1.12 Complex numbers in Cartesian form (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

The number i

The equation $x^2 = -1$ has no real solution, so we define the imaginary unit $\mathrm{i}$ with $\mathrm{i}^2 = -1.$ Powers of $\mathrm{i}$ repeat in a cycle of four: $\mathrm{i}^1 = \mathrm{i}$ , $\mathrm{i}^2 = -1$ , $\mathrm{i}^3 = -\mathrm{i}$ , $\mathrm{i}^4 = 1$ , $\mathrm{i}^5 = \mathrm{i}$ , …

This lets us solve every quadratic: $x^2 + 4x + 13 = 0$ gives $x = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3\mathrm{i}$ .

Key termsimaginary unit

Common mistake

Writing $\mathrm{i}^2 = 1$ when expanding brackets. Every $\mathrm{i}^2$ becomes $-1$ .

Section 2

Cartesian form, real and imaginary parts

A complex number in Cartesian form is $z = a + b\mathrm{i}$ with $a, b \in \mathbb{R}$ .

$a = \mathrm{Re}(z)$ is the real part.
$b = \mathrm{Im}(z)$ is the imaginary part — a real number, so $\mathrm{Im}(3 + 4\mathrm{i}) = 4$ , not $4\mathrm{i}$ .

Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. This gives two equations from one, which is how you solve equations such as $z + 2z^{*} = 9 - 4\mathrm{i}$ or find square roots: $(a + b\mathrm{i})^2 = 5 + 12\mathrm{i}$ gives $a^2 - b^2 = 5$ and $2ab = 12$ , so $a + b\mathrm{i} = \pm(3 + 2\mathrm{i})$ .

Add and subtract by combining real and imaginary parts; multiply by expanding brackets and using $\mathrm{i}^2 = -1$ : $(3 - 4\mathrm{i})(1 + 2\mathrm{i}) = 11 + 2\mathrm{i}$ .

Key termscomplex numberreal partimaginary partequating parts

Exam tip

When a question says $a, b \in \mathbb{R}$ , that is your cue to equate real and imaginary parts.

Section 3

The complex conjugate and division

The complex conjugate of $z = a + b\mathrm{i}$ is $z^{*} = a - b\mathrm{i}$ . Key facts: $z + z^{*} = 2a, \qquad z z^{*} = a^2 + b^2 = |z|^2.$ Since $z z^{*}$ is real, we divide by multiplying the top and bottom by the conjugate of the denominator: $\frac{3 - 4\mathrm{i}}{1 + 2\mathrm{i}} = \frac{(3 - 4\mathrm{i})(1 - 2\mathrm{i})}{(1 + 2\mathrm{i})(1 - 2\mathrm{i})} = \frac{-5 - 10\mathrm{i}}{5} = -1 - 2\mathrm{i}.$ Non-real roots of a polynomial with real coefficients come in conjugate pairs, e.g. $-2 \pm 3\mathrm{i}$ .

Key termscomplex conjugateconjugate pair

Common mistake

Multiplying only the denominator by the conjugate. You must multiply top and bottom, otherwise the value changes.

Section 4

The complex plane, modulus and argument

The complex plane (Argand diagram) represents $z = a + b\mathrm{i}$ as the point $(a, b)$ : the horizontal axis is the real axis and the vertical axis is the imaginary axis.

The modulus $|z| = \sqrt{a^2 + b^2}$ is the distance from the origin to the point.
The argument $\arg z$ is the angle from the positive real axis to the line from the origin to the point, measured anticlockwise, usually taken in $-\pi < \theta \le \pi$ (radians).
$z^{*}$ is the reflection of $z$ in the real axis; $-z$ is the rotation of $z$ by $\pi$ about the origin.
$|z - w|$ is the distance between the points representing $z$ and $w$ .

For $z = 5 + 12\mathrm{i}$ : $|z| = 13$ and $\arg z = \arctan\frac{12}{5} \approx 1.18$ .

Key termscomplex planemodulusargument

Common mistake

Using $\arctan\frac{b}{a}$ without checking the quadrant. For $-1 - \mathrm{i}$ , $\arctan 1 = \frac{\pi}{4}$ , but the point is in the third quadrant, so $\arg = -\frac{3\pi}{4}$ .

Exam tip

Locate the quadrant first from the signs of $a$ and $b$ , then find the angle.

Must know

$\mathrm{i}^2 = -1$ ; $z = a + b\mathrm{i}$ with $\mathrm{Re}(z) = a$ , $\mathrm{Im}(z) = b$ .
Equal complex numbers: equate real parts and imaginary parts.
$z^{*} = a - b\mathrm{i}$ ; $z z^{*} = |z|^2$ ; divide by multiplying by the conjugate of the denominator.
$|z| = \sqrt{a^2 + b^2}$ ; find $\arg z$ from the quadrant.
In the complex plane, $|z - w|$ is the distance between two points and $z^{*}$ is a reflection in the real axis.

That's the notes covered.

Carry on to the next subtopic.