Intuitive Understanding Of Euler’s Formula

Euler's identity seems baffling:

$\displaystyle{e^{i\pi} = -1}$

It emerges from a more general formula:

$\displaystyle{ e^{ix} = cos(x) + i sin(x)}$

Yowza -- we're relating an imaginary exponent to sine and cosine! And somehow plugging in pi gives -1? Could this ever be intuitive?

Not according to 1800s mathematician Benjamin Peirce:

It is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth.

Argh, this attitude makes my blood boil! Formulas are not magical spells to be memorized: we must, must, must find an insight. Here's mine:

Euler's formula describes two equivalent ways to move in a circle.

That's it? This stunning equation is about spinning around? Yes -- and we can understand it by building on a few analogies:

Starting at the number 1, see multiplication as a transformation that changes the number (1 * e^(i*pi))
Regular exponential growth continuously increases 1 by some rate; imaginary exponential growth continuously rotates a number
Growing for "pi" units of time means going pi radians around a circle
Therefore, e^(i*pi) means starting at 1 and rotating pi (halfway around a circle) to get to -1

That's the high-level view -- let's dive into the details. By the way, if someone tries to impress you with "e^(i*pi) = -1", ask them about i to the ith power. If they can't think it through, Euler's formula is still a magic spell to them.

Update: While writing, I thought a companion video might help explain the ideas more clearly:

It follows the post -- watch together, or at your leisure.

Understanding cos(x) + i * sin(x)

The equals sign is overloaded. Sometimes we mean "set one thing to another" (like x = 3) and others we mean "these two things describe the same concept" (like sqrt(-1) = i).

Euler's formula is the latter: it gives two formulas which explain how to move in a circle. If we examine circular motion using trig, and travel x radians:

cos(x) is the x-coordinate (horizontal distance)
sin(x) is the y-coordinate (vertical distance)

The statement

$\displaystyle{cos(x) + i \cdot sin(x)}$

is a clever way to smush the x and y coordinates into a single number. The analogy "complex numbers are 2-dimensional" helps us interpret a single complex number as a position on a circle.

When we set x to pi, we're traveling "pi" units along the outside of the unit circle. Because the total circumference is 2*pi, plain old pi is halfway around, putting us at -1.

Neato: The right side of Euler's formula (cos(x) + i*sin(x)) describes circular motion with imaginary numbers. Now let's figure out how the e side of the equation accomplishes it.

What is Imaginary Growth?

Combining x- and y- coordinates into a complex number is tricky, but manageable. But what does an imaginary exponent mean?

Let's step back a bit. When I see "3^4" I think of it like this:

3 is the end result of growing instantly (using e) at a rate of ln(3). 3 = e^ln(3)
3^4 is the same as growing to 3, but then growing for 4x as long. So 3^4 = e^(ln(3) * 4) = 81

Instead of seeing numbers on their own, you can think of them as something e had to "grow to". Real numbers, like 3, give an interest rate of ln(3) [1.1] and that's what e "collects" as its going along, growing continuosly.

Regular growth is simple -- it keeps "pushing" a number in the same (real) direction it was going. 3 × 3 pushes in the original direction, making it 3 times larger (9).

Imaginary growth is different -- the "interest" we earn is in a different direction! It's like a jet engine that's was strapped on sideways -- instead of going forward, we start pushing at 90 degrees.

The neat thing about a constant orthogonal (perpendicular) push is that it doesn't speed you up or slow you down -- it rotates you! Taking any number and multiplying by i will not change its magnitude, just the direction it points.

Intuitively, here's how I see continuous imaginary growth rate: "When I grow, don't push me forward or back in the direction I'm already going. Rotate me instead."

But Shouldn't We Spin Faster and Faster?

I wondered that too. Regular growth compounds in our original direction, so we go 1, 2, 4, 8, 16, multiplying 2x each time and staying in the real numbers. We can consider this e(ln(2)*x): grow instantly at a rate of ln(2) for "x" seconds.

And hey -- if our growth rate was twice that (2*ln(2)), it would look the same as growing for twice as long (2x vs x). The magic of e lets us swap rate and time; 2 seconds at ln(2) is the same growth as 1 second at 2*ln(2).

Now, imagine we have some imaginary growth rate (R*i) which rotates us: e^R*i becomes imaginary and grows to "i". Well, if we double that, we get i^2 (-1) and as we keep going we just spin around the circle.

Now imagine we double that rate (2R*i). Would that spin us off the circle? Nope! Having a rate of 2R*i means we just spin twice as fast, or alternatively, spin at a rate of R for twice as long.

