# What the heck is an imaginary exponent?March 16, 2011 11:11 PM   Subscribe

What does it mean to raise a number to an imaginary power?

Perhaps the most elegant equation in all of mathematics is "Euler's identity", which you've probably seen before:

e^(πi) + 1 = 0

I can appreciate the astonishing simplicity of this equation, and I can even follow the derivations I've seen, mostly. What I can't exactly wrap my brain around is what that i up in the exponent means, in terms of a transformation of a number I can understand, like e.

I feel like I can understand what e^π means, sort of. It's a multiplication of e by itself π times, which I understand as being partway between cubing it and raising it to the 4th power. It's a particular point on the graph of e^x = y.

It's also easy algebra to see that i^2 = e^(πi), but somehow that doesn't help me to fundamentally grasp what an imaginary exponent means.

This may be a ridiculous question. I don't know why an imaginary exponent seems harder to grasp than the very concept of imaginary numbers in the first place, or transcendental numbers for that matter, neither of which you can count like the number of beans on a plate. But it does.

If anyone has any English to offer to help me grasp this concept on a visceral level, I'd appreciate it. I just feel like I'm missing something in comprehension here, even though the math itself is something I can work through. It just kind of feels like mouthing a phonetic version of a language I don't know. Even though I could come up with the right answer to a math problem with imaginary exponents, I wouldn't actually know what I was talking about, you know?
posted by slappy_pinchbottom to Science & Nature (29 answers total) 21 users marked this as a favorite

Use logarithmic transformations to move the exponent somewhere else. Then the i just becomes a straight multiplier.

Also you can use a power series expansion for e^x which is the standard way to prove that e^(ix) = cos x + i sin x.

The wikipedia page for Euler's Formula covers both of these somewhat tersely. Probably the power series expansion of e^x is the easiest to understand IMO.
posted by GuyZero at 11:22 PM on March 16, 2011 [1 favorite]

I think that you're thinking about this in the wrong way. As russm asks - do you really have a great mental picture of what i means, or are you just so accustomed to it that you accept it?

The way I deal with i is by just accepting it as a symbol that comes with a set of self-consistent rules about how it should be manipulated. It happens that our rules for how real and imaginary numbers interact have the same properties as some natural phenomena. So, for me, if i has some meaning, it must be captured in its representational power. Maxwell's equations and Nyquist plots jump to my too-tired mind ...

I look forward to everybody else's answers!
posted by Metasyntactic at 12:15 AM on March 17, 2011 [1 favorite]

For me the fastest way to understanding e^s (where s is some non trivial value) is to look at the taylor/mclauren/power series expansion. Once you accept that e^s (and cos and sin) can be written as the infinite sum of polynomials, the answer to what e^s is equivalent to is right in your face.

This idea of functions as the sum of infinite polynomials really lead to the ground breaking idea that functions can be written as the infinite sum of sines and cosines. And that, my friends, has made modern engineering possible. Fully wrap your head around that tidbit and this whole e^s stuff isn't so mysterious.

do you really grok what i is? that'd be the first question I reckon

I spent many, many hours in college getting close and personal with i. It no longer holds any mysteries for me. But can I explain it? Not clearly... because to me the truth is rather mundane: i is just a placeholder.

So in my studies (electrical engineering) you are always looking at some "real valued input" and trying to model or achieve the "real valued output". Makes sense: I can't generate "imaginary" signals with my radio. My inputs and outputs have to exist.

Where i comes in handy is two places: either (1) we come across a sqrt(-1) situation or (2) we introduce a sqrt(-1) situation in order to better understand the equation. In either case, the result is the same: i lets us ignore the impossibility of sqrt(-1), work on the -- momentarily -- divided equation, and hope that if we do everything right at the end the i will disappear. If it doesn't then that means you did something wrong or your device will blow up.

Despite it's reputation as being "imaginary" or "complex", it's really just a marker to remind us that part of our problem is doing something different.
posted by sbutler at 12:59 AM on March 17, 2011 [2 favorites]

I don't know if this is helpful to you but I like to view it as a mathematical tool for converting between polar and rectangular coordinates on the complex plane. The multiplier in front of e is the magnitude and the multiplier in the exponent is the angle in radians, and on the other side of the equation you have x and y components. The fact that exp(pi*i) = -1 follows quite naturally if you imagine a vector with magnitude 1 and angle pi (180 degrees). The corresponding x,y coordinates for such a vector are (-1,0) or -1 + 0*i or -1. And likewise exp(pi/2*i) is a vector at 90 degrees pointing straight up so it has x,y values (0,1) or i.
posted by Rhomboid at 1:19 AM on March 17, 2011 [6 favorites]

Yeah, i know what you mean. This may not help, but you can try this line of approach as a sort of next-step-higher:

You have to expand your thinking. You have to accommodate "complex" numbers. When you start dealing with "imaginary" numbers, you're no longer dealing with ordinary numbers. Ordinary single numbers, like 52, 97, x, k, etc. You're dealing with vectors/phasors in a plane. (e.g., 5x + 3y; (5,3); k at p degrees) But instead of an x and y axis, one axis is called "real," the other axis is called "imaginary." i is the unit vector in the imaginary direction.

