An Explanation of Euler’s Identity
Explaining Euler’s Identity 𝑒 2𝜋𝑖 = 1 requires examining several concepts...
1. The idea of imaginary number i.
First, we introduce the imaginary number, 𝑖. If we solve the equation 𝑥 2 + 1 = 0, we get
that 𝑥 2 = −1 . However, we find no solution among the real numbers, so mathematicians
developed the concept of an imaginary number 𝑖, where 𝑖 2 = −1. The resulting number system
is known as the Complex Numbers, where a complex number 𝑧 = 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real
numbers.
Second, let us define a point in the complex plane. We start by illustrating what it means to
multiply by i, geometrically. First, look at the real number line, a geometry with which we are
more familiar.
We have a horizontal line containing all the real numbers, with the positive reals to the right of
the origin and the negative reals the left of the origin. Now let us look at number 𝑛, a positive
real number. If we multiply by -1 we get number – 𝑛, the additive inverse of 𝑛. We know that the
distance from 𝑛 to the origin is the same as the distance from – 𝑛 to the origin, so the points are
symmetric about the origin on the number line, thus if we rotate number 𝑛 180˚ we will get – 𝑛.
So we see that multiplying by -1 is the same as rotating 180˚.
But now let us look at what happens when we want to rotate to −1 by using 2 rotations, so
rotate 90˚then rotate another 90˚ to end at 180˚or multiplying by −1. Let us call a rotation of
90˚ the same as multiplying by 𝑟, just as a rotation of 180˚ is the same as multiplying by −1.
Then if we rotate 𝑟 and then rotate 𝑟 again we get
𝑟 × 𝑟= −1
or
𝑟 2 = −1.
Now to solve for 𝑟 it seems logical to take the square root of both sides, so
r =√−1.
We know that i = √−1 so a rotation of 90˚ must be equal to multiplying by i.
Now back to complex numbers, let us consider 𝑧 = 𝑎 + 𝑏𝑖. When we go to plot this on the
Cartesian plane (𝑥𝑦 plane) we have a problem because both the 𝑥 axis and the 𝑦 axis are real
number lines. There is no place to put a complex number. What happens is that we get a whole
new graphing system, known as the complex plane, where the horizontal axis is the real number
line and the vertical axis is the imaginary number line. Since we know that multiplying by 𝑖 is
the same as rotating 90° if we take the real horizontal axis and multiply the entire line by 𝑖, we
get the imaginary vertical axis.
This means that we can represent any point 𝑧 = 𝑎 + 𝑏𝑖 as a point in the complex plane, where
the 𝑎 value determines the distance moved in the horizontal direction and the value of 𝑏
determines the distance moved in the vertical (imaginary) direction. It is important to note that in
the complex plane each point represents an individual value, whereas a point in the Cartesian
plane represents a coordinate, or two values (an 𝑥 value and a 𝑦 value).
2. The number 𝒆 and the derivative of function 𝒇(𝒙) = 𝒆𝒙 .
Let us look at the number 𝑒 first. The number 𝑒 can be expressed in several different ways.
1 n
For now we define e such that e = limn→∞ �1 + n� .
1 n
What is �1 + n� ?
1 n
Using binomial theorem we expand �1 + n� .
𝑛
𝑛
𝑛
𝑛
𝑛
1 𝑛
1
1
1
1
1
�1 + 𝑛� = � � 1𝑛 (𝑛)0 + � � 1𝑛−1 (𝑛)1 + � � 1𝑛−2 (𝑛)2 + � � 1𝑛−3 (𝑛)3 + ⋯ + � � 10 (𝑛)𝑛
0
3
𝑛
1
2
𝑛
1
= 1 + 1! ∗ (𝑛)1 +
=1+1+
1−
1
𝑛
2!
+
1
𝑛
𝑛(𝑛−1) 1 2
(𝑛 )
2!
2
𝑛
(1− )(1− )
1
3!
+⋯+
Now, when 𝑛 → ∞, 𝑛 → 0. Then
1
1
1
+
(1 + 𝑛)𝑛 = 1 + 1 + 2! + 3! + ⋯ = 𝑒
𝑛(𝑛−1)(𝑛−2) 1 3
(𝑛 )
3!
1
𝑛
2
𝑛
�1− ��1− �…(1−
𝑛!