Once we realize that some exponential growth rate can take us from 1 to i, increasing that rate just spins us faster. We'll never escape the circle.

However, if our growth rate is complex (a+bi vs Ri) then the real part (a) will grow us like normal, while the imaginary part (bi) rotates us. But let's not get fancy: Euler's formula, e^(i*x), is about the purely imaginary growth that keeps us on the circle (more later).

A Quick Sanity Check

While writing, I had to clarify a few questions for myself:

Why e^x -- aren't we rotating 1?

e represents the process of starting at 1 and growing continuously at 100% interest for 1 unit of time.

When we write e we're capturing that entire process in a single number -- e represents all the whole rigamarole of continuous growth. So really, e^x is saying "start at 1 and grow continuously at 100% for x seconds", and starts from 1 like we want.

But what does i as an exponent do?

For normal exponents like 3^4 we ask:

What is the implicit growth rate? It's growing from 1 to 3.
How do we change that growth rate? We make it 4x larger (^4).

Converting to "e" format, we say our instantaneous growth rate is ln(3) and make it 4x larger: ln(3) * 4. The top exponent (4) just scaled our growth.

When the top exponent is i (like 3^i), we get the same growth (ln(3)) but multiply it by i. So instead of growing at ln(3), we're growing at ln(3) * i.

The top exponent modifies the growth rate inherent in the bottom part.

The Nitty Gritty Details

Let's take a closer look. Remember this definition of e:

$\displaystyle{e = \lim_{n\to\infty} \left( 1 + \frac{100\%}{n} \right)^n}$

That 1/n represents the interest we earned in each microscopic period. We assumed the interest was real -- but what if it were imaginary?

$\displaystyle{e = \lim_{n\to\infty} \left( 1 + \frac{100\%\cdot i}{n} \right)^n}$

Now, our newly formed interest adds to us in the 90-degree direction. Surprisingly, this does not change our length -- this is a tricky concept, because it appears to make a triangle where the hypotenuse must be larger. We're dealing with a limit, and the extra distance is within the error margin we specify. This is something I want to tackle another day, but take my word: continuous perpendicular growth will rotate you. This is the heart of sine and cosine, where your change is perpendicular to your current position, and you move in a circle.

We apply i units of growth in infinitely small increments, each pushing us at a 90-degree angle. There is no "faster and faster" rotation - instead, we crawl along the perimeter a distance of |i| = 1 (magnitude of i).

And hey -- the distance crawled around a circle is an angle in radians! We've found another way to describe circular motion!

To get circular motion: Change continuously by rotating at 90-degree angle (aka imaginary growth rate).

So, Euler's formula is saying "exponential, imaginary growth traces out a circle". And this path is the same as moving in a circle using sine and cosine in the imaginary plane.

In this case, the word "exponential" is confusing because we travel around the circle at a constant rate. In most discussions, exponential growth is assumed to have a cumulative, compounding effect.

Some Examples

You don't really believe me, do you? Here's a few examples, and how to think about them intuitively.

Example: e^i

Where's the x? Ah, it's just 1. Intuitively, without breaking out a calculator, we know that this means "travel 1 radian along the unit circle". In my head, I see "e" trying to grow 1 at 100% all in the same direction, but i keeps moving the ball and forces "1" to grow along the edge of a circle:

$\displaystyle{e^i = cos(1) + i \cdot sin(1) = .5403 + .8415i}$

Not the prettiest number, but there it is. Remember to put your calculator in radian mode when punching this in.

Example: 3^i

This is tricky -- it's not in our standard format. But remember, $\displaystyle{3^i = 1 \cdot 3^i}$ -- the real question is "How do we transform 1"?

We want an initial growth of 3x at the end of the period, or an instantaneous rate of ln(3). But, the i comes along and changes that rate of ln(3) to "i * ln(3)":

$\displaystyle{3^i = (e^{ln(3)})^i = e^{ln(3)\cdot i}}$

We thought we were going to transform at a regular rate of ln(3) (a little faster than 100% continuous growth since e is about 2.718). But oh no, i spun us around: now we're transforming at an imaginary rate which means we're just rotating about. If i was a regular number like 4, it would have made us grow 4x faster. Now we're growing at a speed of ln(3), but sideways.

We should expect a complex number on the unit circle -- there's nothing in the growth rate to increase our size. Solving the equation:

$\displaystyle{3^i = e^{ln(3) \cdot i} = cos(ln(3)) + i \cdot sin(ln(3)) = .4548 + .8906i}$

So, rather than ending up "1" unit around the circle (like e^i) we end up ln(3) units around.

