# The Exponential Nature of the Complex Unit Circle

## Introduction

This is an article to hopefully give an understanding to Euler's magnificent equation:

$$e^{i\theta} = cos( \theta ) + i \cdot sin( \theta )$$

This equation is usually proved using the Taylor series expansion for the given functions, but this approach fails to give an understanding to the equation and the ramification for the behavior of complex numbers. Instead an intuitive approach is taken that culminates in a graphical understanding of the equation.

## Complex Numbers

The variable $i$ stands for the square root of negative one. This is the American convention. The European convention uses the letter $j$. An imaginary number is $i$ multiplied by a real scalar. The term imaginary is a little unfortunate in that it is a bit misleading, being synonymous with non-existent. From a Mathematical perspective, $i$ is necessary to bring closure to the square root operation. The first obvious place a need pops up is the employment of the quadratic formula.

A complex number is the sum of a real number and an imaginary number. The complex plane is a representation of the entire set of complex numbers consisting of a Cartesian coordinate system with the horizontal axis representing the real component and the vertical axis representing the imaginary component by traditional convention. The complex number $z = a + b \cdot i$, where $a$ and $b$ are reals, is represented by the point $(a,b)$ on the Cartesian plane. Because of this, complex addition is similar to vector addition, and scalar multiplication is also the same, but this is not true for multiplication. Multiplying two complex numbers is not the same as taking the dot product of the corresponding vectors.

## The Unit Circle

The magnitude of a complex number is the distance from the number on the plane to the origin. Thanks to the Pythagorean theorem it is simple to calculate the magnitude:

$$| a + b \cdot i | = \sqrt{ a^{2} + b^{2} }.$$

The unit circle is the set of complex numbers whose magnitude is one. On the complex plane they form a circle centered at the origin with a radius of one. It includes the value 1 on the right extreme, the value $i$ at the top extreme, the value -1 at the left extreme, and the value $-i$ at the bottom extreme.

## Integer Powers of i

The powers of $i$ form a pattern that start to give a glimpse into the exponential nature of the circle.

$$i^{0} = 1 \\ i^{1} = i \\ i^{2} = -1 \\ i^{3} = i^{2} \cdot i^{1} = -i \\ i^{4} = i^{2} \cdot i^{2} = -1 \cdot -1 = 1 \\ i^{5} = i^{4} \cdot i^{1} = i \\ i^{6} = -1 \\ i^{7} = -i \\ i^{8} = 1$$

The pattern $1, i, -1, -i$ keeps repeating. If this pattern is mapped onto the unit circle it forms a linear scale around the circumference where every unit step on the scale represents a quarter turn on the circle. This pattern also works in the negative direction.

$$i^{-1} = i^{3}/i^{4} = -i \\ i^{-2} = i^{2}/i^{4} = -1 \\ i^{-3} = i^{1}/i^{4} = i \\ i^{-4} = 1/i^{4} = 1$$

Note that the origin on the circumference scale occurs at 1, also called Unity, on the complex plane.

## Some Sample Fractional Powers of i

Raising $i$ to integer powers results in traversing the unit circle in the same number of quarter turns. The next question that arises naturally is if the pattern also applies to fractional values as well. Consider $\sqrt{i}$. Suppose it is $a + b \cdot i$.

$$( a + b \cdot i )^{2} = i \\ a^{2} + 2ab \cdot i + b^{2} \cdot i^{2} = i$$

$$( a^{2} - b^{2} ) + ( 2ab ) \cdot i = 0 + 1 \cdot i$$

In order for two complex numbers to be equal, both the real parts and the imaginary parts must be equal.

$a^{2} - b^{2} = 0$ and $2ab = 1$

$a = ± b$ and $a = { 1 \over {2b} }$

Solving for a,b being real yields two solutions:

$(a,b) = \left( { \sqrt{2} \over 2 }, { \sqrt{2} \over 2 } \right)$ or $\left( - { \sqrt{2} \over 2 }, - { \sqrt{2} \over 2 } \right)$

Both these points are on the unit circle. The first one is half way between 1 and $i$. The second one is on the opposite side of the circle. The first one is exactly where it was expected, at the one half mark on the circumference scale. The second one makes sense if you consider $i = i^{5}$.

$$( i^{5} )^{1/2} = i^{5/2}$$

Five halves on the circumference scale is exactly where the second solution is found. The same results hold for $i = i^{( 1 + 4 \cdot n )}$. No matter what integer value of $n$ is chosen, the result is a value that places the solution on one of the points already found.