And at this point, you have to set aside the =1 version of Euler's Identity, and digest the trigonometric form of it (shown by GZ above), which is the more useful version.

Or, getting back to simplifying things, if you can deal with vectors, this is another sort of vector. This is very much a geometric interpretation.

(Helpful? Or Not? Or makes it worse?)
posted by coffeefilter at 1:29 AM on March 17, 2011 [1 favorite]

Maybe it's easier to think of it as a generalisation. You're happy with e^x where x is a real, and you ask what if there was a function like it but where x could be complex - and you require it has all the important properties of e^x. From there you can (with the right choice of important properties) work out what that function must be.
You could equivalently think about the factorial function - n! = n * (n-1) * (n-2) ... * 1. It's completely non-obvious how that might be extended at first, but you can do it. It doesn't mean that you're somehow doing something with the original definition to extend it - you're finding something with the same important properties, and you're no longer obliged to think about it in the same way.
You're not now obliged to think about exponentiation as the repetition of multiplication - you can for example just think of it as the power series, and recognise that that power series is the repetition of multiplication for natural numbers.
posted by edd at 4:55 AM on March 17, 2011

e^ix is a phasor. Multiplication of phasors is just adding together angles. So, exponentiation by an imaginary involves a rotation in the complex plane.

Looked at another way, e^x for strictly real x is either a damped or growing exponential. e^ix, for real x, is a pure oscillation. e^(x + iy) is a product of both. The functional behaviors---damping, runaway growth, and oscillation---are deeply connected, as anyone who's tried to build a stable feedback loop can testify, because e^z is the archetypical analytic function.
posted by fatllama at 5:41 AM on March 17, 2011

I highly, highly, recommend watching the Khan Academy videos where he proves euler's identity.

It starts here and is 7 videos total.
posted by empath at 5:48 AM on March 17, 2011 [1 favorite]

Like fatllama says, the exponential functional describes a rotation in the complex plane: To really understand what functions with complex domains/ranges are doing, you should probably try to study some complex analysis.
posted by james.nvc at 6:13 AM on March 17, 2011

...are deeply connected...

... I was to expound on this. Analytic functions are not only continuous and differentiable everywhere on the complex plane, they're also infinitely differentiable everywhere as well. This means, if I recall correctly, that knowing the behavior of the function in any arbitrary finite region of the complex plane determines its behavior everywhere else.

Thus, the answer to the question "how the heck can e^-x exponential damping along the real axis become oscillation along the imaginary axis?" is "because it couldn't be any other way." I always found this very satisfying if not immediately intuitive.
posted by fatllama at 6:25 AM on March 17, 2011

Basically, you can define any function f(0) as an infinite series of fractions defined as: f(0) + f'(0)x+(f''(0)/2!)*x^2+ (f'''(0)/3!)*x^3..., to infinity: The MacLaurin Series.

One definition of 'e' is that the derivative of e^x is also e^x, so that f'(x) = f (x).

Since any number raised to the 0th power of 0, and every derivative of e^x at every point is the value of e^x at that point, we can just replace every derivative of f(0) with 1.

So you have e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!.... to infinity..

Replace x with ix and you have:

e^ix = 1 + ix + i^2*x^2/2! + i^3*x^3/3!....

i follows the following cycle of powers:

i = i

i^2 = -1

i^3 = -i

i^4 = 1

i ^5 = i

Substituting back in you get:

1+ix - x^2/2! - i*x^3/3! + x^4/4! + i*x^5/5!...

Now, separate the elements with 'i's from the ones that without them:

(1 - x^2/2! + x^4/4! - x^6/6!...) + i(x - x^3/3! + x^5/5! - x^7/7!...)

