Argand diagrams and modulus-argument formAQA A-Level Further Maths: Revision notes

An Argand diagram represents $z=x+iy$ as the point $(x,y)$, with the real part on the horizontal real axis and the imaginary part on the vertical imaginary axis. The number can also be seen as the position vector from the origin $O$ to that point.

• Adding complex numbers adds the vectors: $z_1+z_2$ is the fourth vertex of the parallelogram formed by $O$, $z_1$, $z_2$.
• The vector from the point $P$ (representing $p$) to $Q$ (representing $q$) represents $q-p$.
• The conjugate $z^*=x-iy$ is the reflection of $z$ in the real axis.

The modulus $|z|$ is the distance from the origin to the point: $|z|=\sqrt{x^2+y^2}.$ Examples: $|3+4i|=5$, $|-1+i\sqrt3|=2$. The distance between the points representing $p$ and $q$ is $|q-p|$.

The argument $\theta=\arg z$ is the angle from the positive real axis to the vector, measured anticlockwise, in radians. The principal argument lies in $(-\pi,\pi]$. Find the acute reference angle $\alpha=\tan^{-1}\left|\frac{y}{x}\right|$, then use the quadrant:

• first quadrant: $\theta=\alpha$; second: $\theta=\pi-\alpha$;
• third: $\theta=-(\pi-\alpha)$; fourth: $\theta=-\alpha$. Examples: $\arg(-1+i\sqrt3)=\frac{2\pi}{3}$, $\arg(-2-2i)=-\frac{3\pi}{4}$. Also $\arg(-1)=\pi$ and $\arg(-i)=-\frac{\pi}{2}$.

With $r=|z|$ and $\theta=\arg z$: $z=r(\cos\theta+i\sin\theta).$ To convert from Cartesian form, find $r$ and $\theta$. To convert back, use $x=r\cos\theta$ and $y=r\sin\theta$. Examples: $1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$ and $2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)=-\sqrt3+i$. Use exact values: $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$, $\sin\frac{\pi}{6}=\frac12$, $\cos\frac{\pi}{3}=\frac12$, $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$.

For $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$: $z_1z_2=r_1r_2\big(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big),$ $\frac{z_1}{z_2}=\frac{r_1}{r_2}\big(\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\big).$ So $|z_1z_2|=|z_1||z_2|$ and $\arg(z_1z_2)=\arg z_1+\arg z_2$; for a quotient, divide moduli and subtract arguments. Proof of the product: $(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)=(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2)$, which is $\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$ by the compound angle formulae. Geometrically, multiplying by $z_2$ enlarges by $|z_2|$ and rotates anticlockwise by $\arg z_2$. Multiplying by $i$ rotates by $\frac{\pi}{2}$.

Take $z_1=1+i\sqrt3=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)$ and $z_2=-\sqrt3+i=2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$. Product: modulus $4$, argument $\frac{\pi}{3}+\frac{5\pi}{6}=\frac{7\pi}{6}$, which is $-\frac{5\pi}{6}$ in $(-\pi,\pi]$. So $z_1z_2=-2\sqrt3-2i$. Quotient: modulus $1$, argument $\frac{\pi}{3}-\frac{5\pi}{6}=-\frac{\pi}{2}$, so $\frac{z_1}{z_2}=-i$. The two vectors are perpendicular. Always bring the final argument back into $(-\pi,\pi]$ by adding or subtracting $2\pi$.