A more dramatic fraction example is finding the cube roots of one.

$$z^{3} = 1 \\ z^{3} - 1 = 0 \\ ( z - 1 ) \cdot ( z^{2} + z + 1 ) = 0$$ $z - 1 = 0$ or $z^{2} + z + 1 = 0$

The first equation gives $z = 1$. The second equation can be solved using the quadratic formula.

$$z = { { -1 ± \sqrt{ 1^2 - 4 \cdot 1 \cdot 1 } } \over {2 \cdot 1} } = { { -1 ± \sqrt{ 1 - 4 } } \over {2} }$$

$z = -{ {1} \over {2} } + i \cdot { \sqrt{ 3 } \over {2} }$ or $z = -{ {1} \over {2} } - i \cdot { \sqrt{ 3 } \over {2} }$

All three solutions lie on the unit circle. They are equally spread out and when connected with line segments form an inscribed equilateral triangle. The cube roots of one can be found a different way using the exponential method shown above along the circumference.

$$1^{1/3} = ( i^{0} )^{1/3} = i^{0} = 1 \\ 1^{1/3} = ( i^{4} )^{1/3} = i^{4/3} \\ 1^{1/3} = ( i^{8} )^{1/3} = i^{8/3} \\ 1^{1/3} = ( i^{12} )^{1/3} = i^{4} = 1$$

The pattern can be continued in both directions, and like the square root example above, the solutions always fall on the same set of points. This set of points are exactly the same ones found in the Cartesian solution.

$$i^{4/3} = -1/2 + i \cdot \sqrt{ 3 }/2 \\ i^{8/3} = -1/2 - i \cdot \sqrt{ 3 }/2$$

## Exponential Behavior

Since it works for several fractional values, it is not a stretch to assume that it will work for all fractional values. This is not mathematically rigorous, but it appears that raising $i$ to a scale distance along the circumference to a point yields the complex number at that point. Therefore any point on the unit circle can be reached by raising $i$ to the appropriate power. This power is not unique, any displacement of an integer multiple of four will also be a solution.

$$z = i^{p} = i^{( p + 4 \cdot n )}$$

This is why the complex unit circle can be seen as being exponential. Furthermore, if two complex numbers on the unit circle are multiplied, the resulting number is located at the sum of the circumference scale values of the two numbers on the unit circle.

$$z_{1} \cdot z_{2} = i^{p_{1}} \cdot i^{p_{2}} = i^{( p_{1} + p_{2} )} = z_{3}$$

The consequence of this is that any arbitrary scale can be used. Any point on the unit circle can be defined as one on a scale along the circumference. Then when that complex value is squared it will occur at distance two, cubed will be at distance 3 and so on.

## Roots of Unity

When the point is at $1/N$ of the circumference, the set of points defined by its integer powers form a special set called the Roots of Unity. Each value at each point, when raised to the $Nth$ power will be one. When $N=2$ you have the square roots of one: 1 and -1. When $N=3$ you get the points derived earlier. When $N=4$, the roots are $1, i, -1,$ and $-i$. It follows from this that $i$ raised to any rational number is actually a Root of Unity and that $i$ raised to an irrational number can never be a Root of Unity.

Another very special case of an arbitrary scale is when the unit point is set at one radian. When this is the case, the distance units along the circumference are the same as the distance units of the complex plane and the power scale coincides with radian angle measures. Suppose we call the value at one radian $u$. The Cartesian value is easily determinable.

$$u = cos( 1 ) + i \cdot sin( 1 )$$

Since the power scale coincides with radian angles the following equation will also be true.

$$u^{\theta} = cos( \theta ) + i \cdot sin( \theta )$$

The right hand side of this equation is identical to the right hand side of Euler's equation. It gives the value of the complex number on the unit circle for a given angle of $\theta$. The left hand side of the equation gives the location of the point on the unit circle defined by the distance along the circumference.

From the left hand side of Euler's equation:

$$u^{\theta} = e^{i\theta} = ( e^{i} )^{\theta}$$

This reveals that the real meaning of Euler's equation is that it functions as a translation of a distance along the circumference of a point on the complex unit circle to the complex value of the point on the complex plane.