+ ⋯+
(𝑛−1)
)
𝑛
𝑛(𝑛−1)(𝑛−2)…..3∗2∗1 1 𝑛
(𝑛 )
𝑛!
1
∴ lim𝑛→∞ (1 + 𝑛)𝑛 = 𝑒
1
1
1
If we let 𝑚 = 𝑛 then we get lim𝑚→0 (1 + 𝑚)𝑚 = lim𝑛→∞ (1 + 𝑛)𝑛 . That means
1
𝑒 = lim𝑚→0 (1 + 𝑚)𝑚
Now we define the derivative of 𝑓(𝑥) = 𝑏 𝑥 . According to the definition of derivative,
derivation is a measure of how a function’s values change as its input changes. Let 𝑦 = 𝑓(𝑥).
Then
∆𝑦
∴∆𝑥 =
∆𝑦 = 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = 𝑏 𝑥+∆𝑥 − 𝑏 𝑥 = 𝑏 𝑥 (𝑏 ∆𝑥 − 1)
𝑏 𝑥 (𝑏 ∆𝑥 −1)
∆𝑥
∆𝑦
∴ 𝑦 ′ = lim∆𝑥→0 ∆𝑥 = lim∆𝑥→0
𝑏 𝑥 (𝑏 ∆𝑥 −1)
∆𝑥
Let 𝑏 ∆𝑥 − 1 = 𝛽. Then 𝑏 ∆𝑥 = 𝛽 + 1
= 𝑏 𝑥 lim∆𝑥→0
𝑏 ∆𝑥 −1
∆𝑥
----------------(1)
When 𝑏 = 𝑒, then do 𝑙𝑛 on both sides.
ln 𝑒 ∆𝑥 = ln(𝛽 + 1)
∴∆𝑥 = ln(𝛽 + 1)
Therefore, 𝛽 → 0 when ∆𝑥 → 0.
So lim∆𝑥→0
lim𝛽→0
𝑒 ∆𝑥 −1
∆𝑥
1
1
ln(𝛽+1)𝛽
=
𝛽
1
= lim𝛽→0 ln(𝛽+1) = lim𝛽→0 ln(𝛽+1)/𝛽 = lim𝛽→0 1
1
𝛽
1
ln(𝛽+1)
= lim𝛽→0
1
1
ln(𝛽+1)𝛽
=
1
ln lim𝛽→0 (𝛽+1)𝛽
1
1
From the above (𝑒 = lim𝑚→0 (1 + 𝑚)𝑚 ), we know lim𝛽→0 (𝛽 + 1)𝛽 = 𝑒. So
lim∆𝑥→0
𝑒 ∆𝑥 −1
∆𝑥
1
1
= ln 𝑒 = 1 = 1 -----------(2)
Now we turn back to (1). 𝑦 ′ = 𝑏 𝑥 lim∆𝑥→0
𝑏 ∆𝑥 −1
∆𝑥
, but we change the base b to e. Then we have
𝑦 ′ = 𝑒 𝑥 lim∆𝑥→0
𝑒 ∆𝑥 −1
∆𝑥
.
Plug in (2) to (1), then we have
𝑦′ = 𝑒 𝑥 ∗ 1 = 𝑒 𝑥
Thus the derivative of 𝑒 𝑥 is 𝑒 𝑥 . This is probably the most important property of (𝑥) = 𝑒 𝑥 , it is
the only exponential function whose derivative is the same as the function itself.
But what happens when we generalize this real valued function to also take in complex
numbers? We want to preserve the property that the function has its own derivative. Now when
dealing with functions of real numbers, we can only vary the value of the input variable, the
independent variable, by changing the value of that input 𝑥. However when we instead look at
function whose input are complex numbers, we can change look at how the function varies as 𝑎,
the real part, changes, or as 𝑏, the imaginary part changes. So if we have a complex number
𝑑
𝑧 = 𝑎 + 𝑏𝑖 we can have either 𝑑𝑎 �𝑒 𝑎+𝑏𝑖 � the exponential property (𝑎𝑚+𝑛 = 𝑎𝑚 𝑎𝑛) also holds
𝑑
𝑑
𝑑
for complex and imaginary numbers so 𝑒 𝑎+𝑏𝑖 = 𝑒 𝑎 𝑒 𝑏𝑖 and thus 𝑑𝑎 �𝑒 𝑎+𝑏𝑖 � = 𝑑𝑎 𝑒 𝑎 × 𝑑𝑎 𝑒 𝑏𝑖 by
properties of derivative. Now since 𝑒 𝑏𝑖 does not depend on 𝑎, as 𝑎 changes 𝑒 𝑏𝑖 does not change
𝑑
𝑑
at all, it is constant and thus 𝑑𝑎 𝑒 𝑏𝑖 = 1. Now 𝑑𝑎 𝑒 𝑎 is saying what is the derivative of 𝑒 𝑎 where 𝑎
is a real number. This is what we showed above, and we said it is the function itself. So
𝑑
𝑑𝑎
𝑒𝑎 = 𝑒𝑎.