Example: i^i

A few months ago, this would have had me tears. Not today! Let's break down the transformations:

$\displaystyle{i^i = 1 \cdot i^i}$

We start with 1 and want to change it. Like solving 3^i, what's the instantaneous growth rate represented by i as a base?

Hrm. Normally we'd do ln(x) to get the growth rate needed to reach x it the end of 1 unit of time. But for an imaginary rate? We need to noodle this over.

In order to start with 1 and grow to i we need to start rotating at the outset. How fast? Well, we need to get 90 degrees (pi/2 radians) in 1 unit of time. So our rate is "i * pi/2". Remember our rate must be imaginary since we're rotating, not growing! Plain old "pi/2" is about 1.57 and results in regular growth.

This should make sense: to turn 1.0 to i at the end of 1 unit, we should rotate pi/2 radians (90 degrees) in that amount of time. So, to get "i" we can use e^(i * pi/2).

$\displaystyle{i = e^{i * \frac{\pi}{2}}}$

Phew. That describes i as the base. How about the exponent?

Well, the other i tells us to change our rate -- yes, that rate we spent so long figuring out! So rather than rotating at a speed of i * pi/2, which is what a base of i means, we transform the rate to:

$\displaystyle{\frac{\pi}{2}i \cdot i = \frac{\pi}{2} \cdot -1 = -\frac{\pi}{2}}$

The i's cancel and make the growth rate real again! We rotated our rate and pushed ourselves into the negative numbers. And a negative growth rate means we're shrinking -- we should expect i^i to make things smaller. And it does:

$\displaystyle{i^i = e^{- \frac{\pi}{2}} \sim .2}$

Tada! (Search "i^i" on Google to use its calculator)

Take a breather: You can intuitively figure out how imaginary bases and imaginary exponents should behave. Whoa.

And as a bonus, you figured out ln(i) -- to make e^x become i, make e rotate pi/2 radians.

$\displaystyle{ln(i) = i \cdot \frac{\pi}{2}}$

Example: (i^i)^i

A double imaginary exponent? If you insist. First off, we know what our growth rate will be inside the parenthesis:

$\displaystyle{i^i = (e^{\frac{\pi}{2}i})^i = e^{-\frac{\pi}{2}}}$

We get a negative (shrinking) growth rate of -pi/2. And now we modify that rate again by i:

$\displaystyle{{(i^i)^i = (e^{-\frac{\pi}{2}})^i = e^{-\frac{\pi}{2}i}}$

And now we have a negative rotation! We're going around the circle a rate of -pi/2 per unit time. How long do we go for? Well, there's an implicit "1" unit of time at the very top of this exponent chain; the implied default is to go for 1 time unit (just like e = e^1). 1 time unit gives us a rotation of -pi/2 radians (-90 degrees) or -i!

$\displaystyle{(i^i)^i = -i}$

And, just for kicks, if we squared that crazy result:

$\displaystyle{((i^i)^i)^2 = -1}$

It's "just" twice the rotation: 2 is a regular number so doubles our rotation rate to a full -180 degrees in a unit of time. Or, you can look at it as applying -90 degree rotation twice in a row.

At first blush, these are really strange exponents. But with our analogies we can take them in stride.

Complex Growth

We can have real and imaginary growth at the same time: the real portion scales us up, and the imaginary part rotates us around:

A complex growth rate like (a + bi) is a mix of real and imaginary growth. The real part a, means "grow at 100% for a seconds" and the imaginary part b means "rotate for b seconds". Remember, rotations don't get the benefit of compounding since you keep 'pushing' in a different direction -- rotation adds up linearly.

With this in mind, we can represent any point on any sized circle using (a+bi)! The radius is e^a and the angle is determined by e^(b*i). It's like putting the number in the expand-o-tron for two cycles: once to grow it to the right size (a seconds), another time to rotate it to the right angle (b seconds). Or, you could rotate it first and the grow!

Let's say we want to know the growth amount to get to 6 + 8i. This is really asking for the natural log of an imaginary number: how do we grow e to get (6 + 8i)?

Radius: How big of a circle do we need? Well, the magnitude is sqrt(6^2 + 8^2) = sqrt(100) = 10. Which means we need to grow for ln(10) = 2.3 seconds to reach that amount.
Amount to rotate: What's the angle of that point? We can use arctan to figure it out: atan(8/6) = 53 degrees = .93 radian.
Combine the result: ln(6+8i) = 2.3 + .93i

That is, we can reach the random point (6 + 8i) if we use e^(2.3 + .93i).

Why Is This Useful?

Euler's formula gives us another way to describe motion in a circle. But we could already do that with sine and cosine -- what's so special?

