Intuitive Understanding Of Euler's Formula

July 19, 2010 by kalid

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 as fast (2*ln(2) vs 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 a regular exponent like 3^4 we ask:

What is the implicit growth rate? We're growing from 1 to 3 [the bottom of the exponent].
How do we change that growth rate? We scale it by 4x (^4, the top of the exponent).

We can convert our growth into "e" format: our instantaneous rate is ln(3), and we increase it to ln(3) * 4. Again, the top exponent (4) just scaled our growth rate.

$\displaystyle{3^4 = e^{ln(3) \cdot 4} = (e^{ln(3)})^4}$

When the top exponent is i (as in 3^i), we just multiply our implicit growth rate by i. So instead of growing at plain old ln(3), we're growing at ln(3) * i.

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

The top part of the exponent modifies the implicit growth rate of 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

90 thoughts on “Intuitive Understanding Of Euler's Formula”

Pingback: Tweets that mention New Post: Intuitive Understanding Of Euler’s Formula -- Topsy.com
Pingback: Euler’s formula explained « Stii ce vrei ?
Pingback: Intuitive Understanding of Sine Waves | BetterExplained
Pingback: An intuitive appproach to mathematics « socialempirics

Aditya on July 19, 2010 at 11:48 pm said:

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.
Brandon Reinhart on July 20, 2010 at 2:53 pm said:

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.
Kalid on July 21, 2010 at 12:22 am said:

@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.
wererogue on July 25, 2010 at 9:22 am said:

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.
Kalid on July 26, 2010 at 9:56 am said:

@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. Our velocity is always perpendicular to our position (and acceleration perpendicular to velocity) which gives us a circle. It’s funny how much overlap there is between math and physics .
Mariano on July 27, 2010 at 8:26 pm said:

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)
Mithun on August 1, 2010 at 8:22 pm said:

Great article, Keep going on….
Alex on August 4, 2010 at 3:12 am said:

Great article and great video!

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

Cheers from Romania,
Alex
stuart on August 17, 2010 at 3:02 pm said:

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.
mark on September 4, 2010 at 10:26 pm said:

Thanks for making me excited about math again.
Sebastian Marquez on September 5, 2010 at 10:30 am said:

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
Kalid's Friend on September 7, 2010 at 6:21 am said:

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 on September 13, 2010 at 2:23 pm said:

@Kalid’s Friend: It really bothered me for a long time also — Euler’s formula was used everywhere but I didn’t have a gut feel for it! I’m really happy it was able to help
Kalid on September 13, 2010 at 6:50 pm said:

@Mariano, @Mithun: Thanks for the kind words!

@stuart: Yes, I think everything should be understood / explained intuitively, and not accepted as mystic.

@mark: You’re more than welcome.

@Sebastian: Haha, I like that phrase too — whatever it takes to make it click .
lewikee on September 14, 2010 at 9:16 am said:

Kalid,

Thank you for this!

Would it be accurate to say that if you traced out the complex growth curve just as you did the real and imaginary growths, you would get a spiral?

By “tracing out”, I simply mean that if you are given e^(ax+bi), you simply put points at specified intervals to the answer. For example the following would give you 3 points on the way to your answer:

e^((a/4)x+(b/4)i)
e^((a/2)x+(b/2)i)
e^((3a/4)x+(3b/4)i)
lewikee on September 14, 2010 at 9:18 am said:

Oops, on my previous post, please assume x=1.
Kalid on September 14, 2010 at 10:54 am said:

@lewikee: Yep, you got it — imaginary exponential growth rotates you in a circle, and regular exponential growth grows you (e^a/4, e^a/2, etc.). You end up spinning around the circle but getting further and further away, making a complex spiral.

