On the real number line, we draw “tick” marks to mark off distances of \(1\) to the left and right of \(0\). On the real number line, we have two direction vectors: \(-1\) and \(1\). They establish a unit length in each direction. Every real number is a positive scalar multiple of one of these.

Can we do the same thing for the Complex numbers?

\(\blacktriangleright \) Polar(Circular) Coordinates

Of course, this is easy with polar coordinates. Unit vectors look like \((1, \theta )\).

Every complex number can be described as a positive scalar multiple of a polar unit vector:

Can we describe each complex number as a scalar multiple of a unit vector in rectangular coordinates?

\(\blacktriangleright \) Rectangular Coordinates

Currently, we can describe every complex number as a sum of two scalar multiples of two direction vectors:

The Cartesian plane has two axes, so two sets of tick marks on two lines. The left/right tick marks are marking off multiples of \(\langle 1, 0 \rangle \) on the real axis. The up/down tick marks are marking off multiples of \(\langle 0, 1 \rangle \) on the imaginary axis.

We can view our vectors as constructed as a sum of perpendicular pieces, each of size \(1\).

\[ \langle 5, 7 \rangle = 5 \cdot \langle 1, 0 \rangle + 7 \cdot \langle 0, 1 \rangle \]

But, we want just a single vector, not a combination of two vectors.

Our idea is to take the vector representing a complex number and factor out its length. The resulting vector is a unit vector. We will have a unit vector pointing in the right direction (a direction vector) and a positive scalar, which is the length of the vector.

Each complex number can be represented by a vector. This vector can then be written as a scalar (number) times a unit vector.

\[ 3 + 4 \, i = \langle 3, 4 \rangle = 5 \cdot \left \langle \frac {3}{5}, \frac {4}{5} \right \rangle \]

The modulus of \(3 + 4 \, i\) is \(| 3 + 4 \, i | = \sqrt {3^2 + 4^2} = \sqrt {25} = 5\).

We factor out this length. \(5 \cdot \left \langle \frac {3}{5}, \frac {4}{5} \right \rangle \).

The result is the length times a unit vector

\[ \left | \cdot \left \langle \frac {3}{5}, \frac {4}{5} \right \rangle \right | = 1 \]

Substitute “modulus” for “length” and we can do the same for complex numbers in standard form.

Any complex number can be factored by dividing the real and imaginary parts by the modulus of the complex number. This modulus is the scalar out front.

We have factored \(a + b \, i \) into a product of a positive real number times a “unit” number.