𝑑
𝑑
Now plugging this into 𝑑𝑎 𝑒 𝑎 × 𝑑𝑎 𝑒 𝑏𝑖 = 𝑒 𝑎 × 1 = 𝑒 𝑎 .
𝑑
𝑑
𝑑
𝑑
Now if we look at 𝑑𝑏 �𝑒 𝑎+𝑏𝑖 � = 𝑑𝑏 (𝑒 𝑎 𝑒 𝑏𝑖 ) = 𝑑𝑏 𝑒 𝑎 × 𝑑𝑏 𝑒 𝑏𝑖 . Since 𝑒 𝑎 does not depend on 𝑏,
𝑑
as 𝑏 varies 𝑒 𝑎 will remain the same and thus 𝑑𝑏 𝑒 𝑎 = 1. Now since we know that in the real
numbers
𝑑
𝑑𝑏
𝑑
𝑑
𝑑𝑎
𝑒 𝑎𝑥 = 𝑎𝑒 𝑥 and since
𝑒 𝑎 × 𝑑𝑏 𝑒 𝑏𝑖 = 1 × 𝑖𝑒 𝑏𝑖 = 𝑖𝑒 𝑏𝑖 .
𝑑
𝑑𝑥
𝑒 𝑥 = 𝑒 𝑥 in the complex plane,
𝑑
𝑑𝑏
𝑒 𝑏𝑖 = 𝑖𝑒 𝑏𝑖 . So thus
We know from above that the multiplying by 𝑖 is the same as rotating90˚. So if we look at
the derivative, it means that the value of the derivative is the value of the function at a 90˚
rotation. Since we are always moving at an 90˚ angle, we are never getting further away or closer
to where we started, instead we just move in a constant circle.
3. The unit circle, radians, and 𝟐𝝅.
The Unite circle in the Cartesian plane and in the Complex Plane
The unite circle on the Cartesian plane is a circle with radius 1.
We know that the distance formula is 𝑑 = �(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 . Because the radius is 1,
𝑑 = 1. Therefore, any line that starts at the origin and ends at a point on the circle must be length
one.
On the other hand, we know that for any angle θ in a right triangle (where 𝜃 is not the right
angle), the cosine function returns the ratio of the length of adjacent side to the length of the
adjacent
hypotenuse. That is cos(θ) = hypotenuse. By the same measure, the sine function returns the
opposite
ratio of the length of opposite side to the length of hypotenuse. That is sin(x) = hypotenuse. As
the pictures showing below, when a point is on the unit circle, the hypotenuse of the right
triangle which is radius, is equal to 1. Therefore, the x-coordinate of a point on the unit circle is
equal to the cosine of the angle and likewise the 𝑦-coordinate is equal to the since of the angle,
for any angle.
The same property holds in the complex plane. We can rewrite all points 𝑎 + 𝑏𝑖 as (𝑎, 𝑏) and
thus (𝑟 cos 𝜃, 𝑟 sin 𝜃) 𝑜𝑟 𝑟 cos 𝜃 + 𝑟𝑖 sin 𝜃 since we know that every coordinate in the complex
plane represents one number. That is known as polar form.
Let us look back at the complex number 𝑒 𝑎+𝑏𝑖 .
We said before that the complex number 𝑒 𝑎+𝑏𝑖 = 𝑒 𝑎 𝑒 𝑏𝑖 and thus this must be equivalent to the
value in polar form. So
𝑒 𝑎 𝑒 𝑏𝑖 = 𝑟 cos 𝜃 + 𝑟𝑖 sin 𝜃 = 𝑟 (cos 𝜃 + 𝑖 sin 𝜃 ) .