If you put

parametric plot (e^(t/4) * cos(t), e^(t/4) * sin(t))

plot on wolfram alpha

into wolfram alpha you can see an example (I separated e^ix into cos(x) + i*sin(x) to get the x and y components, and put them separately into the plot — I also scaled down the regular exponential growth to make a tighter spiral).
nschoe on September 24, 2010 at 12:56 pm said:

Wow, you are *amazing*! You explain things a way *nobody* has ever done (around me I mean). It is unbelievable: it’s like you enter my head, grab the questions inside and then explain each of them with a 100% chance of understanding!
Very nice and useful article, explains what I used to consider trivial, wow!
I love the site, please keep on publishing such wonderful articles!
Thanks a lot.
Kalid on September 27, 2010 at 9:41 pm said:

@nschoe: Thanks, really glad it helped! Yeah, I really feel it’s important to answer those “huh?” questions we get in our heads as they come up, instead of pretending that learning is one smooth process from A to B. We need to explain why we don’t want to go to C and D .

Appreciate the support!
PaulH on October 19, 2010 at 2:40 am said:

Hi Kalid, thanks for all these articles, I’m finding them illuminating.

I’m still struggling with this one a bit, though. I’m confused by the use of the phrase “growing instantly”. E.g. “3 is the end result of growing instantly (using e) at a rate of ln(3), i.e. 3 = e^ln(3)”

If something grows instantly, then its rate of growth is infinite, isn’t it? I don’t understand how something can grow “instantly” at a finite rate.
And then “3^4 is the same as growing to 3, but then growing for 4x as long” – but if the growth was instant before, then “4x as long” is also instant. Four times zero is zero.

And what does “using e” mean in this instance? Starting with a value of e? Why are we doing that? I thought e was a sort of fundamental growth rate, not a starting value? In your Expand-o-tron analogy in another article, the base is the desired growth rate per unit time, and the exponent is the number of units of time to grow for. But here, the exponent seems to be the rate, while the base is the starting value.
Kalid on October 23, 2010 at 6:43 pm said:

@PaulH: Thanks, great question (I need to setup a FAQ for exponents since there are so many parts that I still have to remind myself also).

You’re right — perfectly instantaneous growth would indeed be infinite. Getting at this idea of “infinitely small” is one of the problems of Calculus actually. A better phrasing may be “Growing at a compounding interval that appears perfectly smooth to us”, i.e. we cannot see the sudden, punctuated growth that pops up with compounding on an interval (like making a staircase so fine-grained that it looks like a smooth curve). For example’s sake, we can imagine compounding every nanosecond (billionth of a second).

“Using e” means figuring out the net effect of compounding as fast as possible (every nanosecond, let’s say). So taking the examples you mentioned:

>> E.g. “3 is the end result of growing instantly (using e) at a rate of ln(3), i.e. 3 = e^ln(3)”

We can think of 3 in two ways. The first is the traditional way — at the end of the year, we have 3 times as much (we start with a dollar, get 3 dollars at the end). The other way is to say “3 is the result of starting at 1.0 and growing instantly (compounding as quickly as possible) at some rate.”

If our rate is 100% and we compound as quickly as possible, we end up with 2.718… (e). But we want to get to “3″, so we need to grow a bit faster than 100% per year. The actual number is ln(3) = 109.8%, so we can say:

“We can get to 3.0 at the end of the year if we start at 1.0 and grow at 109.8% per year, compounding as fast as possible”. e is the way we work out what super-fast compounding gives us, so we can write it

e^1.098 = 3

So, we do start at 1.0, but e^x gives us the impact of starting at 1.0 and compounding at x% return as fast as possible.

In other words, e isn’t the starting point because we want to start at 2.718. We start at e because we want to compound 1.0 as fast as possible, and e is the shortcut for that (i.e., the price after shipping, handling, and taxes if you get my drift).

