Waste less time on Facebook — follow Brilliant.

A Long, but Easily Accessible Proof of \(e^{i\pi}=-1\)

This is directly adopted from the rough script for my presentation. Although it lacks visuals, I feel like the proof is in sufficient detail to be understood. The goal of my proof was to teach students with no knowledge of mathematics beyond Algebra 1 how imaginary powers work, and why \(e^{i\pi}=0\). In this proof, it is key to remember that I am referencing e as an arbitrary number with no significance that simply represents this "ugly" number 2.718...

Understanding e to the pi i

Watch this video first. Some of it will most likely make sense and 3Brown1Blue does an amazing job of explaining how this works. However, to keep the video less than half an hour long, he leaves out certain details which, without prior knowledge of the subject, make his video hard to understand. In this paper (this is a rough script for my presentation), I fill in the holes in his video (such as why e is the base and it’s not just “natural”) and give a thorough, incredibly detailed explanation of real numbers to imaginary powers and conclude with why \(e^{i\cdot \pi}=-1\)

Definition of terms: \(i=\sqrt{-1}\)


Real numbers: the stuff u worked with since preschool unless you grew up in China, in which case you dealt with real and complex numbers in preschool.

Imaginary numbers: also called complex numbers. They are numbers of the form a+b\cdot i where a and b are real numbers. For simplicities sake of this presentation, assume all complex numbers take the form b\cdot i.


Properties: \(x^{a+b}=x^{a}\cdot x^b\)



\(~ \)

\(\huge{\text{Why does e^(i*pi)=-1?}}\\ \color{grey}{\text{A proof that requires knowledge Alg 1 and minimal knowledge of complex numbers}}\)

Note: this proof only explains how to think of imaginary powers. It does not tell you how to solve the equation \(x^{i\cdot\pi}=-1\)

Forget adding as adding and forget multiplication as repetitive adding.

Imagine that all numbers were on a number line and that multiplication and addition were ACTIONS. Adding SHIFTS a number line (or plane) and multiplying STRETCHES a number line (or plane). If we add, say 1, then we shift the line right one. We call 1 and ADDER. If we multiply a number by two, then we stretch a graph to double its original length while keeping 0 in the same place.

Imagine graph of \(2^x\). It goes way way way up. But the graph of \(1^x\) is a flat line. Thus the absolute value of 1 to any power is 1. Keep this in mind as you go through the proof.

Now, let’s go back to the number line. Imagine 0 as a fixed point and we place a variable point at 1. Imagine multiplying one by \(2^q\). Well \(2^0=1\), and \(2^a \cdot 2^b=2^{a+b}\). Therefore, \(1\cdot 2^q=2^0\cdot 2^q= 2^{0+q}\). q is an ADDER to 0. Therefore, raising a number to the power of a variable turns that number from an adder into a multiplier.

Go to the complex plane. What is this? Well it has real numbers on the “\(x\) axis” and imaginary numbers (of the form \(n\cdot i\)) on the “\(y\) axis”. Adding a real shifts the plane left by some factor, and multiplying scales the plane by some factor. Adding \(i\) shifts the plane up (up to \(i\)), so multiplying does what? Multiplying 1 by any multiplier scales 1 to that number. Well if \(1\cdot i=i\), and i is a multiplier, then it must scale 1 to \(i\), and 2 to 2\(i\), and 3 to 3\(i\)... This means that it rotates the plane by some amount. Now say that we multiply i\cdot i, which is -1. i is once again a multiplier and hence we rotate the plane one more time. Notice that we have rotated the plane 180o, or \(\pi\) radians... notice the correlation? \(e^{i\cdot\pi}=-1\)... rotating the plane \(\pi\) radians... But that’s not a complete proof, the only people who would consider this to be a proof are 1) Physicists and 2) Fermat.

So now we go back to changing adders to multipliers by raising them to a certain power.
\(x^a\cdot x^b=x^{a+b}\) and thus \(x^0\cdot x^i=x^{0+i}\). This means that if we take ANY number x and raise it to the i, then we should scale x to somewhere on the vertical axis. So \(2^i\) should be mapped somewhere on the vertical axis and \(4^i\) should be mapped somewhere on the vertical as well... right? Well not exactly.