## The Point at One Radian

Raising both sides to $1/\theta$ (or setting $\theta = 1$) leads to:

$$u = e^{i}$$

By setting $\theta = \pi/2$ in the first $u^{\theta}$ equation it becomes:

$$u^{\pi/2} = i$$

Raising both sides to $2/\pi$ gives:

$$u = i^{2/\pi}$$

Equating the two expressions for $u$.

$$e^{i} = i^{2/\pi}$$

Isn't that interesting? A real number raised to an imaginary power that equals an imaginary number raised to a real power in a simple equation involving the three fundamental constants of math: $e, i,$ and $\pi$. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):

$$e^{i\cdot( 1 + 2\pi\cdot m )} = i^{( 2/\pi + 4n )}$$

All the generalization does is introduce multiple representations for the same point.

## The Traditional Proof of Euler's Equation

Euler's equation is traditionally proved using the Taylor series expansions for the included functions. They are:

$$e^x = 1 + x + { x^2 \over 2 } + { x^3 \over {3!} } + { x^4 \over {4!} } + { x^5 \over {5!} } + { x^6 \over {6!} } + { x^7 \over {7!} } + ....$$ $$cos(x) = 1 - { x^2 \over 2 } + { x^4 \over {4!} } - { x^6 \over {6!} } + { x^8 \over {8!} } + ....$$ $$sin(x) = x - { x^3 \over {3!} } + { x^5 \over {5!} } - { x^7 \over {7!} } + { x^9 \over {9!} } + ....$$

The similarities in the three formulas should be readily apparent. Cosine is an even function, so all its powers are even. Likewise, Sine is an odd function so all its powers are the odd numbers.

The proof is straightforward. Substitute $i\theta$ for $x$ in the first series.

$$e^{i\theta} = 1 + i\theta + { (i\theta)^2 \over 2 } + { (i\theta)^3 \over {3!} } + { (i\theta)^4 \over {4!} } + { (i\theta)^5 \over {5!} } + { (i\theta)^6 \over {6!} } + { (i\theta)^7 \over {7!} } + ....$$

Next, factor out the powers of $i$ and reduce them to their simplest form.

$$e^{i\theta} = 1 + i\theta - { {\theta}^2 \over 2 } - i \cdot { {\theta}^3 \over {3!} } + { {\theta}^4 \over {4!} } + i \cdot { {\theta}^5 \over {5!} } - { {\theta}^6 \over {6!} } - i \cdot { {\theta}^7 \over {7!} } + ....$$

Separate the real and the imaginary terms.

$$e^{i\theta} = \left[ 1 - { {\theta}^2 \over 2 } + { {\theta}^4 \over {4!} } - { {\theta}^6 \over {6!} } + .... \right] + i \cdot \left[ \theta - { {\theta}^3 \over {3!} } + { {\theta}^5 \over {5!} } - { {\theta}^7 \over {7!} } + .... \right]$$

Finally, substitute back in the function definitions for the series.

$$e^{i\theta} = cos( \theta ) + i \cdot sin( \theta )$$

Quite easily done. This proof doesn't give any hint of the relationship of the equation to the complex unit circle, but it does have the advantage of proving the equation is valid for all values.

## Rescaling to the i Scale

Euler's equation is radian based. It can be converted to the $i$ scale with some simple substitutions.

$$e^{i\theta} = (e^{i})^{\theta} = (i^{2/\pi})^{\theta} = i^{{2 \over {\pi}} \cdot \theta}$$

This suggests the following substitution.

$$p = {2 \over {\pi}} \cdot \theta$$

Solve for $\theta$.

$$\theta = {{\pi} \over 2} \cdot p$$

Substitute these expressions into Euler's formula to get an equivalent $i$ scale based formula.

$$i^p = cos( {{\pi} \over 2} \cdot p ) + i \cdot sin( {{\pi} \over 2} \cdot p )$$

This formula validates the assumption made earlier that the $i$ based circumference power scale works for fractional values. Since $\pi/2$ is a right angle, it also confirms that each unit step on the power scale represents a quarter turn on the complex unit circle.

## Conclusion

Euler's magnificent equation relates the distance along the circumference of the complex unit circle to the underlying complex value. That this relationship is essentially exponential is not really intuitive. Knowing that the complex unit circle is inherently exponential in its behavior and that the radian scale, used by Euler's equation, is just one scale of many, makes the equation more understandable. Anytime $e^{ i \cdot something }$ appears in a formula, it means somehow the formula is referring to a point on the complex unit circle in some context.