Hope this helps!
Rafay on December 8, 2010 at 1:40 pm said:

How would you go about solving this problem, i was asked to put ln( pi*i -ln (-e^w)) into the for a + bi, and the question states that w is a complex number. Need help for a final tomorrow.
Flo on December 12, 2010 at 7:34 pm said:

Fantastic!!! You should do more explanations like that, you’re as good (or perhaps even better) than Sal Khan from Khan Academy!
Kalid on January 6, 2011 at 9:57 am said:

@Rafay: I can’t really comment on specific homework problems, but break it into groups: you can see that ln(e^w) should just be “w” (since ln and e are inverses) so you get something like ln(pi * i – w). From the rules of logarithms, this equals ln(pi * i) / ln(w). The natural log of i is explained above — that should help get you started.

@Flo: Thanks! I plan to keep creating more .
mark ptak on January 8, 2011 at 2:37 pm said:

Nicely done. Another internet applied math whiz is discovered….
kain on January 11, 2011 at 10:49 pm said:

Wow, the last time my mind was blown this hard was when I did LSD. Only this time I can remember what it was that had caused the mind-blow (lol?) the next day.
Mark on January 30, 2011 at 5:05 pm said:

Kalid,

I’m not sure, but I think your description of 3=e^ln(e) is a very interesting way of converting discrete mathematics to continuous. It is like saying, there is a finite rate with which to generate a number (3), you grow so much per unit time. You can use this method to compute new numbers, by “growing them”. Brilliant.
Mark on January 30, 2011 at 5:06 pm said:

^ I meant 3=e^ln(3) .
Kalid on January 31, 2011 at 12:09 pm said:

@Mark: Glad you liked it! Yes, it’s interesting to see regular whole numbers (like 3, 4, etc.) as just “some amount” of growth that e takes you to.
Jorge on February 25, 2011 at 11:29 pm said:

You are simply brilliant !
Please keep updating your web with new and interesting problems.
Thanks
Nasser on March 1, 2011 at 1:32 am said:

It deserves to be called a “phenomenal” article.

Hope, to see Fourier Transform/Laplace Transform/Z-transform articles. If that happens, thousands of students would change the way they learn (I think they will like it more).
Kalid on March 2, 2011 at 3:03 pm said:

@Nasser: Thanks! I’m planning on doing the Fourier transform soon — the others, I still need to learn about .
steve on March 9, 2011 at 8:44 pm said:

I’m a math major studying topology, and really really sympathize with the attitude that math is about the underlying structure, the geometry and simplicity of the problem; with the analytic formulas being a good book-keeping device, but unenlightening unless you have a picture in your head. I can’t say enough how well-written I found this article (and the one about “e”), I have never enjoyed complex numbers because it always feels like a bunch of tedious algebraic laws, but this is a wonderful explanation of euler’s formula; and after reading the small bit in “Visual Complex Analysis; I picked up a copy and its awesome.
Keep up the good work; and I also look forward to the fourier transform/laplace transform installment!
Kalid on March 9, 2011 at 9:22 pm said:

@Steve: Thanks! I really like Visual Complex Analysis too — even though I wasn’t a math major (maybe in another life ) I have a severe interest in it and want to study it deeper. Appreciate the encouragement!
Anonymous on March 11, 2011 at 1:55 pm said:

Just a comment I always considered e^ix identical to e^x, except the mapping of e^ix to the 3D complex plane created a helix that turned the runaway exponential to spiral around the x axis, but still always increasing in size.

The projection on the real and imaginary axis being the cosine and sine made it look like a simple rotation, but looking at the complete 3D picture this is not so.
C on March 14, 2011 at 6:17 pm said:

This may be the most insightful article I have ever read. Finally someone care about making formulas intuitive! Very VERY well explained. Genius!
Kalid on March 15, 2011 at 11:15 am said:

@C: Thank you — really made my day!
Kalid on March 15, 2011 at 11:23 am said:

@Anonymous: That’s a great insight! This site has a cool diagram showing the spiral interpretation: http://www.songho.ca/math/euler/euler.html.
Regan on April 25, 2011 at 1:18 am said:

