Complex roots of quadratic equationsEdexcel International A Level Further Maths: Revision notes

For $az^2+bz+c=0$ with real $a,b,c$, the discriminant is $\Delta=b^2-4ac$. If $\Delta>0$ there are two distinct real roots; if $\Delta=0$ one repeated real root; if $\Delta<0$ there are no real roots, but there are two complex roots. For $z^2-6z+13=0$, $\Delta=36-52=-16<0$. The square root of a negative number uses $\sqrt{-n}=\mathrm{i}\sqrt n$, so $\sqrt{-16}=4\mathrm{i}$.

Use $z=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. When $\Delta<0$, write $\sqrt{\Delta}=\mathrm{i}\sqrt{|\Delta|}$. Example: $z^2-6z+13=0$ gives $z=\frac{6\pm4\mathrm{i}}{2}=3\pm2\mathrm{i}$. For $z^2+4z+29=0$, $\Delta=16-116=-100$ and $z=\frac{-4\pm10\mathrm{i}}{2}=-2\pm5\mathrm{i}$. For $2z^2-6z+5=0$, $\Delta=36-40=-4$ and $z=\frac{6\pm2\mathrm{i}}{4}=\frac32\pm\frac12\mathrm{i}$. Divide both parts of the numerator by $2a$.

Completing the square is often quicker when $a=1$ and $b$ is even. $z^2-4z+13=0$ becomes $(z-2)^2+9=0$, so $(z-2)^2=-9$ and $z-2=\pm3\mathrm{i}$, giving $z=2\pm3\mathrm{i}$. In general $z^2-4z+k=0$ gives $(z-2)^2=4-k$, so when $k>4$ the roots are $2\pm\mathrm{i}\sqrt{k-4}$.

When a quadratic has real coefficients its two complex roots are complex conjugates: $p+q\mathrm{i}$ and $p-q\mathrm{i}$. They have the same modulus and are reflections of each other in the real axis, so on an Argand diagram the distance between them is $2|q|$. For $z^2-4z+20=0$ the roots are $2\pm4\mathrm{i}$; each has modulus $\sqrt{20}=2\sqrt5$ and the two points are $8$ apart.

A condition such as 'no real roots' means $\Delta<0$. For $z^2+kz+13=0$ this gives $k^2<52$, so $-2\sqrt{13}<k<2\sqrt{13}$. To check a root by substitution, expand carefully: for $z=-2+5\mathrm{i}$, $z^2=4-20\mathrm{i}+25\mathrm{i}^2=-21-20\mathrm{i}$, then $z^2+4z+29=-21-20\mathrm{i}-8+20\mathrm{i}+29=0$.