If you found this article interesting, you may like its followup: Angle Addition Formulas from Euler's Formula

[ - ]
Comment by March 17, 2015
"That this relationship is essentially exponential is not really intuitive." -- The intuition is that to multiply complex numbers we add their angles. What transforms multiplication to addition? Taking a log. So in some sense the angle of a complex number (its argument) is some sort of logarithm. That's the key.
[ - ]
Comment by March 10, 2015
I tend to visualise Euler's equation this way (for a given angle):
cos = (.5cos + .5cos) + (j.5sin - j.5sin)
if we then draw all above rectangular components on complex plane we get equivalent two polar forms:
cos = (e^j + e^-j)/2

Kaz
[ - ]
Comment by January 2, 2023
An overview of all my articles can be found at https://www.dsprelated.com/showarticle/1490.php

This is the first article in the "Fundamentals" section.  The next article is:
[ - ]
Comment by August 17, 2023

i,j --> are they American and Europian notations? I was under the impression that they are Mathematicians vs Electrical Engineering notations respectively.

[ - ]
Comment by August 17, 2023
Yes, I believe you are correct about that.  Thank you.  I am clearly a Mathematician, also I use $S_n$ and $Z_k$ instead of $x[n]$ and $X[k]$ in my DFT articles.

To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
The text passage is a detailed mathematical article that explores the exponential nature of the complex unit circle, including the derivation and explanation of Euler's equation. It covers topics such as complex numbers, the unit circle, integer and fractional powers of \( i \), and the relationship between the exponential form and the Cartesian form of complex numbers.

### Analysis:
multiple_questions_with_answers

the passage contains multiple mathematical questions and their corresponding answers, such as:
1. **finding the square root of \( i \)**:
   - **question**: what are the values of \( a \) and \( b \) such that \( (a + b \cdot i)^2 = i \)?
   - **answer**: the solutions are \( (a, b) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \) or \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

2. **finding the cube roots of 1**:
   - **question**: what are the cube roots of 1?
   - **answer**: the cube roots of 1 are \(
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# The Exponential Nature of the Complex Unit Circle

## Introduction

This is an article to hopefully give an understanding to Euler's magnificent equation:

$$e^{i\theta} = cos( \theta ) + i \cdot sin( \theta )$$

This equation is usually proved using the Taylor series expansion for the given functions, but this approach fails to give an understanding to the equation and the ramification for the behavior of complex numbers. Instead an intuitive approach is taken that culminates in a graphical understanding of the equation.

## Complex Numbers

The variable $i$ stands for the square root of negative one. This is the American convention. The European convention uses the letter $j$. An imaginary number is $i$ multiplied by a real scalar. The term imaginary is a little unfortunate in that it is a bit misleading, being synonymous with non-existent. From a Mathematical perspective, $i$ is necessary to bring closure to the square root operation. The first obvious place a need pops up is the employment of the quadratic formula.

A complex number is the sum of a real number and an imaginary number. The complex plane is a representation of the entire set of complex numbers consisting of a Cartesian coordinate system with the horizontal axis representing the real component and the vertical axis representing the imaginary component by traditional convention. The complex number $z = a + b \cdot i$, where $a$ and $b$ are reals, is represented by the point $(a,b)$ on the Cartesian plane. Because of this, complex addition is similar to vector addition, and scalar multiplication is also the same, but this is not true for multiplication. Multiplying two complex numbers is not the same as taking the dot product of the corresponding vectors.

## The Unit Circle

The magnitude of a complex number is the distance from the number on the plane to the origin. Thanks to the Pythagorean theorem it is simple to calculate the magnitude:

$$| a + b \cdot i | = \sqrt{ a^{2} + b^{2} }.$$

The unit circle is the set of complex numbers whose magnitude is one. On the complex plane they form a circle centered at the origin with a radius of one. It includes the value 1 on the right extreme, the value $i$ at the top extreme, the value -1 at the left extreme, and the value $-i$ at the bottom extreme.

