The exponential form of complex numbers

Writing a complex number as $z = \rho e^{i\theta}$ looks like a mere convenient notation. In fact it hides a question worth untangling: why is $e^{i\theta}$ a complex number, and where does the exponential come from?

The idea is to see a rotation as a sum of many small steps. Rotating by an angle $\theta$ amounts to multiplying by the factor $\left(1 + \dfrac{i\theta}{n}\right)$ repeated $n$ times: each step is a small displacement perpendicular to the current position, because multiplying by $i$ rotates by $90°$. Starting from the number $1$ and applying this factor $n$ times gives the polygon

$$\left(1 + \frac{i\theta}{n}\right)^n.$$

As $n$ grows, the polygon gets finer and settles onto the arc of the unit circle: its final modulus tends to $1$ and the accumulated angle tends to $\theta$. That limit is, by definition, $e^{i\theta}$ — a complex number, obtained as a limit of complex numbers — and its value is

$$e^{i\theta} = \cos\theta + i\sin\theta.$$

This is Euler’s formula, which here is not assumed but emerges.

e^iθ as the limit of a polygon

e^iθ = lim_n→∞ (1 + iθ/n)ⁿ

steps n

5

angle θ

final modulus |z_n|

1.000

target: 1

final argument arg(z_n)

90.0°

polygon (1+iθ/n)ⁿtarget arce^iθ

Move the two sliders: with only a few steps the polygon is jagged and its final vertex lands outside the circle, with modulus $|z_n| > 1$ and angle smaller than $\theta$. As you increase $n$, the two numbers below converge to $1$ and $\theta$, and the vertex settles onto the point $e^{i\theta}$. For an arbitrary number $z = \rho e^{i\theta}$ you then just scale the circle by the modulus $\rho$.