Complex numbers extend the reals with i where i squared equals minus one, and are represented as points or vectors on the Argand plane.

What this dot point is asking

The equation $x^2=-1$ has no real solution, so mathematicians define the imaginary unit $i$ with $i^2=-1$. A complex number in Cartesian (rectangular) form is $z=a+bi$, where $a=\operatorname{Re}(z)$ is the real part and $b=\operatorname{Im}(z)$ is the imaginary part.

Arithmetic in Cartesian form

Addition and subtraction act componentwise:

Multiplication uses the distributive law and $i^2=-1$:

Division removes $i$ from the denominator by multiplying numerator and denominator by the conjugate of the denominator:

The Argand plane

The Argand plane represents each complex number $z=a+bi$ as the point $(a,b)$, with the horizontal axis real and the vertical axis imaginary. Equivalently $z$ is the position vector from the origin to that point.

This geometric picture makes operations visual:

• Addition of $z_1$ and $z_2$ follows the parallelogram (tip-to-tail vector) rule.
• The conjugate $\bar z$ is the reflection of $z$ in the real axis.
• The modulus $|z|=\sqrt{a^2+b^2}$ is the distance from the origin to the point, the length of the vector. Note $|z|^2=z\bar z$.

Powers of i and equality of complex numbers

The powers of $i$ cycle with period four: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and then the pattern repeats. To simplify a high power, divide the exponent by $4$ and use the remainder; for example $i^{27}=i^{24}\cdot i^3=1\cdot(-i)=-i$. Two complex numbers are equal only when both their real parts and their imaginary parts match, so an equation between complex numbers gives two simultaneous real equations - the technique behind the first worked exam question above and behind solving for unknowns in complex equations generally.

Loci on the Argand plane

Because $|z-w|$ is the distance between the points $z$ and $w$, modulus equations describe geometric shapes. $|z-c|=r$ is a circle of radius $r$ centred at $c$; $|z-a|=|z-b|$ is the perpendicular bisector of the segment joining $a$ and $b$, since it is the set of points equidistant from both. Recognising a modulus condition as a distance lets you translate between the algebra of complex numbers and the geometry of the plane, which is exactly what the circle question above tests.

Common errors

Why it matters

Cartesian arithmetic and the Argand plane are the foundation for the rest of Topic 2. Multiplication and division become far simpler once you convert to polar form and apply De Moivre's theorem, and the geometric view is essential when finding roots of complex numbers.