NZ NCEA Level 3 Mathematics and Statistics

The imaginary unit

No real number squares to a negative, so the equation $x^2 = -1$ has no real solution. The imaginary unit $i$ is defined to be a number with

which lets every negative number have a square root: $\sqrt{-16} = 4i$ and $\sqrt{-3} = \sqrt{3}\,i$. Take the factor $i$ out before doing anything else with a root of a negative number, because the usual rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only safe when at least one of $a$ and $b$ is positive.

Powers of $i$ repeat in a cycle of four:

so $i^{n}$ depends only on the remainder of $n$ divided by 4. For example $i^{27} = i^{24}\cdot i^{3} = -i$.

Rectangular form

A complex number has the form

where $x$ and $y$ are real. The number $x$ is the real part, written $\operatorname{Re}(z)$, and $y$ is the imaginary part, written $\operatorname{Im}(z)$. The imaginary part is the real number $y$, so it does not include the $i$. For $z = 5 - 2i$, $\operatorname{Re}(z) = 5$ and $\operatorname{Im}(z) = -2$.

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal. This is what lets one complex equation be split into two real equations, and it is used constantly when solving for unknown constants.

A number with $y = 0$ is real, so every real number is also a complex number. A number with $x = 0$ and $y \neq 0$ is purely imaginary.

Worked example: equating parts

Find real $a$ and $b$ if $(a + 2) + (3b - 6)i = 7 - 9i$.

The complex conjugate

The conjugate of $z = x + iy$ is the number with the sign of the imaginary part reversed:

It is sometimes written $z^*$. Three results follow at once and are used repeatedly:

The first is always real, the second is always purely imaginary, and the third is always real and never negative. The third is the reason conjugates are used to remove $i$ from a denominator.

Conjugation also respects arithmetic: the conjugate of a sum is the sum of the conjugates, and the conjugate of a product is the product of the conjugates. A complex number equals its own conjugate exactly when it is real.

Modulus

The modulus of $z = x + iy$ is its distance from the origin on the Argand diagram:

It is never negative and it is zero only for $z = 0$. For $z = 3 - 4i$:

The modulus is written $r$ when a number is in polar form. A useful consequence is $|z|^2 = z\bar{z}$, so $|z|^2$ can be found without taking a root.

The Argand diagram

An Argand diagram is a plane in which the complex number $x + iy$ is plotted as the point $(x, y)$. The horizontal axis is the real axis and the vertical axis is the imaginary axis. A complex number can equally be drawn as a vector from the origin to the point, which is how its modulus (the length) and its argument (the direction, described in polar form) are read off.

Several operations have a direct picture:

The last row can be checked directly: if $z = x + iy$ then $iz = ix + i^2 y = -y + ix$, which is the point $(x, y)$ rotated $90°$ anticlockwise to $(-y, x)$.

The distance between the points for $z$ and $w$ is $|z - w|$. This is the idea behind loci such as $|z - a| = r$, a circle of radius $r$ about the point $a$.

Worked example: the diagram in use

Let $u = 2 + i$ and $w = -1 + 3i$. Then $u + w = 1 + 4i$, which is the fourth vertex of the parallelogram with vertices at the origin, $u$ and $w$. The vector from $u$ to $w$ is

so the distance between the two points is

Worked example: conjugate and modulus together

Given $z = 4 + 3i$, find $z\bar{z}$ and confirm it equals $|z|^2$.

and $|z|^2 = 4^2 + 3^2 = 25$. The cross terms cancel in every product of a number with its conjugate, which is why the result is always real.

Real and imaginary parts from an equation

A single equation in a complex unknown often hides two real equations. Writing $z = x + iy$ and equating real and imaginary parts turns it into a pair of simultaneous equations.

Solve $z + 2\bar{z} = 9 + 6i$.

Substituting $z = x + iy$ gives $(x + iy) + 2(x - iy) = 3x - iy$. So $3x - iy = 9 + 6i$, and equating parts gives $3x = 9$, so $x = 3$, and $-y = 6$, so $y = -6$. Therefore $z = 3 - 6i$.

Check: $z + 2\bar{z} = (3 - 6i) + 2(3 + 6i) = 9 + 6i$, as required.