I love that. any number can be written in terms of e ‘growing’ at a certain rate for a certian amount of time! brilliant!
Kalid on April 25, 2011 at 2:37 am said:

@Regan: Awesome, glad it helped!
Loui on April 27, 2011 at 3:10 pm said:

Kalid, you are my new bestfriend.
Kalid on April 27, 2011 at 3:55 pm said:

@Loui: Math brings everyone together!
Anonymous on May 25, 2011 at 3:49 pm said:

thanks!
Kalid on May 27, 2011 at 10:40 pm said:

@Anonymous: You’re welcome!
Paul on May 30, 2011 at 1:13 pm said:

Second paragraph under “But Shouldn’t We Spin Faster and Faster?”: mustn’t it be “if our growth rate was twice that (2*ln(2)), it would look the same as growing twice as fast (2x vs x)” instead of “if our growth rate was twice that (2*ln(2)), it would look the same as growing for twice as long (2x vs x)”? Thank you.
Anonymous on June 3, 2011 at 9:25 am said:

Really helpful for the project that my group is doing in Math class! Thanks, I understand this so much more clearly now!
Sam on June 3, 2011 at 12:05 pm said:

the comment above is from me, i forgot to write my name but i just wanted all of you math lovers to know that i LOVE math!
Jamie Seagren on June 3, 2011 at 12:06 pm said:

math is a wonderful thing
math is a really cool thing
so get off your ass, lets do some math
math math math math mathhhhh
Kalid on June 5, 2011 at 8:42 pm said:

@Paul: Great question. I clarified the sentence a little — e^rt has the cool property that increasing the growth rate (2*ln(2) vs ln(2)) would have the same effect as increasing the amount of time spent (2x vs x).

@Sam: Awesome, glad it helped!

@Jamie: Math rocks .
Rick on June 30, 2011 at 11:53 am said:

This article is amazing! Reading it makes you feel like a genius.
Peter on July 21, 2011 at 8:06 am said:

In the section “what does i as an exponent do” it seems like you are saying that 3^4 is equivalent to ln(3) * 4. Obviously, this is wrong.
Peter on July 21, 2011 at 8:30 am said:

Nevermind I see what I was missing.

3^4 = (e^ln(3))^4

3^i = (e^ln(3))^i
Benedikt on July 22, 2011 at 2:49 pm said:

Kalid, the intuitions your articles provide are of unmatched quality. Thanks, and keep writing!
I really like the physics interpretation wererogue mentioned (angular and uniform acceleration), which is an excellent metaphor. But for me (who thought about this topic like 19^th century Benjamin), the most important inside you provided is that (-1)^n becomes a less awkward special case if you have complex numbers. The e^(i pi n) model now feels more elegant and obviously superior
I need to read more (all) of your articles, but in case you haven’t yet: Would you please demystify residues and contour integrals ?
Benedikt on July 22, 2011 at 4:29 pm said:

Doh, two! wrong words in the analogy. Although the idea of viewing the imaginary part as kind of centripetal force seems still nice, the real part is of course not uniform but proportional to the position. So x” = x works perfectly, but the analogy (with one acceleration growing exponentially, and the other one staying constant, if I got it right this time) is maybe not that nice after all.
Kalid on July 23, 2011 at 11:33 am said:

@Rick: Thanks! I love sharing those aha moments with people

@Peter: Thanks for the comment — I’ve updated the article to be more clear in that section.

@Benedikt: Wow, appreciate the note! I’d love to do those topics as I learn more about them!

For the analogy, yes, you have to
@Rick: Thanks! I love sharing those aha moments with people

@Peter: Thanks for the comment — I’ve updated the article to be more clear in that section.

@Benedikt: Wow, appreciate the note! I’d love to do those topics as I learn more about them!

For the analogy, yes, you have to sort of pick the portions where it works best. There’s clearly some issues with scaling and rotating at once — I prefer to split them up.

