+a_0=0.$$ This tells us that the non-real roots of real polynomials always occur in conjugate pairs: if z is a root, then so is its conjugate. This is a highly useful fact. Suppose, for example, that we know that 2- 3 i is one of the roots of the quartic $$z^4 - 7 z^3 + 27 z^2 - 47 z + 26 = 0.$$ Well, then we instantly know, also, that 2 + 3 i is another root. Now, $$(z - [2 - 3 i]) (z - [2 + 3 i]) = z^2 - 4 z + 13,$$ and therefore, by the Factor Theorem, $$z^4 - 7 z^3 + 27 z^2 - 47 z + 26 $$ is the product of $$z^2 - 4 z + 13$$ and another quadratic factor: $$z^4 - 7 z^3 + 27 z^2 - 47 z + 26 = ( z^2 - 4 z + 13) (p z^2 + q z + r),$$ say.
It's not hard to work out that p = 1 and r = 2, by comparing the coefficients of z_4 and the constant terms respectively. It's a little more involved to work out, by (say) comparing coefficients of z, what q is. Here, $$-47 = 13 q - 4r = 13 q - 8, $$ and therefore q = -3. So we have $$z^4 - 7 z^3 + 27 z^2 - 47 z + 26 = ( z^2 - 4 z + 13) (z^2 - 3 z + 2), $$ and the roots are therefore 2 - 3 i, 2 + 3 i, 1 and 2. Note that this quartic has four roots. If non-real roots are allowed, this is always true: the number of roots of a polynomial is equal to its degree. This is called the Fundamental Theorem of Algebra.