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

An Argand diagram plots $z=x+iy$ as the point $(x,y)$, with the real axis horizontal and the imaginary axis vertical. $z$ is also the vector from the origin to that point.

• Sum: $z_1+z_2$ is the fourth vertex of the parallelogram with sides $Oz_1$ and $Oz_2$ (vector addition).
• Difference: $z_1-z_2$ is the vector from $z_2$ to $z_1$, so $|z_1-z_2|$ is the distance between the two points.
• The conjugate $z^*$ is the reflection of $z$ in the real axis. For $z_1=3+i$ and $z_2=1+2i$: $z_1+z_2=4+3i$ and $|z_1-z_2|=|2-i|=\sqrt5$.

Every non-zero $z$ can be written $z=r(\cos\theta+i\sin\theta)$, where $r=|z|$ and $\theta=\arg z$. The principal argument lies in $-\pi<\theta\le\pi$. To convert from $x+iy$: $r=\sqrt{x^2+y^2}$, find the acute angle $\alpha=\tan^{-1}\left|\frac yx\right|$, then place it in the correct quadrant: first $\theta=\alpha$, second $\pi-\alpha$, third $-(\pi-\alpha)$, fourth $-\alpha$. To convert back: $x=r\cos\theta$ and $y=r\sin\theta$. Example: $-1+i\sqrt3$ has $r=2$, $\alpha=\frac{\pi}{3}$, second quadrant, so $\theta=\frac{2\pi}{3}$: $2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)$.

For $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$, using the compound angle formulae: $z_1z_2=r_1r_2\left(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right),\quad \frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right).$ So $|z_1z_2|=|z_1||z_2|$, $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$, $\arg(z_1z_2)=\arg z_1+\arg z_2$ and $\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2$. Multiplying multiplies the lengths and adds the angles. Adjust results outside $-\pi<\theta\le\pi$ by adding or subtracting $2\pi$.

Let $z_1=1+i$ and $z_2=\sqrt3-i$. Then $z_1=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$ and $z_2=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)$. Hence $z_1z_2=2\sqrt2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)$. Expanding in Cartesian form gives $(\sqrt3+1)+(\sqrt3-1)i$, and equating real parts gives $\cos\frac{\pi}{12}=\frac{\sqrt6+\sqrt2}{4}$. Geometrically, multiplying by $z_2$ rotates by $\arg z_2$ and enlarges by $|z_2|$. Multiplying by $i$ (modulus 1, argument $\frac{\pi}{2}$) is a rotation by $90^\circ$ anticlockwise. This explains why two vectors with $\frac zw=\pm i$ are equal in length and perpendicular.