The Argand diagram and geometry - TCE Mathematics Specialised (Tasmania)

What this dot point is asking

This dot point asks you to think of complex numbers as points or vectors in a plane rather than as algebraic symbols. The Argand diagram, named after Jean-Robert Argand, places the real part on the horizontal axis and the imaginary part on the vertical axis. Every algebraic operation then has a picture, and those pictures unlock the harder loci and proof questions.

Points and vectors

A complex number $z = x + iy$ is plotted at the coordinates $(x, y)$. It is equally useful to think of $z$ as the vector arrow from the origin to that point. This vector viewpoint is what makes addition geometric.

Addition as the parallelogram rule

Adding $z_1$ and $z_2$ corresponds to adding their position vectors. If you draw $z_1$ and $z_2$ from the origin, then $z_1 + z_2$ is the fourth corner of the parallelogram they span. Subtraction $z_1 - z_2$ is the vector that points from $z_2$ to $z_1$.

Multiplication as rotation and scaling

Multiplying by a complex number $w = r\operatorname{cis}\phi$ scales the modulus by $r$ and rotates by the angle $\phi$ anticlockwise. In particular, multiplying by $i$ rotates a point exactly $90^\circ$ anticlockwise without changing its distance from the origin, since $i = \operatorname{cis}\tfrac{\pi}{2}$. This is the geometric reason arguments add when you multiply.

Conjugate and negative

The conjugate $\bar z$ is the reflection of $z$ in the real axis: the real part stays, the imaginary part flips sign. The negative $-z$ is the reflection through the origin, a $180^\circ$ rotation.

Modulus and the triangle inequality

The modulus $|z| = \sqrt{x^2 + y^2}$ is the length of the vector from the origin to $z$. Because moduli behave like vector lengths, they satisfy the triangle inequality:

Equality holds only when $z_1$ and $z_2$ point in the same direction. This mirrors the geometric fact that one side of a triangle cannot exceed the sum of the other two.

Geometric proof of conjugate facts

Because the conjugate reflects in the real axis, the midpoint of the segment from $z$ to $\bar z$ always lies on the real axis at $\operatorname{Re}(z)$, and the segment itself is vertical. This is the picture behind $z + \bar z = 2\operatorname{Re}(z)$. Drawing such relationships rather than only manipulating symbols often shortens a proof to one or two lines.

Rotation about a point

Multiplication by $\operatorname{cis}\theta$ rotates a vector about the origin. To rotate a point $z$ by angle $\theta$ about a different centre $c$, first translate so the centre is at the origin, rotate, then translate back:

This single formula handles every "rotate $B$ about $A$ by $90^\circ$" question. For a $90^\circ$ anticlockwise turn, $\operatorname{cis}\dfrac{\pi}{2} = i$, so $z' = c + i(z - c)$, which is fast enough to do by hand. The geometric reading of multiplication as rotate-and-scale is what makes these transformations one-liners instead of coordinate-geometry slogs.

Why this matters

Every locus and region question in the complex plane is solved fastest by translating it into geometry on the Argand diagram. Build the habit now of sketching $z$, $\bar z$ and key differences before reaching for algebra, and remember that multiplication encodes rotation while addition encodes translation.