## Integer Powers of i

The powers of $i$ form a pattern that start to give a glimpse into the exponential nature of the circle.

$$i^{0} = 1 \\ i^{1} = i \\ i^{2} = -1 \\ i^{3} = i^{2} \cdot i^{1} = -i \\ i^{4} = i^{2} \cdot i^{2} = -1 \cdot -1 = 1 \\ i^{5} = i^{4} \cdot i^{1} = i \\ i^{6} = -1 \\ i^{7} = -i \\ i^{8} = 1$$

The pattern $1, i, -1, -i$ keeps repeating. If this pattern is mapped onto the unit circle it forms a linear scale around the circumference where every unit step on the scale represents a quarter turn on the circle. This pattern also works in the negative direction.

$$i^{-1} = i^{3}/i^{4} = -i \\ i^{-2} = i^{2}/i^{4} = -1 \\ i^{-3} = i^{1}/i^{4} = i \\ i^{-4} = 1/i^{4} = 1$$

Note that the origin on the circumference scale occurs at 1, also called Unity, on the complex plane.

## Some Sample Fractional Powers of i

Raising $i$ to integer powers results in traversing the unit circle in the same number of quarter turns. The next question that arises naturally is if the pattern also applies to fractional values as well. Consider $\sqrt{i}$. Suppose it is $a + b \cdot i$.

$$( a + b \cdot i )^{2} = i \\ a^{2} + 2ab \cdot i + b^{2} \cdot i^{2} = i$$

$$( a^{2} - b^{2} ) + ( 2ab ) \cdot i = 0 + 1 \cdot i$$

In order for two complex numbers to be equal, both the real parts and the imaginary parts must be equal.

$a^{2} - b^{2} = 0$ and $2ab = 1$

$a = ± b$ and $a = { 1 \over {2b} }$

Solving for a,b being real yields two solutions:

$(a,b) = \left( { \sqrt{2} \over 2 }, { \sqrt{2} \over 2 } \right)$ or $\left( - { \sqrt{2} \over 2 }, - { \sqrt{2} \over 2 } \right)$

Both these points are on the unit circle. The first one is half way between 1 and $i$. The second one is on the opposite side of the circle. The first one is exactly where it was expected, at the one half mark on the circumference scale. The second one makes sense if you consider $i = i^{5}$.

$$( i^{5} )^{1/2} = i^{5/2}$$

Five halves on the circumference scale is exactly where the second solution is found. The same results hold for $i = i^{( 1 + 4 \cdot n )}$. No matter what integer value of $n$ is chosen, the result is a value that places the solution on one of the points already found.

A more dramatic fraction example is finding the cube roots of one.

$$z^{3} = 1 \\ z^{3} - 1 = 0 \\ ( z - 1 ) \cdot ( z^{2} + z + 1 ) = 0$$ $z - 1 = 0$ or $z^{2} + z + 1 = 0$

The first equation gives $z = 1$. The second equation can be solved using the quadratic formula.

$$z = { { -1 ± \sqrt{ 1^2 - 4 \cdot 1 \cdot 1 } } \over {2 \cdot 1} } = { { -1 ± \sqrt{ 1 - 4 } } \over {2} }$$

$z = -{ {1} \over {2} } + i \cdot { \sqrt{ 3 } \over {2} }$ or $z = -{ {1} \over {2} } - i \cdot { \sqrt{ 3 } \over {2} }$

All three solutions lie on the unit circle. They are equally spread out and when connected with line segments form an inscribed equilateral triangle. The cube roots of one can be found a different way using the exponential method shown above along the circumference.

$$1^{1/3} = ( i^{0} )^{1/3} = i^{0} = 1 \\ 1^{1/3} = ( i^{4} )^{1/3} = i^{4/3} \\ 1^{1/3} = ( i^{8} )^{1/3} = i^{8/3} \\ 1^{1/3} = ( i^{12} )^{1/3} = i^{4} = 1$$

The pattern can be continued in both directions, and like the square root example above, the solutions always fall on the same set of points. This set of points are exactly the same ones found in the Cartesian solution.

$$i^{4/3} = -1/2 + i \cdot \sqrt{ 3 }/2 \\ i^{8/3} = -1/2 - i \cdot \sqrt{ 3 }/2$$

## Exponential Behavior

Since it works for several fractional values, it is not a stretch to assume that it will work for all fractional values. This is not mathematically rigorous, but it appears that raising $i$ to a scale distance along the circumference to a point yields the complex number at that point. Therefore any point on the unit circle can be reached by raising $i$ to the appropriate power. This power is not unique, any displacement of an integer multiple of four will also be a solution.