The first seriesis the same as a similar expansion for cos(x), and the second series is the expansion of sin(x).

so now you know that e^ix = cos(x)+i*sin(x)

From there it's trivial to plug in pi for x and get the Euler identity.
posted by empath at 6:45 AM on March 17, 2011

Since any number raised to the 0th power of 0

I meant: Since any number raised to the 0th power is 1
posted by empath at 6:47 AM on March 17, 2011

empath: You can't define just any function as a Maclaurin expansion (or more generally a Taylor series), but it does work for the exponential function. Taylor's Theorem tells you when you can do this.
posted by edd at 7:08 AM on March 17, 2011 [1 favorite]

Right, I actually meant to say 'certain functions'.
posted by empath at 7:11 AM on March 17, 2011

I actually didn't even finish pre-calc in college. I've just been teaching myself calculus for the past year or so using itunesU, wolfram and wikipedia, so I'm never super confident I have any idea what I'm talking about.
posted by empath at 7:14 AM on March 17, 2011

Here is how I was taught to understand imaginary numbers: don't try. They are black-box problem solving tokens. They were invented to make some equation work. You are raising a real number to the power of apples, because at some point you are going to divide the apples back out again.

If you have to visualize them, just think of them as alternate dimensional spaces. The domain of real numbers is a string of the numbers in a row, which are infinitely multiplyable and divisible. (Sort of like a linear fractal- you have a ruler that is 1M. Then you zoom in a power and it is marked in cm. Then mm. Then nm, etc.) The things you do to the numbers change your view of them, and their placement in the "grid" of the real number system. If you square the ruler, you have a plane instead of a line. If you cube it, you have a cube. If you raise it to the 6th power, you have a cube of cubes.

As long as you remain in the domain of real numbers, they will fit into the real universe (dimensional space) of "positions".

Imaginary numbers move you off into some other universe temporarily. The key is temporarily: math solves problems. The ultimate answer to the problem you are solving will come back out of the imaginary numbers and become real again.
posted by gjc at 7:15 AM on March 17, 2011

What does it really mean to "understand" anything? There's a feeling, and there's a cognitive part. Metaphorically, it is akin to finding where a piece of a jigsaw puzzle fits in. It has all these pseudopods and then there's this space which exactly matches it!

People are giving you rotations in the complex plane and taylor series to provide the space in which to fit it all in, but that's not the context you want unless you deal with that kind of stuff all the time. I think you "understand" exponents as things multiplied by themselves and want to see an exponent of i behave something like that. You already "get" real numbers as exponents because you started with integer exponents and moved to rational exponents and see reals as being near the rationals (exponentiation being a continuous function.) So, x raised to the sqrt(2) power is right near x raised to the 1.414 power. But you can also see,
from (x^m)^n being x^(m*n), how x^sqrt(2) can then be raised to the sqrt(2) and then you get x squared. Raising something to the square root of 2 power is kind of multiplying that something by itself half way to squaring it.

So if x^(-1) is the reciprocal of x, x^sqrt(-1) is half way to taking a reciprocal. If we think of a reciprocal as turning something upside down--x becoming 1/x, think of a power of i as a turning x halfway there--a quarter turn on the way to being upside down.

Russel said "Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true," but not any old mathematical statement works aesthetically. It has to work and play well with the rest of mathematics. That's why division by zero remains undefined. There's no definition that will not just make a big mess of all the other rules. Using i as an exponent can be defined in a way that fits in nicely with everything else. We abstract and generalize the concept of exponent in a way that preserves the feel aesthetically and allows it to function usefully in the real world, but ultimately, it's all metaphorical. Read Lakoff on this.
posted by Obscure Reference at 7:50 AM on March 17, 2011 [2 favorites]

Where were you jackasses when I was in high school? I just learned more from this thread than any teacher I ever had.
posted by Cool Papa Bell at 8:22 AM on March 17, 2011

So if x^(-1) is the reciprocal of x, x^sqrt(-1) is half way to taking a reciprocal. If we think of a reciprocal as turning something upside down--x becoming 1/x, think of a power of i as a turning x halfway there--a quarter turn on the way to being upside down.

That kind of doesn't work as intuition, at least for me, because then you'd think that 2^i would equal something like .25 (1/2 of 1/2)
posted by empath at 8:22 AM on March 17, 2011

Where were you jackasses when I was in high school? I just learned more from this thread than any teacher I ever had.

posted by empath at 8:24 AM on March 17, 2011

Here is how I was taught to understand imaginary numbers: don't try.

As you can probably guess, I bristle at this attitude. Russel's quote above is closer to the point: with mathematics and numbers, the reality of the subject is as much as you'd like to assign to it.

Look.

Say all that you know is the + operator. Then the non-negative integers 0, 1, 2,... form a complete set of numbers for you. That is, for any x and y that are in the set, z = x+y is also in the set.

