# Can you explain $(1 + iX/n)^{n}$ without using e, sin, or cos?

In my ongoing strugle to understand $e^{\pi i}$ I managed to narrow down my conceptual difficulty. I'm having intuitive trouble understanding why $(1 + iX/n)^{n}$ is conceptually the same as a rotation by X radians about a unit circle as n approaches infinity.

The relationship between e, cos, and sin is what I set out to understand in the first place, so any explanation relying on the fact that $e^{xi} = cos(x) + i*sin(x)$ leaves me right back where I started. Furthermore, I'm looking for an explanation, not a proof. I've seen the taylor series proof of the above equality, I'm looking for an intuitive explanation of why $\lim_{n\to\infty}(1+iX/n)^{n}$ is the same as a rotation of X radians.

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The following argument is very informal, but I just want to see if this is the kind of reasoning you're looking for. So suppose we want to 'rotate' the number 1 in the complex plane an infinitessimal bit, then the first thing to notice is that the movement of our number 1 is going to be orthogonal to its location relative to 0, the centre of rotation. Thus, infinitessimaly we can say that 1 rotates to 1+x*i for some infinitessimal angle x. If we want to do this for a non-infinitessimal angle X, we just turn over an angle of X/n n times, for n to infinity and we get (1+iX/n)^n. –  gfes Jun 5 '11 at 21:54
That helped a lot, thanks. In my own terms, the key point that I was missing was that taking $\lim_{n\to\infty}$ negates the drift off the unit circle that you see in $1+iX/n$, but does not negate the rotation caused by $1+iX/n$. It still astounds me that by taking the limit in this situation you negate the drift but not the rotation - that's an incredibly beautiful and powerful idea. :) –  Nate Jun 6 '11 at 2:49

This is not a rigorous argument — arguing about things being "close" in the manner below, without actual error estimates, can easily lead to mistakes — but a demonstration that may be helpful.

Consider $(1+iX/n)$, where $n$ is sufficiently large so that $X/n$ is small. On the complex plane, it corresponds to the point $(1, X/n)$. The arc (red in figure above) subtended by this point and the real line is of length approximately $X/n$ units when $X/n$ is small. So the angle at the origin, which in radians is the same as the arc length, is also approximately $X/n$. Further, the modulus/absolute value of $(1+iX/n)$ (its distance from origin), is roughly $1$. I presume you know that multiplying by a complex number which has unit modulus and angle/argument $\theta$ is rotation by $\theta$. So here, multiplying by $(1+iX/n)$ is roughly a rotation by $X/n$. Multiplying by $(1+iX/n)^n$ is therefore roughly rotating by $X/n$ $n$ times, that is, rotation by $X$. As $n \to \infty$, this becomes exact.

Here's an illustration showing $(1+iX/n)^n$ (and also $(1+iX/n)^k$ for $k < n$) as $n$ increases. (This illustration is for $X=1$, but it's similar for any $X$.) You can see that as $n$ becomes large, $(1+iX/n)^n$ aproaches the point that on the circle that is at angle $X$, i.e. the point that correponds to rotation by $X$. (Although you didn't want $e$, $\sin$ or $\cos$ used in the explanation, it may be worth mentioning for future reference that this point on the circle is the same as $e^{iX}$ or $\cos X + i\sin X$.)

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Here's an animation from Wikipedia that shows the first $n$ powers of $1+\dfrac{i\pi}{n}$ for larger and larger and larger values of $n$:

The idea is that, if $\theta$ is a small angle, then $1 + i\theta$ has a modulus of approximately 1 and an argument of approximately $\theta$. This follows from the linear approximations $$\arctan\theta \approx \theta \qquad\text{and}\qquad \sqrt{1+\theta^2} \approx 1$$ which become more and more accurate the smaller the value of $\theta$. It follows that the powers of $1 + i\theta$ all have modulus approximately $1$, and have arguments that are multiples of $\theta$.

In particular, since $1+\dfrac{iX}{n}$ has an argument of approximately $X/n$ and a modulus of approximately $1$, the quantity $\left(1+\dfrac{iX}{n}\right)^n$ has an argument of approximately $X$ and a modulus of approximately $1$. As you can see from the animation, this approximation becomes more and more accurate as $n$ increases.