Note that \((a^b)^c\) is the same as \((a^c)^b\).


\(\ln \left((a^b\right)^c)=c\cdot ln(a^b)=c\cdot b\cdot ln(a)\)

\(\ln((a^c)^b)=b\cdot ln(a^c)=c\cdot b\cdot ln(a)\)

If you have no idea what that meant, don’t worry. Here’s another proof



Let’s assume that ANY number to the power of i exists somewhere on the vertical axis. This means that any number to the power of i takes the form k\cdot i for some real number k. Because we can choose any number, let’s choose 2. If our logic is true, then \(2^i=ki\) and \(4^i=i\cdot k^2\)

(\(k\) is squared because 4 is 2 squared).

\(2^i=k\cdot i\).

Then, squaring both sides,

\((2^i)^2=(k\cdot i)^2=k^2(-1)=-k^2\)

Thus \((2^i)^2=-k^2\)

Therefore, using our above property \((2^i)^2=(2^2)^i\)


Thus \(4^i=-k^2\)

But that contradicts our previous assumption where we stated that ANY number to the power of i equals to \(k\cdot i\) for some number \(k\). This means that NOT ALL values to the power of i take the form \(k \cdot i\) (note that this does not mean that NO values satisfy this condition, hehe logic).

Well then what value of \(x\) satisfies the equation \(x^i=i\)? Well it turns out that there are infinitely many solutions, but not every solution works. The exact values of the solutions are impossible to calculate without a computer, so here is the smallest possible positive solution: 4.8104773809653516554730..... Lets approximate it to \(sqrt(23)\) for future use (they’re only 0.01464 apart). This number is transcendental (explain meaning). Why don’t all values work and why are there infinitely many solutions? I’ll get to that in a second.

Fact: If we consider i as a multiplier that rotate points in the complex plane, and we consider \(a^i \cdot a^i \cdot a^i \cdot ....= \color{blue}{a}^{\color{red}{b}\cdot i}\) (where b is the number of \(a^i\) on the left side of the equation), the greater a and b are, the more we rotate the the point. Let me explain.

First we prove that the larger the value of b, the more it rotates. Using our estimation of \(a=\sqrt{23}\) and the properties listed above; (NOTE: I approximate it to \(\sqrt{23}\) because it is easier to demonstrate how the power (\(b\)) affects the rotation degree).

\({\color{blue}{ \sqrt{23}}}^{\color{red}{0}\cdot i}=1\)

\({\color{blue}{ \sqrt{23}}}^{\color{red}{1}\cdot i}=i\)

\({\color{blue}{ \sqrt{23}}}^{\color{red}{2}\cdot i}=\left(\color{blue}{ \sqrt{23}}^{i}\right)^{\color{red}{2}}=i^{\color{red}{2}}=-1\)

\({\color{blue}{ \sqrt{23}}}^{{\color{red}{3}}\cdot i}=i^{\color{red}{3}}=i\cdot i^2=-i\)

\({\color{blue}{ \sqrt{23}}}^{\color{red}{4}\cdot i}=i^4=(-1)^2=1\).

We’ve gone one full rotation, or \(2\pi\) radians, and are back where we started! We can keep doing this an infinite number of times. Note that if we trace out the path of the point as we rotate it around zero, we get a circle of radius 1. Even though we can do this one more time, let’s go one more turn and see what happens.

\(\color{blue}{ \sqrt{23}}^{\color{red}{5}\cdot i}\approx i^{\color{red}{5}}= i\).

Remember how I said that \(x^i=i\) had an infinite number of solutions? Well this is the second solution, approximately \(\color{blue}{ \sqrt{23}}^{5}\). I can no longer “legally” use the equal sign (technically I couldn’t use it in the first place, but then I would have nothing to replace the current \(\approx\) signs with) because of the gigantic exponent. Whereas our approximation used to be 0.0146 off, it is now a gigantic 38.98 off! Here is a table of the first few positive solutions:

Will fix table tomorrow (too tired lol)

\[\begin{array}{|c|c|c|} \hline \text{Solution Number} & \text{value}\\ \hline 1 & 4.8104773809\\ \hline 2 & 2,575.9704965975\\ \hline 3&1,379,410.705805983402\\ \hline 4&7.386629225006301x10^8\\ \hline 5&3.955478312446234x10^{11}\\ \hline 100,000,000&5.26991655718763x10^{272,875,268}\\ \hline \end{array}\]

Next we will show that the larger \(a\) is in \({\color{green}{a}}^{{\color{red}{b}}\cdot i}\), the QUICKER we rotate the plane. Let us look at the case where \(a=23=\sqrt{23}^2\)

\({\color{green}{23}}^{\color{red}{0}\cdot i}=\left(\sqrt{\color{green}{23}}^2\right)^{\color{red}{0}\cdot i}=1\)

\({\color{green}{23}}^{\color{red}{1}\cdot i}=\left(\sqrt{\color{green}{23}}^2\right)^{\color{red}{1}\cdot i}\)

Using the property \(\left(a^b\right)^c=\left(a^c\right)^b\)

\(~~\left(\sqrt{\color{green}{23}}^2\right)^{\color{red}{1}\cdot i}=\left(\sqrt{\color{green}{23}}^{\color{red}{1}\cdot i}\right)^2=i^2=-1\)

\(23^{\color{red}{2}\cdot i}=\left(\sqrt{\color{green}{23}}^2\right)^{\color{red}{2}\cdot i}=\left(\sqrt{\color{green}{23}}^{\color{red}{2}\cdot i}\right)^2\approx (-1)^2=1\)

We’ve already gone around the circle once, yet our value of \(b\) is HALF of what it was when we worked with \(a=\sqrt{23}\). This clearly shows that the larger \(a\) is, the quicker we rotate the plane.

One day, some genius named Euler looked at this equation and thought, “Well, how about we define this function in terms of the central angle.” (show audience what central angle is).

Let’s call this angle \(\theta\). I will switch back to using the exact value of the first solution to \(x^i=i\), 4.81047.... You’ll see why in just a second. We need to check for a correlation between \(\theta,~b,\) and \( {\color{blue}{a}}^{{\color{orange}{b}}\cdot i}\)

When \(\theta=0\), \({\color{orange}{b}}={\color{orange}{0}}\) and \({\color{blue}{4.81047}}^{{\color{orange}{0}}\cdot i}=1\).

When \(\theta=\frac{\pi}{2}\), \({\color{orange}{b}}=\color{orange}{1}\) and \({\color{blue}{4.81047}}^{{\color{orange}{1}}\cdot i}=i\).


When \(\theta=2\pi\), \({\color{orange}{b}}=\color{orange}{4}\) and \({\color{blue}{4.81047}}^{{\color{orange}{4}}\cdot i}=1\).

An obvious correlation between \(\theta\) and \(b\) is emerging where \(\theta=\dfrac{\pi}{2}\cdot b\)... or \(\dfrac{2}{\pi}\cdot\theta=b\). Plugging our new value of \(b\) into our old function. We get the following equation.

\( ({\color{blue}{4.81047...}})^{{\color{orange}{b}}\cdot i}=({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\cdot\theta\right)\cdot i}\)

Moving the parentheses around

\(({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\cdot\theta\right)\cdot i}=({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\right)\cdot \left(i\theta\right)}\)

Using the property \(a^{(b)c}=\left(a^{(b)}\right)^c\)

\(({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\right)\cdot \left(i\theta\right)}=\left(({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\right)}\right)^{i\theta}\)

And this is this is where the kids work, this is where champions are separated from adults, and this is where babies are made... or some permutation of that. Can anyone guess what number \(({\color{blue}{4.81047...}})^{\left(\frac{2}{\pi}\right)}\) is?


And finally, when \(\color{orange}{b}=\color{orange}{2}\), \(\theta=\pi\) and \({\color{blue}{4.81047}}^{{\color{orange}{2}}\cdot i}=-1\).

Meaning that

\({\color{blue}{4.81047}}^{{\color{orange}{2}} \cdot i}=e^{i\pi}=-1\).

Note by Trevor Arashiro
1 year, 2 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)


Sort by:

Top Newest

Nice observations made here.

Check out the Complex Algebra exploration, which builds up to this result.

Calvin Lin Staff - 1 year, 2 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...