So, if your exponent is e^(a+bi) then I see that as a combination of real exponential growth and imaginary exponential growth (rotation). It’s really e^a * e^bi and you get the best of both . I’m not sure if this answers your question? If it doesn’t, let me know.
Jon on August 16, 2011 at 10:19 am said:

I’m still trying to grasp the concepts, and have some basic ideas.

One analogy to this movement is like a pen moving in the x direction and writing on a rotating flat disk below the pen (the disks diameter is parallel to the x axis). A (2-dimensional) spiral will be created. The spiral will be perpendicular to the pen then. This is just a basic analogy, perhaps a rotating cylinder centered on the x axis would be another analogy. The drawn figure would be a 3-dimensional spiral.

If anyone knows a similar type of analogy, let us know.
Kalid on August 17, 2011 at 8:52 am said:

Hi Jon! You might like the spiral diagrams on this page: http://www.songho.ca/math/euler/euler.html.
Jon on August 17, 2011 at 6:00 pm said:

Thanks Kalid. Here is a movie that shows the sine and cosine waves at right angles to each other. It seems a bit different that the others kind of descriptions.

http://upload.wikimedia.org/wikipedia/commons/8/8e/Sine_and_Cosine_fundamental_relationship_to_Circle_%28and_Helix%29.gif

The movie was found on a page for Eulers Formula: http://en.wikipedia.org/wiki/Euler%27s_formula
Stephen on September 16, 2011 at 7:40 pm said:

Hi Kalid,

Thank you for this wonderful explanation. One question for you, however, or for anyone else who might be able to answer it: I still can’t seem to understand why it makes intuitive sense that imaginary growth is orthogonal. Could anyone explain this?

Thanks in advance, and again, a really awesome website.

–Stephen
Stephen on September 16, 2011 at 8:47 pm said:

I mean I see why the shape of the imaginary growth is a general curve, but how do you know it’s circular (ie the growth is orthogonal)?
Kalid on September 16, 2011 at 10:13 pm said:

@Jon: Thanks, that’s a great diagram! Seeing a helix is another way to interpret the formula.

@Stephen: Great question, and thanks for the kind words! For me, the key to imaginary numbers is to see an equation like

x^2 = -1

and break it down to

1 * x * x = -1

That is “What transformation x, when applied twice, will turn 1 to -1?”. A rotation of 90 degrees is one such interpretation; as long as the rotation is perfectly orthogonal, then two such rotations will result in a mirror image.

If imaginary growth had a small component in your current direction (a 89 degree rotation, say) then

1) Two imaginary rotations would not perfectly flip your direction (89 + 89 = 178)
2) Accumulating imaginary rotations could slowly grow you as you added imaginary interest (in effect, you are multiplying by a complex number, not a purely imaginary number)

But, a key principle in imaginary multiplication is that 1 = i^4 = i^8 = i^12, i.e. every set of 4 perfectly cancels. In my head, I think “multiplying by an imaginary number cannot give you any components in your current direction, otherwise that ‘boost’ could accumulate over time.”

I hope this helps! Let me know if it didn’t, I love really getting at the heart of what makes these analogies click.
Stephen on September 16, 2011 at 11:26 pm said:

Hi Kalid,

Thanks so much for your really quick reply! It really means a lot to me. You don’t know how much this concept has been bugging me haha. You’re website has really made me think deeply over the past few days….

My question doesn’t so much revolve around imaginary multiplication, but rather the complex number interest multiplication that shifts the vector, starting from 1 on the real axis, as outlined by e^(ix).

I get your point about imaginary multiplication and that because it cycles back to 1, there can be no net growth in the vector magnitude, and from there it’s reasonable to conclude that all the change is orthogonal. That make sense.