Edit: Just to clarify the precise reasoning for the modulus and the argument. From basic trigonometry, we can see that $$\arg \left(1+\frac{iX}{n}\right) \;=\; \arctan(X/n) \qquad\text{and}\qquad \left|1+\frac{iX}{n}\right| \;=\; \sqrt{1+(X/n)^2}$$ It follows that $$\arg\left(1+\frac{iX}{n}\right)^n \;=\; n\arctan(X/n) \qquad\text{and}\qquad \left|\left(1+\frac{iX}{n}\right)^n\right| \;=\; \left(1+\frac{X^2}{n^2}\right)^{n/2}.$$ Thus, the statement that $e^{iX}$ has argument $X$ and modulus $1$ follows from the equations $$\lim_{n\to\infty}\; n\arctan(X/n) \;=\; X \qquad\text{and}\qquad \lim_{n\to\infty} \left(1+\frac{X^2}{n^2}\right)^{n/2} =\; 1.$$ Both of these limits should be intuitively clear, and can be proven rigorously using L'Hospital's rule.

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The reasoning about the modulus could be precised: the modulus of the $k$th power of $1+i\theta$ stays close to $1$ if and only if $k\theta^2\ll 1$. In your case $\theta$ is of order $1/n$ and $k=n$, hence you win but... –  Did Jun 5 '11 at 23:28
@Didier: That's true. I have added a bit at the end to make the argument more precise. –  Jim Belk Jun 6 '11 at 1:30
Thanks! I don't quite understand the image above: I understand the first frame represents $1 + i*\pi$, but I don't understand the following frames. I know I'm being a bit dense, but could you break frame 3 down for me into what each segment represents? –  Nate Jun 6 '11 at 2:17
@Nate: The segments are immaterial -- they're just there to make it easier to see where the endpoints are. The endpoints of the segments are the first $n$ powers of $(1+i\pi/n)$. For example, in the third frame, the three points are $(1+i\pi/3)$, $(1+i\pi/3)^2$, and $(1+i\pi/3)^3$. –  Jim Belk Jun 6 '11 at 2:40

First of all, $$\left( 1 + \left(\frac{iX}{n}\right)\right)^n \neq \lim_{n \to \infty}\left(1 + \left(\frac{iX}{n}\right)\right)^n$$

It is the right hand side that is equivalent to $e^{iX}$. That is, $$\lim_{n \to \infty}\left(1 + \left(\frac{iX}{n}\right)\right)^n = e^{iX},\qquad\qquad (1)$$ where $X$ is in radians. This follows from the definition of the limit of $e^x$:

It's just that in our case $x = iX$.

Note that in the special case in which $x = 1$, we can define $e$:

Are you comfortable with your understanding $e$ and $e^x$ in terms of the limits given above? If so, perhaps now you can make a connection between the right- and left-hand sides of equation (1) above. Because in this case, the numerator of the fraction for which we are finding a limit is $iX$. $iX$ is the exponent of $e$ in the resulting limit. The fact that $X$ is multiplied by $i$ in $e^{iX}$ can be understood as $\cos(X) - i\sin(X)$, although I take it you're not comfortable with that formulation of $e^{ix}$, whatever $x$ may happen to be. I guess I'm not quite understanding precisely where your confusion lies.

One problem may be that $e^{iX}$ and the role played by $X$, is best understood in its relation to the complex plane and the unit circle. The exponential function $e^x$ is perhaps best characterized by its graph in the Cartesian plane. Given the different coordinate systems, it is not surprising that making a connection may prove difficult.

It is always a good exercise to devote some time to exploring the graphs of such functions in different coordinate systems and comparing how each varies in response to a variation in the exponent's value. The better you're able to visualize what's happening, the more sense it will make.