Once you discover or invent the - operator, you've got to include negative integers -1, -2, -3, ... to have a complete set of numbers.

Suppose you want the division /. Well, some fractions are reduced to integers, 4/2 = 2, but most are not, like 1/2. So you round up all the rational fractions of integers to make your set complete.

Multiplication * is just short hand for addition, so far so good, and exponentiation is just a short hand for multiplication... our set (integers and their rational fractions) is still complete. But if we want to do the inverse operations, roots, we're in deep trouble.

Consider: the word surd was given to be a root of a number. If that number turned out to be irrational, like sqrt(2), it was called ab surd, an etymological root for our English absurd. Merely asking what the diagonal length across a square whose sides are length 1 leads one to absurdity because the answer is not in the set of rational fractions! Reflect, though, that this evocative jargon is just as descriptive as imaginary because these days we seem to have no fundamental mental trauma while dealing with these infinitely precise numbers as the real number system.

Anyway, if we want the inverse operator of the exponent, the root, we need "irrational" numbers added to our set for it to be complete.

But what to do with the solution to the equation x^2 + 1 = 0? Well, apparently, for our set to be complete we need another absurd number, sqrt(-1) = i. This is getting exhausting!

Is our set complete now? One of the great questions one can ask is this: is that it? Will there ever be a need for some other crazy number j or k to make our set complete? Is there some sort of hypercomplex number that looks like a + ib + jc + kd + ... where the a, b, c are just real coefficients for the crazyweird numbers i, j, k,...? The answer is sort of. These "hyper complex" numbers truly live in a higher dimensional space of group theory. They are, in many senses, reducible to structures (e.g. vectors, matrices) of real and complex numbers. For a single dimension of "number" (a U(1) group), complex numbers are the whole of the frontier. There will be no more crazyweird definitions just to make the set complete. We're done, and that fact that we're done is so important.

If you think that real numbers alone are reality and that imaginary numbers are "just some tool we invent to make circuit theory work", I say you are living in Flatland. God, by every account we know, does his work in the complex plane and nothing "higher". Evidence available by request.

This isn't to say that complex numbers are useful to most people. They aren't. They aren't even useful for a great deal of very beautiful mathematics. But if your work is physics, or certain brands of engineering, you essentially swim in the complex plane for the rest of your life.

FWIW, the best lay approach to these topics---completeness in numbers as well as the larger problem of completeness in logical systems---is offered in David Foster Wallace's Oblivion.

posted by fatllama at 8:44 AM on March 17, 2011 [5 favorites]

Apparently my "close bold tags" are imaginary as well. Apologies.
posted by fatllama at 8:46 AM on March 17, 2011 [1 favorite]

I hate my life: the book is Everything and More, not Oblivion of course.
posted by fatllama at 8:48 AM on March 17, 2011

empath: That kind of doesn't work as intuition, at least for me, because then you'd think that 2^i would equal something like .25 (1/2 of 1/2)

Your metaphors need some tightening up. Here! Use this wrench! (2^i)^i = 1/2 doesn't make 2^i into 1/4 (25 cents). It's not just halving it. It's not even square-rooting it. It's giving it a quarter turn. The wrench will let you turn it that way.
posted by Obscure Reference at 9:01 AM on March 17, 2011

fatllama: Is our set complete now? One of the great questions one can ask is this: is that it? Will there ever be a need for some other crazy number j or k to make our set complete?

I'm asking my class that on their midterm (for extra credit). In particular, now that we can take a square root of -1, thanks to i, what do we need to do to take a square root of i? I'm hoping some of them will figure out that (sqrt(2)/2)(1 + i) will do the trick..
posted by Obscure Reference at 9:07 AM on March 17, 2011

On a quasi-related note, I believe it was in Asimov on Numbers where i was described as being badly mis-named. sqrt(-1) is not more imaginary than 1/2 because you can no more draw 1/2 than you can i. Maybe try to find an old copy of that book.
posted by GuyZero at 9:45 AM on March 17, 2011

Obscure Reference, you are likely an awesome teacher. I loved these sorts of questions.
posted by fatllama at 10:18 AM on March 17, 2011

Multiplying by i means rotating by 90 degrees counterclockwise in the complex plane. Try it with a few examples to convince yourself.

You already know (hopefully, if you're dealing with exponentials) that the derivative of e^nz with respect to z is ne^nz. So by definition the derivative of e^iz = ie^iz.

