How Do You Write a Complex Number in Cartesian Form (a + bi)?

A complex number is in Cartesian form when it is written as $a+bi$, where $a,b\in\mathbb{R}$ and $i^2=-1$. Simplify the expression, collect all real terms into $a$, and collect all multiples of $i$ into (bi).

This is IB Mathematics: Analysis and Approaches HL, sub-topic AHL 1.12: Complex numbers in Cartesian form and the Argand diagram. The algebra is particularly suitable for a Paper 1 question, although it can be tested on Paper 2.

The Method

1. Expand brackets and simplify fractions where necessary.
2. Replace every occurrence of $i^2$ with $-1$.
3. Combine the real terms.
4. Combine the terms containing $i$.
5. Write the final result in the order $a+bi$.

For example,

$$2(1+i)-(3-4i)=2+2i-3+4i=-1+6i.$$

Therefore, $a=-1$ and $b=6$.

Multiplication uses the same process: expand first and then reduce powers of $i$. For example,

$$(2+3i)(4-i)=8-2i+12i-3i^2=11+10i.$$

The product is therefore $11+10i$. Notice that the two terms involving $i$ combine, while $-3i^2$ becomes the real number $3$.

Before simplifying, it can help to label each contribution as real or imaginary. Addition and subtraction only combine like components; multiplication may create $i^2$, which changes an imaginary-looking product into a real value. After division, check that no $i$ remains in the denominator and that both coefficients are real numbers. This check catches sign errors.

For division, remove the complex number from the denominator by multiplying the numerator and denominator by its complex conjugate. The conjugate of $1-i$ is $1+i$:

$$\frac{3+2i}{1-i}\times\frac{1+i}{1+i} =\frac{(3+2i)(1+i)}{(1-i)(1+i)}.$$

Using $i^2=-1$,

$$\frac{3+5i+2i^2}{1-i^2} =\frac{1+5i}{2} =\frac{1}{2}+\frac{5}{2}i.$$

On an Argand diagram, the Cartesian number $a+bi$ corresponds to the point ((a,b)): $a$ is the real coordinate and $b$ is the imaginary coordinate.

A common misconception is that the imaginary part of $a+bi$ is (bi). The imaginary part is the real coefficient $b$, while (bi) is the imaginary term.

Exam Technique

Show the substitution $i^2=-1$ and, for division, show multiplication by the conjugate. Examiners award method marks for these steps, so a correct final answer without working may not receive full marks.