Now 𝑒 𝑎 is a real number since 𝑎 is a real number, thus 𝑒 𝑎 = 𝑟, the only real part in polar form.
So 𝑒 𝑏𝑖 = cos 𝜃 + isin 𝜃. Now in the complex number we are interested in 𝑒 2𝜋𝑖 , 𝑎 = 0 and thus
the radius, 𝑟 = 𝑒 0 = 1 for all possible values of 𝑏.
This means that for every value of 𝑒 𝑏𝑖 we are always 1 unit away from the origin of the circle.
That means that the values of 𝑏 map 𝑒 𝑏𝑖 onto a circle in the complex plane with radius 1.
Another way to think about this is to look again at when 𝑏 = 0 so at 𝑒 0𝑖 = 1. Since the rate
of change is 𝑖𝑒 𝑏𝑖 , which is equal to 𝑖, there is not a component that brings it closer to the origin,
just a rotation. Thus it stays a constant distance, 1, from the radius. This implies that we are
looking at a circle with radius 1 and that the magnitude of 𝑒 𝑏𝑖 and 𝑒 𝑏𝑖 is 1 for all values of 𝑏.
Since 𝑏 is always changing, we can assume that every point along the circle represents some 𝑏
value and as 𝑏 changes we move around the circle. But how do we know what relation the 𝑏
value has to the location on the circle? Since the derivative of 𝑒 𝑏𝑖 is 𝑖𝑒 𝑏𝑖 and the length of 𝑒 𝑏𝑖 is
1 for all values of 𝑏, the length of 𝑖𝑒 𝑏𝑖 must also be 1 just at a 90˚angle from 𝑒 𝑏𝑖 . This implies
that the rate of change has the same magnitude, or length, of 𝑒 𝑏𝑖 , just moving at a 90° angle. So
this means that the rate at which the function 𝑒 𝑏𝑖 is changing, or the magnitude of the derivative,
is the same as the amount the independent variable has changed. So we know that we need to go
all the way around a circle with radius 1, or 2π units around the circle. Since we know that the
rate or the speed we are moving at is the same as the amount 𝑏 has changed, we know 𝑏 must be
the number of radians we have rotated in the counterclockwise direction from the positive x-axis.
Another way to think about this is to take the real number line of all possible values of 𝑏. Now
also make a circle with radius 1 centered about the origin of the complex plane. For every 1 unit
you move on the real line, also move 1 unit around the circle. Since the circle has radius 1, the
circumference is equal to 2πr, or 2π units so we have to move 2π units around the circle, and thus
2π units on the real number line. Since 𝑏 is our only input value, it must correspond directly to
the angle measure. Since there are 2π radians in a circle and we must move 2π units. So the
change of distance in the range must correspond to the change of distance in the domain.
Radians and 2π
Let us look at what a radian is. Taking any circle with center at the vertex of the angle, a
radian measure of the angle is equal to the ratio of the length of the subtended arc to the radius of
the circle.
What happen when we look at the entire circle? How can we calculate the arc length of the
circle? Well, the arc length is equal to the circumference and the circumference of a circle is
equal to 2𝜋𝑟. If the circle is a unit circle, which means the radius is 1, then the circumference is
equal to 2π. According to the definition of radian, the number of radians is equal to the arc length,
which is 2π.
Euler’s Identity 𝐞𝟐𝛑𝐢 = 𝟏
Let us back to we are interested complex number 𝑒 2𝜋𝑖 . From the pictures above we see that
after rotating from the positive horizontal axis 2𝜋 radians, we get the value of 𝑒 2𝜋𝑖 .
In above we got 𝑒 𝑏𝑖 = cos 𝜃 + isin 𝜃. Now we know that 𝜃 is equal to out input 𝑏. So this
means that 𝑒 𝑏𝑖 = cos 𝑏 + isin 𝑏. If we plug in our desired 𝑏 value 2𝜋, then we get
𝑒 2𝜋𝑖 = cos(2𝜋) + 𝑖 sin(2𝜋) = 1 + 𝑖(0) = 1.
PS: if you are interested in another way to prove Euler’s identity using calculus, you can go
here: http://en.wikipedia.org/wiki/Euler%27s_formula.