$$z = i^{p} = i^{( p + 4 \cdot n )}$$

This is why the complex unit circle can be seen as being exponential. Furthermore, if two complex numbers on the unit circle are multiplied, the resulting number is located at the sum of the circumference scale values of the two numbers on the unit circle.

$$z_{1} \cdot z_{2} = i^{p_{1}} \cdot i^{p_{2}} = i^{( p_{1} + p_{2} )} = z_{3}$$

The consequence of this is that any arbitrary scale can be used. Any point on the unit circle can be defined as one on a scale along the circumference. Then when that complex value is squared it will occur at distance two, cubed will be at distance 3 and so on.

## Roots of Unity

When the point is at $1/N$ of the circumference, the set of points defined by its integer powers form a special set called the Roots of Unity. Each value at each point, when raised to the $Nth$ power will be one. When $N=2$ you have the square roots of one: 1 and -1. When $N=3$ you get the points derived earlier. When $N=4$, the roots are $1, i, -1,$ and $-i$. It follows from this that $i$ raised to any rational number is actually a Root of Unity and that $i$ raised to an irrational number can never be a Root of Unity.

Another very special case of an arbitrary scale is when the unit point is set at one radian. When this is the case, the distance units along the circumference are the same as the distance units of the complex plane and the power scale coincides with radian angle measures. Suppose we call the value at one radian $u$. The Cartesian value is easily determinable.

$$u = cos( 1 ) + i \cdot sin( 1 )$$

Since the power scale coincides with radian angles the following equation will also be true.

$$u^{\theta} = cos( \theta ) + i \cdot sin( \theta )$$

The right hand side of this equation is identical to the right hand side of Euler's equation. It gives the value of the complex number on the unit circle for a given angle of $\theta$. The left hand side of the equation gives the location of the point on the unit circle defined by the distance along the circumference.

From the left hand side of Euler's equation:

$$u^{\theta} = e^{i\theta} = ( e^{i} )^{\theta}$$

This reveals that the real meaning of Euler's equation is that it functions as a translation of a distance along the circumference of a point on the complex unit circle to the complex value of the point on the complex plane.

## The Point at One Radian

Raising both sides to $1/\theta$ (or setting $\theta = 1$) leads to:

$$u = e^{i}$$

By setting $\theta = \pi/2$ in the first $u^{\theta}$ equation it becomes:

$$u^{\pi/2} = i$$

Raising both sides to $2/\pi$ gives:

$$u = i^{2/\pi}$$

Equating the two expressions for $u$.

$$e^{i} = i^{2/\pi}$$

Isn't that interesting? A real number raised to an imaginary power that equals an imaginary number raised to a real power in a simple equation involving the three fundamental constants of math: $e, i,$ and $\pi$. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):

$$e^{i\cdot( 1 + 2\pi\cdot m )} = i^{( 2/\pi + 4n )}$$

All the generalization does is introduce multiple representations for the same point.

## The Traditional Proof of Euler's Equation

Euler's equation is traditionally proved using the Taylor series expansions for the included functions. They are:

$$e^x = 1 + x + { x^2 \over 2 } + { x^3 \over {3!} } + { x^4 \over {4!} } + { x^5 \over {5!} } + { x^6 \over {6!} } + { x^7 \over {7!} } + ....$$ $$cos(x) = 1 - { x^2 \over 2 } + { x^4 \over {4!} } - { x^6 \over {6!} } + { x^8 \over {8!} } + ....$$ $$sin(x) = x - { x^3 \over {3!} } + { x^5 \over {5!} } - { x^7 \over {7!} } + { x^9 \over {9!} } + ....$$

The similarities in the three formulas should be readily apparent. Cosine is an even function, so all its powers are even. Likewise, Sine is an odd function so all its powers are the odd numbers.

