Complex numbers and the Argand diagramEdexcel International A Level Further Maths: Revision notes

A complex number is a number $z=a+b\mathrm{i}$ with $a,b$ real and $\mathrm{i}^2=-1$. $a=\text{Re}(z)$ is the real part and $b=\text{Im}(z)$ is the imaginary part (a real number, without the $\mathrm{i}$). Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal. This turns one complex equation into two real equations: $(2x+y)+(x-3y)\mathrm{i}=7+7\mathrm{i}$ gives $2x+y=7$ and $x-3y=7$, so $x=4$, $y=-1$. The same number can be written in modulus-argument form $z=r(\cos\theta+\mathrm{i}\sin\theta)$, where $r=|z|$ and $\theta=\arg z$. Then $a=r\cos\theta$ and $b=r\sin\theta$.

The conjugate of $z=a+b\mathrm{i}$ is $z^*=a-b\mathrm{i}$: the sign of the imaginary part is reversed. The modulus is the distance from the origin, $|z|=\sqrt{a^2+b^2}$. It is always a non-negative real number. The product of a number and its conjugate is real: $zz^*=(a+b\mathrm{i})(a-b\mathrm{i})=a^2+b^2=|z|^2$. For $z=-3+4\mathrm{i}$, $zz^*=9+16=25$ and $|z|=5$. For two complex numbers, $|z_1z_2|=|z_1||z_2|$. With $z_1=5-12\mathrm{i}$ ($|z_1|=13$) and $z_2=1+\sqrt3\,\mathrm{i}$ ($|z_2|=2$), $|z_1z_2|=26$.

On an Argand diagram the complex number $a+b\mathrm{i}$ is the point $(a,b)$: the horizontal axis is the real axis and the vertical axis is the imaginary axis. The conjugate $z^*$ is the reflection of $z$ in the real axis. The modulus $|z|$ is the length $OP$ from the origin to the point $P$. The distance between the points for $z_1$ and $z_2$ is $|z_1-z_2|$. Points with the same imaginary part, such as $1+\mathrm{i}$ and $-\sqrt3+\mathrm{i}$, are a horizontal distance $1+\sqrt3$ apart.

The argument $\arg z=\theta$ is the angle between the positive real axis and the line $OP$, measured anticlockwise. The principal argument satisfies $-\pi<\theta\le\pi$. Find the acute angle $\alpha=\tan^{-1}\left|\frac{b}{a}\right|$, then use the quadrant: first quadrant $\theta=\alpha$; second $\theta=\pi-\alpha$; third $\theta=-(\pi-\alpha)$; fourth $\theta=-\alpha$. Example: $z=-3+4\mathrm{i}$ is in the second quadrant, $\alpha=\tan^{-1}\frac43=0.927$, so $\arg z=\pi-0.927=2.21$. For $z=5-12\mathrm{i}$ (fourth quadrant) $\arg z=-\tan^{-1}\frac{12}{5}=-1.18$.

Write $z_2=-\sqrt3+\mathrm{i}$ in the form $r(\cos\theta+\mathrm{i}\sin\theta)$. $r=\sqrt{3+1}=2$. $z_2$ is in the second quadrant with $\alpha=\tan^{-1}\frac{1}{\sqrt3}=\frac{\pi}{6}$, so $\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$. So $z_2=2\left(\cos\frac{5\pi}{6}+\mathrm{i}\sin\frac{5\pi}{6}\right)$. Check: $2\cos\frac{5\pi}{6}=-\sqrt3$ and $2\sin\frac{5\pi}{6}=1$. With $z_1=1+\mathrm{i}=\sqrt2\left(\cos\frac{\pi}{4}+\mathrm{i}\sin\frac{\pi}{4}\right)$, the angle between the two points at the origin is $\frac{5\pi}{6}-\frac{\pi}{4}=\frac{7\pi}{12}$, and the triangle they form with $O$ has area $\frac12(\sqrt2)(2)\sin\frac{7\pi}{12}$.