It's all about perspective. Sine and cosine describe motion in terms of a grid, plotting out horizontal and vertical coordinates.

Euler's formula uses polar coordinates -- what's your angle and distance? Again, it's two ways to describe motion:

Grid system: Go 3 units east and 4 units north
Polar coordinates: Go 5 units at an angle of 71.56 degrees

Depending on the problem, polar or rectangular coordinates are more useful. Euler's formula lets us convert between the two to use the best tool for the job. Also, because e^(ix) can be converted to sine and cosine, we can rewrite formulas in trig as variations on e, which comes in very handy (no need to memorize sin(a+b), you can derive it -- more another day). And it's beautiful that every number, real or complex, is a variation of e.

But utility, schmutility: the most important result is the realization that baffling equations can become intuitive with the right analogies. Don't let beautiful equations like Euler's formula remain a magic spell -- build on the analogies you know to see the insights inside the equation.

Happy math.

Appendix

The screencast was fun, and feedback is definitely welcome. I think it helps the ideas pop, and walking through the article helped me find gaps in my intuition.

References:

Brian Slesinsky has a neat presentation on Euler's formula
Visual Complex Analysis has a great discussion on Euler's formula -- see p. 10 in the Google Book Preview

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Posted July 19, 2010, under Calculus, Math
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Comments

This is a phenomenal article! Took me quite a bit of time and multiple readings to get my head around it, but now I get it. I think.

Aditya — July 19, 2010 @ 11:48 pm
Hey cool that makes a lot of sense. I already thought of imaginary numbers that way, but the growth pulling in a circle is very straightforward.

Are there complex equations that move in this way that intersect the real set of numbers in cool ways? Like I could imagine some complex equation that loops around, intersecting the real set of numbers (the real numberline) to create the set of primes or something like that.

Brandon Reinhart — July 20, 2010 @ 2:53 pm
@Aditya: Thanks! Yes, it took me a while to really see the equation, there may be a nicer way to go back and streamline how it was presented — I’d like to avoid the need for people to have multiple readings .

@Brandon: That’s an interesting question — actually, the Mandelbrot set is like that to some degree, where there is a complex (2d) function which gives rise to some pretty amazing patterns. I don’t know of any others off the top of my head though.

Kalid — July 21, 2010 @ 12:22 am
I <3 BetterExplained – keep 'em coming!

This gets easier if you've already got the hang of the physics concept that to move in a circle you must keep accelerating. If you accelerate in one direction, you will get faster and faster, but if you keep accelerating in a new relative direction, your speed will be the same, but you'll move in a circle (your velocity changes.)

Acceleration is a kind of growth, and so it logically follows that if you grow in a relative direction, you'll rotate but not speed up.

wererogue — July 25, 2010 @ 9:22 am
@wereogue: Thanks for the support! Yes, the physics interpretation definitely helps see this relationship, and I like the way you put it — our growth/change is really an acceleration, and we are constantly accelerating in a perpendicular direction which gives us a circle. It’s funny how much overlap there is between math and physics .

Kalid — July 26, 2010 @ 9:56 am
Thank you for showing us that Maths can be easy and simple. You are an example to today’s mathematicians, what you are doing is really inspiring. I always try to think how to make things easier to teach, learn and do. But you really make it so well!

Congratulations!
Keep it doing it!
you can go very far!

Mariano
(Excuse my English)

Mariano — July 27, 2010 @ 8:26 pm
Great article, Keep going on….

Mithun — August 1, 2010 @ 8:22 pm
Great article and great video!

I was wondering about what you used to generate the graphics, they look great .

Cheers from Romania,
Alex

Alex — August 4, 2010 @ 3:12 am
I didnt bother to follow your argument because the topic doesnt interest me but I like your attitude that math is just logic and “common sense” and theres way too much hocus pocus and ‘mysticism” that often creeps in imho.

stuart — August 17, 2010 @ 3:02 pm
Thanks for making me excited about math again.

mark — September 4, 2010 @ 10:26 pm
Kalid,

I’m still digesting it all, but i just have to say: “continuous perpendicular growth will rotate you” is just plain sexy. Wow, it makes sooo much sense to me Keep it up sir!

Sebastian

Sebastian Marquez — September 5, 2010 @ 10:30 am
Kalid, this is extremely impressive. I’ve been trying to understand this for a long long time. I found it difficult to see past the numbers and symbols and wanted to understand it ‘visually’ so I knew what was trying to be achieved. You have described it all beautifully and for the first time I am really understanding how it all fits together. I so wish you were teaching me in school 25 years ago. Thanks again.

Kalid's Friend — September 7, 2010 @ 6:21 am

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