Euler's Formula for Complex Numbers

(There is another "Euler's Formula" about Geometry,
this page is about the one used in Complex Numbers)

First, you may have seen this famous equation:

eiπ + 1 = 0

It seems absolutely magical that such a neat equation combines:

But if you want to take an interesting trip through mathematics, then read on to find out why it is true.

Euler's Formula

It actually comes from Euler's Formula:

eix = cos x + i sin x

When we calculate it for x = π we get:

eiπ = cos π + i sin π
eiπ = −1 + i × 0   (because cos π = −1 and sin π = 0)
eiπ = −1
eiπ + 1 = 0

So eiπ + 1 = 0 is just a special case of a much more useful formula that Euler discovered.


It was around 1740, and mathematicians were interested in imaginary numbers.

An imaginary number, when squared gives a negative result

imaginary squared is negative

This is normally impossible (try squaring any number, remembering that multiplying negatives gives a positive), but just imagine that you can do it, call it i for imaginary, and see where it carries you:

i2 = -1

Euler was enjoying himself one day, playing with imaginary numbers (or so I imagine!), and he took this Taylor Series (which was already known):

 e^x = Sigma n=0 to infinity of x^n/n! = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...

(You can use the Sigma Calculator to play with this.)

And he put i into it:

 e^ix = 1 + ix + (ix)^2/2! + (ix)^3/3! + (ix)^4/4! + (ix)^5/5! + ...

And because i2 = -1, it simplifies to:

 e^ix = 1 + ix - x^2/2! - ix^3/3! + x^4/4! + ix^5/5! - ...

Now gather the terms with i and put them at the end:

taylor series e^ix = (1-x^2/2! + x^4/4! - ...) + i (1-x^3/3! + x^5/5! - ...)

And here is the miracle ...

  • the first group is the Taylor Series for cos
  • the second group is the Taylor Series for sin


taylor series cos(x) = 1-x^2/2! + x^4/4! - ..., and  sin(x) = 1-x^3/3! + x^5/5! - ...

He must have been so happy when he discovered this!

The result is:

e^ix = cos(x) + i sin(x)

Example: when x = 3

eix = cos x + i sin x
e3i = cos 3 + i sin 3
e3i = −0.990 + 0.141 i   (to 3 decimals)

Note: we are using radians, not degrees.

The answer is a combination of a Real and an Imaginary Number, which together is called a Complex Number.

We can plot such a number on the complex plane (the real numbers go left-right, and the imaginary numbers go up-down):

graph real imaginary -0.990 + 0.141i
Here we show the number −0.990 + 0.141 i

Which is the same as e3i

A Circle!

In fact, putting Euler's Formula on that graph produces a circle:

e^ix = cos(x) + i sin(x) on circle
ix produces a circle of radius 1

And we can turn any point (such as 3 + 4i) into reix form (by finding the correct value of x and the radius, r, of the circle)

Example: the number 3 + 4i

To turn 3 + 4i into reix form we do a Cartesian to Polar conversion:


So 3 + 4i can also be 5e0.927 i

3+4i = 5 at 0.927


There are many cases (such as multiplication) where it is easier to use reix than a+bi

Plotting eiπ

Lastly, here is the point created by eiπ (where our discussion began):

e^ipi = -1 + i on circle

eiπ = −1