So what does that mean? It mean that for f(z) = e^iz, f'(z) = i * f(z). That is, the derivative is perpendicular (on the complex plane) to the value. And what happens if you keep on going perpendicularly to your current location (relative to the origin)? You travel in a circle (relative to the origin). If you let z range from 0 to π, you'll start at e^(i*0) = 1, and move counterclockwise in a circle until you end at e^(i*π) = -1.

That's super handwavy, but the whole argument is explained really well and in much more detail in Tristan Needham's amazing book Visual Complex Analysis.
posted by dfan at 6:17 PM on March 17, 2011 [1 favorite]

I think everything said by everyone above is helpful, but maybe hasn't directly provided what you are looking for: a simple visual intuition for what an imaginary exponent is actually doing. I'll try to share mine with you.

Take multiplication all by itself first with no exponents. What is the visual intuition for multiplying by some real positive number? Well, imagine the number line with a big bold arrow running from zero to one. Multiplying by x is growing that arrow by some amount. We multiply by three, visualize the length of that arrow tripling. Duh, right? Now bring in an exponent. As you point out, we can think of exponentiation as repeated multiplication. So if we multiply by 3^3, we are taking the length of our original arrow and tripling it, tripling it again, and tripling it a third time.

It sounds like you probably have the visual intuition already for complex numbers, specifically what it means to multiply by i. Instead of growing our arrow, multiplying by i rotates it, pulls it off the real number line into the complex plane. We'll need this intuition in a minute.

Now, let's think about the e in your equation. e is a real positive number, but is is a very special one. Maybe you already have the visual intuition I am about to describe for e, but I'll go over it again for good measure. Go back to our unit length arrow running from zero to one. Let's multiply it by two: i.e. we double its length. Hmm, what happens if we break this doubling into two parts: we grow the original arrow by half, and then we grow that new arrow by half again? We get an arrow of length 2.25 (1*1.5*1.5), cool it is bigger than just our doubled arrow. What if take our doubling and chop it up into 3 steps, growing our arrow by 1/3 each time? Oooh, even longer at 2.37. How about 4 steps of 1/4 each? 2.44 this time. If we keep chopping up our original doubling into smaller and smaller pieces, can we make this arrow grow as big as we want? Nope. If you break the original doubling into an infinite number of infinitely tiny pieces, you will grow the unit arrow to be e units long, that is why e is such a special number. This is the visual intuition behind the formula e = lim n->inf (1+1/n)^n.

Now the above intuition for e assumed you started by wanting to double the unit arrow, i.e. grow its length by 1. What if you wanted to grow it by more? What if you wanted to grow its length by some value x (so final length was x+1)? All of the above still applies, but our final formula is e^x = lim n->inf (1+x/n)^n and e^x is how long you will make the arrow by chopping up your single-step growth of x into an infinite number of infinitely tiny pieces.

We're almost there. The intuition for e^x above assumed that x was some positive real number, i.e. we assumed you wanted to grow the arrow and see how chopping that growth up into many tiny pieces would change things. Rather than starting with some desired growth x, what if we want to leave its length unchanged and instead rotate it? We know how to rotate our arrow, we just multiply it by i to rotate it pi/2. Why can't we chop up this rotation into many tiny bits just like we did the growth above? Let's see what happens: e^i = lim n-> inf (1+i/n)^n. Ignore the infinity part for a bit and just think of n as a really big number. What is happening to our unit arrow as it is transformed by one of these (1+i/n) factors? Well, we get back our unit arrow (the multiplication by 1) and then we add to it a very small version of our arrow (the 1/n factor) which has been rotated by pi/2 (the i factor). So visualize our unit arrow with another tiny little vector hanging off the end of it and pointing perpendicular to it. When you add these together, the net effect is rotating our unit arrow by a little bit, pulling it up into the complex plane just a bit. Now we do this n times, which pulls our unit arrow way up into the complex plane. How far up? Well, we took the original pi/2 (90 deg) rotation (multiply by i) and chopped it up by plugging it into our formula for i, but we actually end up with a smaller rotation of only 1 radian (57 deg). Interesting! chopping up the growth (multiplication by real) made the total growth bigger, but chopping up rotation (multiplication by imaginary) makes the total rotation smaller.

So, what have we learned? e raised to an imaginary power is really a rotation in the complex plane just like e raised to a real power is really a way to grow something. Both are special in that they are taking a simple, one-time growth/rotation and chopping it up into infinite little bits of growth/rotation. If you understand how e raised to a real represents continuous compound interest, you can also think of e raised to an imaginary as representing continuous compound imaginary interest.
posted by stoffer at 9:31 PM on March 17, 2011 [3 favorites]

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