Complex conjugates and polynomial rootsAQA A-Level Further Maths: Revision notes

The complex conjugate of $z=x+iy$ is $z^*=x-iy$ (also written $\bar z$): same real part, imaginary part with the sign reversed. Key facts: $zz^*=x^2+y^2$ is real, and $z+z^*=2x$ is real. Conjugating also preserves sums and products: $(z_1+z_2)^*=z_1^*+z_2^*$ and $(z_1z_2)^*=z_1^*z_2^*$. Example: the conjugate of $3-2i$ is $3+2i$, and $(3-2i)(3+2i)=13$.

Theorem. If a polynomial has real coefficients and $z_0$ is a non-real root, then $z_0^*$ is also a root. Reason: if $p(z_0)=0$ then, because conjugation preserves sums and products and real coefficients are unchanged, $p(z_0^*)=\overline{p(z_0)}=0$. So non-real roots come in conjugate pairs. A cubic with real coefficients therefore has at least one real root, and a quartic has $0$, $2$ or $4$ real roots.

For a conjugate pair $a\pm bi$: $\big(z-(a+bi)\big)\big(z-(a-bi)\big)=(z-a)^2+b^2=z^2-2az+a^2+b^2.$ Example: roots $2\pm i$ give $z^2-4z+5$. This quadratic has real coefficients and discriminant $-4<0$.

When one root is known, find the other roots by factorising. Real root given: $z^3-5z^2+17z-13=0$ with root $z=1$. Divide by $(z-1)$, or compare coefficients in $(z-1)(z^2+pz+q)$, to get $z^2-4z+13$. Its roots are $2\pm3i$. Complex root given: $z^3-4z^2+6z-4=0$ with root $1+i$. Then $1-i$ is a root, the quadratic factor is $z^2-2z+2$, and $(z^2-2z+2)(z+c)$ with $2c=-4$ gives the third root $z=2$. Substituting a root into the equation is a valid check.

For a real quartic you are given one complex root or a quadratic factor. Given a complex root $a+bi$: write down $a-bi$, form the real quadratic factor, and divide to find the other quadratic. Example: $f(z)=z^4-2z^3+9z^2-8z+20$ has factor $z^2-2z+5$. Comparing coefficients, the other factor is $z^2+4$. Then $z^2-2z+5=0$ gives $1\pm2i$ and $z^2+4=0$ gives $\pm2i$. Always list all four roots (counting repeats).

State why the conjugate is a root: ‘the coefficients are real’. Use exact values in surd or $a+bi$ form. Check by expanding your factors back to the original polynomial, or by substituting one root. Questions with unknown real coefficients $a,b$ usually find them by expanding the factorised form and comparing coefficients.