But as you said, when you’re looking at the multiplication you’re doing for e^(ix) (to achieve growth along the circle), you’re in fact multiplying by a complex number on each infinitesimal step (I guess your very first bit of interest would be all imaginary, that is, vertical). If I’m not mistaken, the multiplication would look something like (1+ix/n)(1+ix/n)(1+ix/n)….n times, with really small n’s, (please correct me if I’m wrong.) Each (1+ix/n) would cause a perpendicular change to the vector and would result in a rotation. So is there anyway to directly see how this multiplication changes the vector in a specifically perpendicular direction?

I mean, I don’t know what other path you’d take to achieve the transformation you’re looking for, other than a circular path. It makes sense…I might be chasing nothing here, but I guess I’m looking for some way to see that multiplying by a small component of (1+ix/n) with a really small n, guarantees a change in a perpendicular direction specifically, not just in a general upwards direction. After all, there is a real component to the complex multiplication we’re doing here, so does the imaginary number multiplication logic hold up here? Couldn’t we end up with say…a spiral? idk

Thanks, Kalid, for bearing with me. I really, really appreciate it.

Take care
Kalid on September 17, 2011 at 12:39 am said:

@Stephen: You’re more than welcome, these are really fun to think about.

Ah, I think I see what you’re getting at! Yes, it’s interesting how those little minute changes add up to a perfectly circular rotation… check out this page:

http://www.cut-the-knot.org/arithmetic/algebra/Scott.shtml

There are some diagrams halfway through, but he’s plotted an example of (1 + 2i/10)^10 [i.e. taking steps of 10]. You can see how it converges on a circle. As you make the exponent higher (n=10, n=100, etc.) you can see how it “wraps” the circle more tightly.

He has more formal arguments about why the magnitude is 1 (no scaling) and the angle is exactly theta, which I need to work through and understand intuitively for myself .
Stephen on September 19, 2011 at 12:23 am said:

Thanks a bunch, Kalid. The website is very helpful…I will have to go over it a few times though haha. A bit complicated….

Appreciate your help.
Kalid on September 22, 2011 at 12:53 pm said:

@Stephen: No problem. Yep, there’s a lot of gritty math there, but I found the diagrams helpful.
erik wolf on October 15, 2011 at 1:41 am said:

Hi,

Is it a right “transcription” of the original formula:

exp(pi) = ‘(square root of -1) root’ of -1

?
Kalid on October 16, 2011 at 10:53 pm said:

@erik: Hrm, I’m not sure what you mean by transcription… do you mean “How would you put the formula into a sentence?”
bill solomon on October 31, 2011 at 7:03 pm said:

really good do you have any ideas about tetration to an imaginary number or irrational number?
Bill Anderson on November 12, 2011 at 10:28 am said:

Hi,

I love this site! Thank you!

Quick question: if e^i*pi means going pi radians around a circle (in this case 1/2 circle), does that mean I can always think of the coefficient of i as radians? Is it correct to think about it this way?

Thanks again for a great site.
kalid on November 14, 2011 at 12:47 am said:

@Bill: Thanks for the comment! I consider the coefficient on i as how “long” you go around the unit circle, assuming unit speed. So, pi “seconds” takes you halfway around. Another way is to think about it in terms of distance, i.e. halfway around directly (speed and time are “equivalent” here since it’s a unit speed… but I like thinking e^rate*time where the “rate” is i and the time is pi).

We’re using radians because we want measurements from the perspective of the mover (vs. degrees, which are an arbitrary measure from the perspective of the observer).
Yonggook on November 14, 2011 at 3:55 am said:

(1+x/n)^n =(1+x/(x*m))^(x*m)=((1+1/m)^m)^x | N=x*m
So taking limit n goes to infinite, also m goes to infinite
So (1+x/n)^n=e^x

Was great insight of you. Thanks for your explanation.
Was great job of you.
Yonggook on November 14, 2011 at 4:33 am said:

(1+x/n)^n =(1+x/(x*m))^(x*m)=((1+1/m)^m)^x | N=x*m
So taking limit n goes to infinite, also m goes to infinite

