Complex Numbers

Complex numbers extend the real number system by introducing $i = \sqrt{-1}$ , so every complex number takes the form $z = x + iy$ where $x, y \in \mathbb{R}$ . The real part is $\operatorname{Re}(z) = x$ and the imaginary part is $\operatorname{Im}(z) = y$ . On the Argand diagram, $z$ is plotted as the point $(x, y)$ , or equivalently as a position vector from the origin.

Syllabus note (important if you use older materials): since the 2025 revision of 9758, this topic is “Introduction to Complex Numbers” — everything happens in cartesian form. Polar (modulus-argument) form, exponential form $re^{i\theta}$ and Argand loci are no longer examined. Older notes, prelim papers and websites still cover them; skip those parts.

What you need to know

Arithmetic in cartesian form: add, subtract, multiply and divide complex numbers, remembering $i^2 = -1$ . To divide, multiply numerator and denominator by the conjugate of the denominator.
The complex conjugate of $z = x + iy$ is $z^* = x - iy$ . Key results: $z + z^* = 2\operatorname{Re}(z) \in \mathbb{R}$ , $z z^* = x^2 + y^2 \in \mathbb{R}$ , and $(z^*)^* = z$ .
The modulus is $|z| = \sqrt{x^2 + y^2} \geq 0$ , so $z z^* = |z|^2$ .
Geometrical effects on the Argand diagram: conjugation reflects $z$ in the real axis; negation rotates $z$ by $180^\circ$ about the origin; adding $w$ translates by the vector representing $w$ ; multiplying by $i$ rotates $z$ by $90^\circ$ anticlockwise about the origin.
Equating real and imaginary parts: two complex numbers are equal exactly when their real parts and imaginary parts are separately equal — the workhorse technique for solving equations in $z$ .
For a polynomial with real coefficients, complex roots occur in conjugate pairs: if $\alpha$ is a root then $\alpha^*$ is also a root.

Worked example 1 — quadratic with complex roots

Problem. Solve $z^2 - 4z + 13 = 0$ and show the roots on an Argand diagram.

Solution. The discriminant is $(-4)^2 - 4(13) = -36 < 0$ , so the roots are complex:

z = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.

The two roots $2 + 3i$ and $2 - 3i$ are a conjugate pair — on the Argand diagram they are mirror images in the real axis, at the points $(2, 3)$ and $(2, -3)$ .

Worked example 2 — finding a polynomial using conjugate-pair roots

Problem. A cubic polynomial $p(x)$ has real coefficients. One root is $z = 1 + 2i$ . Given that $p(x)$ has a real root $x = 3$ , find $p(x)$ as a product of real factors and state all three roots.

Because the coefficients are real, $z^* = 1 - 2i$ is also a root. The quadratic factor from the conjugate pair is

(x - (1+2i))(x - (1-2i)) = (x-1)^2 + 4 = x^2 - 2x + 5.

Including the real root $x = 3$ ,

p(x) = (x - 3)(x^2 - 2x + 5).

The three roots are $x = 3$ , $x = 1 + 2i$ , and $x = 1 - 2i$ .

Worked example 3 — solving $z^2 = 3 + 4i$ in cartesian form

Problem. Find the complex numbers $z$ such that $z^2 = 3 + 4i$ .

Solution. Let $z = x + iy$ with $x, y$ real. Then

z^2 = (x^2 - y^2) + 2xyi = 3 + 4i.

Equating real and imaginary parts:

x^2 - y^2 = 3, \qquad 2xy = 4.

From the second equation $y = \dfrac{2}{x}$ (note $x \neq 0$ ). Substituting into the first:

x^2 - \frac{4}{x^2} = 3 \implies x^4 - 3x^2 - 4 = 0 \implies (x^2 - 4)(x^2 + 1) = 0.

Since $x$ is real, $x^2 = 4$ , so $x = \pm 2$ and correspondingly $y = \pm 1$ :

z = 2 + i \quad \text{or} \quad z = -2 - i.

As a check, the two square roots are negatives of each other, as expected.

Worked example 4 — geometrical effect of multiplying by $i$

Problem. Let $z = 3 + 2i$ . Find $iz$ and describe the transformation that maps $z$ to $iz$ on the Argand diagram.

Solution.

iz = i(3 + 2i) = 3i + 2i^2 = -2 + 3i.

The point $(3, 2)$ maps to $(-2, 3)$ : this is a rotation of $90^\circ$ anticlockwise about the origin. Note that $|iz| = \sqrt{4+9} = \sqrt{13} = |z|$ — multiplication by $i$ preserves distance from the origin.

Common mistakes

Dropping $i^2 = -1$ when expanding. In $(x + iy)^2$ the cross-term is $2xyi$ and the $i^2y^2$ term contributes $-y^2$ to the real part; sign slips here derail the whole question.
Assuming all polynomials have conjugate-pair roots. The conjugate-pair theorem applies only when all coefficients are real. A polynomial with complex coefficients need not have this symmetry.
Errors when dividing. To simplify $\frac{1}{z}$ , multiply numerator and denominator by $z^*$ — and remember the denominator becomes the real number $|z|^2$ .
Revising removed content. Time spent on modulus-argument form, $re^{i\theta}$ or loci from pre-2025 notes and papers is time wasted for the current 9758 exam.