Perform addition, subtraction, multiplication and division of complex numbers in Cartesian form using the conjugate

What this dot point is asking

SCSA Unit 3 opens the complex numbers by defining the imaginary unit $i$ with the single rule $i^2 = -1$. Once that is admitted, every quadratic and indeed every polynomial has a root. This dot point asks you to carry out the four arithmetic operations on numbers of the form $z = x + iy$ fluently in the calculator-free section, where the conjugate is the central tool.

Cartesian form and the imaginary unit

Write $z = x + iy$, where $x = \operatorname{Re}(z)$ is the real part and $y = \operatorname{Im}(z)$ is the imaginary part. Both $x$ and $y$ are real numbers. The defining property is $i^2 = -1$, from which the powers of $i$ cycle with period four:

To simplify any power $i^n$, reduce $n$ modulo $4$. For example $i^{50} = i^{2} = -1$ because $50 = 4 \times 12 + 2$.

Addition, subtraction and multiplication

Addition and subtraction act on the real and imaginary parts separately:

Multiplication uses the distributive law, then collapses $i^2$ to $-1$:

You never need to memorise that formula; expanding the brackets and substituting $i^2 = -1$ always works and is what markers want to see.

Division by realising the denominator

There is no $i$ in a denominator in a finished answer. To divide, multiply top and bottom by the conjugate of the denominator. Because $z\bar z = x^2 + y^2$ is real, this clears the imaginary part below the line:

The denominator $c^2 + d^2$ is the squared modulus of $c + di$. Split the result into real and imaginary parts to finish.

Solving quadratics with complex roots

Cartesian arithmetic immediately lets you solve quadratics that have no real roots. For $z^2 - 4z + 13 = 0$ the discriminant is $\Delta = 16 - 52 = -36 < 0$, so the quadratic formula gives

Here $\sqrt{-36} = \sqrt{36}\,\sqrt{-1} = 6i$. The two roots $2 + 3i$ and $2 - 3i$ are conjugates of each other, which is no accident: when a quadratic has real coefficients, complex roots always occur in conjugate pairs. You can check the roots with the sum and product: the sum is $4 = -\tfrac{b}{a}$ and the product is $(2 + 3i)(2 - 3i) = 4 + 9 = 13 = \tfrac{c}{a}$, both matching the coefficients.

Equating real and imaginary parts

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal. This turns one complex equation into two real equations. To find real $a$ and $b$ with $(a + bi)(2 - i) = 5 + 5i$, expand the left side to $(2a + b) + (2b - a)i$, then equate parts: $2a + b = 5$ and $2b - a = 5$. Solving the pair gives $a = 1$, $b = 3$. This technique, sometimes called the method of equating coefficients, recurs throughout the complex-numbers topic and in the conjugate root theorem.

Why this matters

Cartesian arithmetic underpins everything later in the topic: solving quadratics with complex roots, the conjugate root theorem when factorising polynomials, and converting to polar form for de Moivre's theorem. A single careless sign on $i^2$ propagates through the whole question, so the discipline of writing $i^2 = -1$ explicitly at every multiplication pays off in the exam.