The proof is straightforward. Substitute $i\theta$ for $x$ in the first series.

$$e^{i\theta} = 1 + i\theta + { (i\theta)^2 \over 2 } + { (i\theta)^3 \over {3!} } + { (i\theta)^4 \over {4!} } + { (i\theta)^5 \over {5!} } + { (i\theta)^6 \over {6!} } + { (i\theta)^7 \over {7!} } + ....$$

Next, factor out the powers of $i$ and reduce them to their simplest form.

$$e^{i\theta} = 1 + i\theta - { {\theta}^2 \over 2 } - i \cdot { {\theta}^3 \over {3!} } + { {\theta}^4 \over {4!} } + i \cdot { {\theta}^5 \over {5!} } - { {\theta}^6 \over {6!} } - i \cdot { {\theta}^7 \over {7!} } + ....$$

Separate the real and the imaginary terms.

$$e^{i\theta} = \left[ 1 - { {\theta}^2 \over 2 } + { {\theta}^4 \over {4!} } - { {\theta}^6 \over {6!} } + .... \right] + i \cdot \left[ \theta - { {\theta}^3 \over {3!} } + { {\theta}^5 \over {5!} } - { {\theta}^7 \over {7!} } + .... \right]$$

Finally, substitute back in the function definitions for the series.

$$e^{i\theta} = cos( \theta ) + i \cdot sin( \theta )$$

Quite easily done. This proof doesn't give any hint of the relationship of the equation to the complex unit circle, but it does have the advantage of proving the equation is valid for all values.

## Rescaling to the i Scale

Euler's equation is radian based. It can be converted to the $i$ scale with some simple substitutions.

$$e^{i\theta} = (e^{i})^{\theta} = (i^{2/\pi})^{\theta} = i^{{2 \over {\pi}} \cdot \theta}$$

This suggests the following substitution.

$$p = {2 \over {\pi}} \cdot \theta$$

Solve for $\theta$.

$$\theta = {{\pi} \over 2} \cdot p$$

Substitute these expressions into Euler's formula to get an equivalent $i$ scale based formula.

$$i^p = cos( {{\pi} \over 2} \cdot p ) + i \cdot sin( {{\pi} \over 2} \cdot p )$$

This formula validates the assumption made earlier that the $i$ based circumference power scale works for fractional values. Since $\pi/2$ is a right angle, it also confirms that each unit step on the power scale represents a quarter turn on the complex unit circle.

## Conclusion

Euler's magnificent equation relates the distance along the circumference of the complex unit circle to the underlying complex value. That this relationship is essentially exponential is not really intuitive. Knowing that the complex unit circle is inherently exponential in its behavior and that the radian scale, used by Euler's equation, is just one scale of many, makes the equation more understandable. Anytime $e^{ i \cdot something }$ appears in a formula, it means somehow the formula is referring to a point on the complex unit circle in some context.

If you found this article interesting, you may like its followup: Angle Addition Formulas from Euler's Formula

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Comment by March 17, 2015
"That this relationship is essentially exponential is not really intuitive." -- The intuition is that to multiply complex numbers we add their angles. What transforms multiplication to addition? Taking a log. So in some sense the angle of a complex number (its argument) is some sort of logarithm. That's the key.
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Comment by March 10, 2015
I tend to visualise Euler's equation this way (for a given angle):
cos = (.5cos + .5cos) + (j.5sin - j.5sin)
if we then draw all above rectangular components on complex plane we get equivalent two polar forms:
cos = (e^j + e^-j)/2

Kaz
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Comment by January 2, 2023
An overview of all my articles can be found at https://www.dsprelated.com/showarticle/1490.php

This is the first article in the "Fundamentals" section.  The next article is:
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Comment by August 17, 2023

i,j --> are they American and Europian notations? I was under the impression that they are Mathematicians vs Electrical Engineering notations respectively.

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Comment by August 17, 2023
Yes, I believe you are correct about that.  Thank you.  I am clearly a Mathematician, also I use $S_n$ and $Z_k$ instead of $x[n]$ and $X[k]$ in my DFT articles.

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The text passage is a detailed mathematical article that explores the exponential nature of the complex unit circle, including the derivation and explanation of Euler's equation. It covers topics such as complex numbers, the unit circle, integer and fractional powers of \( i \), and the relationship between the exponential form and the Cartesian form of complex numbers.

### Analysis:
TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

The passage contains multiple mathematical questions and their corresponding answers, such as:
1. **Finding the square root of \( i \)**:
   - **Question**: What are the values of \( a \) and \( b \) such that \( (a + b \cdot i)^2 = i \)?
   - **Answer**: The solutions are \( (a, b) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \) or \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

2. **Finding the cube roots of 1**:
   - **Question**: What are the cube roots of 1?
   - **Answer**: The cube roots of 1 are \(