And n is real number. If x is imag i, m should be -i * absolute (n)

So as n goes infinite (1+i/n)^n become [(1+1/(-i * absolute (n)))^(-i * absolute (n))]^i = [e]^i

So I just wonder the expression n goes infinite applied only in real number direction. But as it shows it also work in complex number set.

as (-i * absolute (n)) goes infinite…

Am I something wrong?
Yonggook on November 14, 2011 at 4:43 am said:

If some number goes to infinite also work in 2D number set (complex number set), what is it really mean?…. ” the some number goes to some direction to the infinite”…
kalid on November 14, 2011 at 7:50 pm said:

@Yonggook: That’s a really good question — check out this page which talks more about the limit http://www.cut-the-knot.org/arithmetic/algebra/Scott.shtml. Right now I’m not sure what it means to have a complex number go to infinity.
Christian on November 22, 2011 at 1:23 pm said:

You claim i^i = e^(-π/2). Is it possible you might be forgetting an infinite amount of possible values? I could be wrong, but hear me out:

In general, sin([4n+1]π/2) = 1 and cos([4n+1]π/2) = 0 for any integer n
So e^(i*[4n+1]π/2) = i
Exponentiating, i^i = {e^(i*[4n+1]π/2)}^i = e^(–[4n+1]π/2)
For example, taking n = 1 yields your answer, whereas taking n = 2 (for which it seems this equation is still valid), yields i^i = e^(-5π/2). So I would conclude i^i doesn’t map to only one place on the real line.
kalid on November 22, 2011 at 2:47 pm said:

@Christian: Good question! When talking about sine, I think about it more like this:

i^i is equal to the computation e^(-pi/2) which is approximately 0.207

The fact that e^(-pi/2) = e^(-5pi/2) is a bit like 4/4 = 2/2 = 1 — different “computations” that approximate the same real number. So, i^i still only has one spot on the number line because e^(5pi/2) = e^(pi/2).
Christian on November 23, 2011 at 9:18 am said:

@Kalid:
e^(5π/2) does not equal e^(π/2)
Dmitry on November 25, 2011 at 12:43 am said:

@Christian is correct:
http://www.math.hmc.edu/funfacts/ffiles/20013.3.shtml
but to be fair @Kalid has been motivating this graphically the whole time so he used:
http://en.wikipedia.org/wiki/Complex_logarithm#Definition_of_principal_value
Alqazzafi a. Ibrahim on November 25, 2011 at 10:17 am said:

i wasted 4 ys of my life in the faculty of science studying magic spells
punch of idiots were there
kalid on November 30, 2011 at 12:50 pm said:

@Christian: Whoops, my mistake! I meant to write e^(-i * 5*pi/2) = e^(-i * pi/2) but that leads right back to your point!

You are correct, if you allow the base “i” to be defined as any value e can be raised to (i*pi/2, i*5pi/2, …) then when you raise it to the “i” power you could have e^(-pi/2), e^(-5*pi/2), etc. which clearly have different values.

@Dmitry: Thanks for the clarification! Google calculator, for example, will treat (e^(i * 5 * pi/2))^i the same as (e^(i*pi/2))^i. More interesting things to read up on!

@Alqazzafi: Funny how much we can study but not really learn, right? Anyway, happy if the article helped.
suhas on December 8, 2011 at 10:25 pm said:

Phenomenal – I thought I was the only person trying to understand intutively – superb
kalid on December 9, 2011 at 11:35 am said:

@suhas: Thank you!
Tariq on December 17, 2011 at 2:35 am said:

Youre a GENIUS!!!
kalid on December 20, 2011 at 9:20 pm said:

@Tariq: Glad you liked the article =).
eaca on January 25, 2012 at 11:37 am said:

Priceless insight, thank you.
kalid on January 27, 2012 at 10:36 am said:

@eaca: Thanks, you’re welcome.

BetterExplained
Learn Right Not Rote