Complex numbers and quadratic, cubic and quartic equationsEdexcel A-Level Further Maths: Revision notes

A complex number has the form $z=x+iy$ where $x$ and $y$ are real and $i^2=-1$. $x=\text{Re}(z)$ is the real part and $y=\text{Im}(z)$ is the imaginary part. Note that Im$(z)$ is the real number $y$, not $iy$. The modulus is $|z|=\sqrt{x^2+y^2}$, the distance from the origin on an Argand diagram. The argument $\arg z$ is the angle between the positive real axis and the line to the point $(x,y)$, measured anticlockwise in radians. For $z=3+2i$: $|z|=\sqrt{13}$ and $\arg z=\tan^{-1}\frac23\approx0.59$.

Add and subtract the real and imaginary parts separately. Multiply by expanding brackets and replacing $i^2$ with $-1$: $(3+4i)(1-2i)=3-6i+4i-8i^2=11-2i$. To divide, multiply the numerator and denominator by the complex conjugate of the denominator, which makes the denominator real: $\frac{3+4i}{1-2i}=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}=\frac{-5+10i}{5}=-1+2i$. Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal; this lets you solve equations by equating real and imaginary parts.

If $b^2-4ac<0$, the quadratic $az^2+bz+c=0$ has no real roots, but has two complex roots from $z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $\sqrt{-k}=i\sqrt{k}$. The roots are always a conjugate pair. Example: $z^2-4z+13=0$ gives $z=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3i$. Completing the square gives the same result: $(z-2)^2=-9$.

If a polynomial $f(z)$ has real coefficients and $z_1$ is a root, then $z_1^*$ is also a root. Non-real roots therefore occur in conjugate pairs. The pair $p\pm qi$ gives the real quadratic factor $z^2-2pz+(p^2+q^2)$. Consequences: a cubic with real coefficients has either three real roots or one real root and one conjugate pair; a quartic has four real roots, two real roots and one pair, or two pairs. The theorem fails if any coefficient is non-real, so check this before using it.

You are given enough information to find one root (cubic), or a complex root or quadratic factor (quartic). Method: use the conjugate to get a real quadratic factor, divide to find the remaining factor, then solve. Example 1: $f(z)=2z^3+5z^2+7z+10$ has factor $z+2$. Dividing: $f(z)=(z+2)(2z^2+z+5)$, so $z=-2$ or $z=\frac{-1\pm\sqrt{1-40}}{4}=\frac{-1\pm i\sqrt{39}}{4}$. Example 2: $g(x)=x^4-x^3+6x^2+14x-20$ with $g(1)=0$ and $g(-2)=0$. Then $(x-1)(x+2)=x^2+x-2$ is a factor and $g(x)=(x^2+x-2)(x^2-2x+10)$, so the other roots solve $x^2-2x+10=0$, giving $x=1\pm3i$. Roots: $1$, $-2$, $1+3i$, $1-3i$. Check by comparing coefficients or by expanding the factors.