For example, if we substitute $\pi$ for $X$ in equation (1) above, we can use the following image to see how the graph of $\displaystyle 1 + \left(\frac{i\pi}{n}\right)^n$ changes as $n$ becomes increasingly larger (approaching infinity):

If we would have selected $\displaystyle \frac{\pi}{2}$, we would have ended up with a vertical line approaching an arc whose angle is $\pi/2$ etc. Whatever value for the angle $X$ that we choose to use, we will end up with a rotation of $X$ about the unit circle in the complex plane.

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This is perhaps not such a trivial result to 'explain,' but we can see what we can do.

The big idea rests on believing firstly that a complex number of modulus (absolute value) 1 lies on the complex unit circle. I noticed in one of your previous [questions] that you believe this, so I will let that just sort of slide by. If we multiply two complex numbers of modulus 1, we clearly see that the new modulus is still one. But if the number is different, then a rotation has happened (why? because it's still on the complex unit circle).

Suppose that we multiply a complex unit, which I will denote by $z_0 = a + bi$, by a complex unit that is very close to 1. That is, we will rotate it by multiplying it by a number like $1 + xi$, where x is very very very small. (Technically, this is not a unit, but it's really close to one). Then this will result in a very small rotation (as it's almost 1, which doesn't change a number, i.e. multiplying a number by 1 does not change the number). It might not be obvious that it's small, but some notion of continuity might allow us to imagine that really small deviations from 1 will result in really small rotations.

Now we hit the key idea: if we want to rotate by a larger angle, say $\theta$, than $1+i \theta$ is not a complex unit. But $1+\frac{\theta }{n}$, for large n, is essentially a complex unit. And as long as we believe that multiplying by the same complex unit results in the same rotation (maybe not obvious without the series definition!), then we could just do this n times to get the original theta.

So we have that $\lim (1 + \theta / n)^n$ acts like a rotation. But perhaps it is only now that we hit the crux - how do we know that the rotation is actually $\theta$ radians? That's really not obvious, to be honest. Here, I need to step back and justify this claim.

Let's look at our complex unit $z = a + bi$. It's a unit, so $a^2 + b^2 = 1$ (to clarify, in case that wasn't already clear). What this means is that the point $(a,b)$ is on the unit circle in the plane. We should ask ourselves, what is the angle with the positive x-axis? Well. imagine the triangle with x-width a and y-height b. Then the angle is given by $\arctan{ \frac{b}{a}}$. (Draw the triangle if that's hard to see - I think it will be readily apparent).

Now let's instead look at our 'near-unit' $1 + \theta / n$. Then the rotation angle will be given by $\arctan { \frac{\theta}{1 * n} }$. But if we look at a graph of arctan, such as below: We can see that for really small values of x, $\arctan(x)$ is about x. So $\arctan {\frac{\theta}{n}}$ is about $\theta / n$. This is more accurate as n gets larger.

I hope that brief mathematical interlude isn't too bad - but ultimately I think that we can now see that multiplying by $1+i \theta / n$ corresponds to a rotation by $\theta / n$ radians, and so doing it n times yields a rotation by $\theta$. Taking the limit as n goes to infinity yields your result.

Is that clear?

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@amWhy: whoops - thanks! –  mixedmath Jun 5 '11 at 23:51
No problem: I usually try not to "tamper" with others' answers, but part of the TeX wasn't rendering, and it was easily fixed by a single "brace"...glad to do it. –  amWhy Jun 5 '11 at 23:57
I get it! Multiplying by $1+ix$ gives us rotation by x but doesn't keep us on the unit circle. However, if we divide by n and raise to the power n, we manage to quell the drift off the circle but maintain the rotation. The drift is lost if we divide by and raise to the power n, but the rotation is not. Looking deeper, is there any field of math concerned with which operations are susceptible to elimination by such limits and which operations are unaffected? –  Nate Jun 6 '11 at 2:28
@Nate: not so directly, but there are many related fields of math. A proper geometer is interested in the isometries of various things, and in a sense this limiting process extracts an isometry from what might not appear to be an isometry. A bit more different, perhaps, but quaternions and Lie Algebra also examine rotations and transformations (but without the limiting process... it's nonetheless absolutely incredible). –  mixedmath Jun 6 '11 at 2:41

EDIT: I realize now, after writing it, that I did not answer the asker's original question at all. Feel free to downvote me, but I sorta put a bit too much time into this answer so I think I'll leave it here if people want to read it, but understand that it doesn't answer the asker's original question. Again, feel free to downvote.

