Complex numbers in rectangular form

Where complex numbers come from

• Some quadratic equations have no real solution. $x^2 + 1 = 0$ needs a number whose square is $-1$, and no real number squares to a negative.
• Mathematicians defined one:

$$i = \sqrt{-1} \quad \text{so that} \quad i^2 = -1$$

• $i$ is called the imaginary unit. Everything in this standard is built from it.
• A complex number is any number of the form $a + bi$, where $a$ and $b$ are real.
  • $a$ is the real part, written $\text{Re}(z)$.
  • $b$ is the imaginary part, written $\text{Im}(z)$.
• Note carefully: the imaginary part is $b$, a real number — not $bi$. For $z = 3 - 4i$, $\text{Im}(z) = -4$, not $-4i$.

Every real number is a complex number

• The 2026 specification states it directly: complex numbers include real numbers.
• A real number such as 7 is the complex number $7 + 0i$ — its imaginary part is zero.
• A number with zero real part, such as $5i$, is called purely imaginary.
• This matters because questions ask things like "find $k$ so that $z$ is purely real". That is an instruction to set the imaginary part equal to zero and solve.

Simplifying powers of $i$

• Powers of $i$ cycle with period 4:

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$$

• Divide the exponent by 4 and use the remainder.
  • $i^{23}$: $23 = 4 \times 5 + 3$, so the remainder is 3 and $i^{23} = i^3 = -i$.
  • $i^{100}$: remainder 0, so $i^{100} = 1$.

Square roots of negative numbers

• Write the negative root as a positive root times $i$:

$$\sqrt{-n} = i\sqrt{n} \quad (n > 0)$$

• $\sqrt{-25} = 5i$, and $\sqrt{-12} = i\sqrt{12} = 2i\sqrt{3}$.
• Convert to $i$ form first, then multiply. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails for negatives: $\sqrt{-4}\,\sqrt{-9}$ is $2i \times 3i = 6i^2 = -6$, not $\sqrt{36} = 6$.

Adding and subtracting

• Add real parts to real parts and imaginary parts to imaginary parts. It is collecting like terms.

$$(a + bi) + (c + di) = (a + c) + (b + d)i$$

• $$(3 + 5i) + (2 - 8i) = 5 - 3i$$
• $(3 + 5i) - (2 - 8i) = 1 + 13i$ — watch the sign on the $-8i$; subtracting it adds.

Multiplying

• Expand the brackets exactly as in ordinary algebra, then replace $i^2$ with $-1$ and collect.

$$(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$$

• Do not memorise that formula — just expand and simplify. Worked through:

$$(3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10i^2$$

$$= 12 - 7i - 10(-1) = 12 - 7i + 10 = 22 - 7i$$

• Squaring works the same way, but you must expand the binomial properly:

$$(2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i$$

• The error the assessors see is writing $(2 + 3i)^2 = 4 + 9i^2$. There is a middle term.

Equating real and imaginary parts

• Two complex numbers are equal only when both their real parts and their imaginary parts match.

$$a + bi = c + di \quad \Longleftrightarrow \quad a = c \;\text{ and }\; b = d$$

• This turns one complex equation into two real equations, and it is the engine behind most Merit and Excellence questions on this standard.

• To use it:

  • Expand and simplify both sides into the form $(\text{real}) + (\text{real})i$.
  • Group every term containing $i$ together, and everything else together.
  • Set the two real parts equal, and the two imaginary parts equal.
  • Solve the resulting pair of simultaneous equations.

• Worked through — find real $x$ and $y$ with $(x + yi)(2 - i) = 5 + 5i$:

  • Expand: $2x - xi + 2yi - yi^2 = (2x + y) + (2y - x)i$
  • Real parts: $2x + y = 5$
  • Imaginary parts: $2y - x = 5$
  • Solving gives $x = 1$, $y = 3$.

Purely real and purely imaginary

• Purely real means $\text{Im}(z) = 0$ — set the imaginary part to zero and solve.
• Purely imaginary means $\text{Re}(z) = 0$ — set the real part to zero and solve.
• The 2025 report lists "demonstrated understanding of purely real or purely imaginary complex numbers, and could form the resulting equations" as a Merit behaviour, so these questions are worth the practice.