The "moving particle" explanation was the final explanation that gave me an intuitive grasp on the subject. It is elaborated in the book Visual Complex Analysis by Tristan Needham, and I recommend looking at it if you ever get the chance.

The basic principle is this: Say that the position of a particle in the complex plane is given by $\mathbf{r} = e^{i \omega t}$, parameterized by $\mathbf{r} = < x(t),\ y(t) >$. If we can find $x(t)$ and $y(t)$, we will be able to find the real and imaginary parts of $e^{it}$.

Let's consider the velocity of this particle, $\mathbf{v} = \dot{\mathbf{r}}$, or $\frac{d\mathbf{r}}{dt}$. This is pretty simple; $i$ is simply a constant number, so $\dot{\mathbf{r}} = i \omega e^{i \omega t}$.

As you know, multiplication by $i$ is the same as a rotation by $\pi/2$ counter-clockwise in the complex plane.

Basically, this tells us that the velocity of our particle is perpendicular to its position vector.

If we find the acceleration of the particle, we get $\ddot{\mathbf{r}} = - \omega^2 e^{i \omega t}$. As you know, multiplcation by $-1$ is the same as a rotation by $\pi$ in the complex plane.

We can rewrite the acceleration as $\ddot{\mathbf{r}} = - \omega^2 \mathbf{r}$. Conceptually, this is saying that the acceleration of our particle is always in the opposite direction of the position vector (that is, towards the origin) with a magnitude proportional to the square of the angular velocity.

What kind of motion do we know behaves this way?

That's right -- motion in a circle around the origin.

This behaves exactly like $\mathbf{r} = < cos(\omega x),\ sin(\omega t) >$. Its velocity is $\dot{\mathbf{r}} = \omega < -sin(\omega x),\ cos(\omega t) >$, which is $\mathbf{r}$ rotated $\pi/2$ radians counter-clockwise. Its acceleration is $\ddot{\mathbf{r}} = -\omega^2 < cos(\omega x),\ sin(\omega t) > = - \omega^2 \mathbf{r}$.

You will find that the solution $\mathbf{r} = < cos(\omega x),\ sin(\omega t) >$ one of the solutions for the system of differential equations $\dot{\mathbf{r}} = i \omega \mathbf{r}$ and $\ddot{\mathbf{r}} = - \omega^2 \mathbf{r}$.

If we throw in our initial condition that $\mathbf{r}(0) = 1$ (that is, $e^{0 i \omega} = 1$), we find that $\mathbf{r} = < cos(\omega x),\ sin(\omega t) >$ is our unique solution to the differential equations. (Please correct me if I'm wrong --- I'm trying to think of other cases. I'm thinking there should be another constraint, but I can't think of it.).

So, $\mathbf{r} = < cos(\omega x),\ sin(\omega t) >$.

Therefore, $e^{i \omega t} = cos(\omega t) + i sin(\omega t)$

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

Where $e^{i \theta}$ parameterizes a particle that has moved through a circle an angular distance $\theta = \omega t$

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You can move this answer to another question... there are lots of questions on this website asking about this equation. :-) –  ShreevatsaR Jun 5 '11 at 22:18

One way is to write $1+iX/n = r_n (\cos \theta_n + i \sin \theta_n)$. Then you get that $r_n = \sqrt{1+\frac{X^2}{n^2}}$ and $\theta_n = \arctan(X/n)$. We can see that $r_n^n = (1+\frac{X^2}{n^2})^{n/2}$ approaches 1 (since $r_n^{2n^2}$ approaches $e^{X^2}$.)

So you want to know that $n\theta_n=n\arctan(X/n)$ approaches $X$ as $n\rightarrow \infty$. You can write this as $\frac{\arctan(hX)}h$ where $h=\frac{1}{n}$, and this is obviously the derivative of $\arctan{Xz}$ at $z=0$, which equals X.

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