I am trying to understand what it means to have an imaginary number in an exponent. What does $x^{i}$ where $x$ is real mean?

I've read a few pages on this issue, and they all seem to boil down to the same thing:

  1. Any real number $x$ can be written as $e^{\ln{x}}$ (seems obvious enough.)
  2. Mumble mumble mumble
  3. This is equivalent to $e^{\cos{x} + i\sin{x}}$

Clearly I'm missing something in step 2. I understand (at least I think I do) how the complex number $\cos{x} + i\sin{x}$ maps to a point on the unit circle in a complex plane.

What I am missing, I suppose, is how this point is related to the natural log of $x$. Moreover, I don't understand what complex exponentiation is. I can understand integer exponentiation as simple repeated multiplication, and I can understand other things (like fractional or negative exponents) by analogy with the operations that undo them. But what does it mean to repeat something $i$ times?

  • $\begingroup$ Nonono, $\exp(ix)=\cos\;x+i\sin\;x$. Have you looked at the other questions tagged complex-numbers? They seem to answer a lot of your questions. And no, you really no longer speak of "repeating $i$ times" in advanced work. $\endgroup$ – J. M. isn't a mathematician Nov 10 '10 at 21:52
  • $\begingroup$ @ friedo: I assume you meant $x^i = e^{ilog(x)} = cos(log(x)) + i sin(log(x))$ $\endgroup$ – user17762 Nov 10 '10 at 21:59
  • 1
    $\begingroup$ The step from real exponentiation to a complex one is seemed more complex compared to the step from integral base/power to rational and then to real numbers. However, the point is just to understand the general step towards complex numbers, it's relatively simple compared to the step from rational to real as it only requires to relieve the notion of order, and allow things to grow into different directions. It's an abstraction problem, I agree. But once you realize the step itself, it should be easier. $\endgroup$ – Asaf Karagila Nov 10 '10 at 22:00
  • $\begingroup$ @Sivaram, I really don't know what I mean at this point. :( $\endgroup$ – friedo Nov 10 '10 at 22:08
  • $\begingroup$ @ friedo: Does $2^{\sqrt{2}}$ make sense? What is your intuition for $2^{\sqrt{2}}$? $\endgroup$ – user17762 Nov 10 '10 at 22:54

Consider a real number $A$, and take it to the power $i$. If our system of complex numbers is to be consistent, then $A^i$ must be a complex number; in other words, there must be two real numbers $x$ and $y$, which depend on $A$, such that:


Furthermore, we can write $A^{-i}=x-iy$ for the same $x$ and $y$. Hence:


We have shown that for any real number $A$, $|A^i|=1$, and therefore $A^i$ corresponds to a complex number which lies some angle $\theta$ along the unit circle.

Now consider the sine and cosine functions for extremely small angles $\epsilon$. A tiny angle $\epsilon$ cuts out a slice of the unit circle, and the curvature of the circumference over this small angle is negligible. We can therefore think of this slice as a right triangle with angle $\epsilon$, and the hypotenuse and adjacent sides are both length one since they correspond to the radius of the unit circle.

Using the formula for the arc length of a circle, it's easy to determine that in the right triangle formed by the small angle approximation, the length of the side opposite to the angle $\epsilon$ is equal to $\epsilon$. We can read off the $(x,y)$ coordinates from this diagram (which are $(cos(\epsilon),sin(\epsilon))$), and therefore we conclude that for very small angles $\epsilon$:

$sin(\epsilon) \approx \epsilon \hspace{10mm} cos(\epsilon) \approx 1$

therefore $cos(\epsilon) + isin(\epsilon) \approx 1+i\epsilon$, and hence for real numbers $A$ which are extremely close to one (so that $lnA$ is small), the complex number $A^i$ lies approximately at an angle $lnA$ along the unit circle, since $A^i=e^{ilnA}\approx 1+i(lnA)$.

| cite | improve this answer | |
  • 3
    $\begingroup$ I actually learned it this way from Feynman volume 1 chapter 22, so you might want to check that out. $\endgroup$ – Matt Calhoun Nov 10 '10 at 23:39
  • 12
    $\begingroup$ Sorry for a silly question, but why is $A^{-i}=x-iy$? $\endgroup$ – Muhammad Alkarouri Nov 11 '10 at 17:49
  • 6
    $\begingroup$ @Muhammad: I don't think this is a silly question at all! I think about it like this: i apply the complex conjugate function to both sides of the equation $A^i=x+iy$, and all this function does is switch the sign of $i$. $\endgroup$ – Matt Calhoun Nov 12 '10 at 17:14
  • 9
    $\begingroup$ But how do we know that the conjugate applied to $A^i$ produces $A^{-i}$? Most functions wouldn't affect the exponent (for example, multiplying both sides by two wouldn't produce $A^{2i}$), so why does it work for the conjugate? $\endgroup$ – tangentstorm Apr 23 '16 at 14:04
  • 4
    $\begingroup$ I don't quite get the first step. A^i = x + iy. Taking reciprocals on both sides, 1/A^i = A^-i = 1/(x + iy) = (x - iy)/(x - iy)(x + iy) = (x - iy)/(x^2 + y^2). What am I missing here? $\endgroup$ – Yogesch Jul 30 '16 at 17:41

Lets say you want to figure out what $x^{a + ib}$ is, then like you mentioned, you start by writing

$x^{a + ib} = e^{a \ln(x) + i b \ln(x)}$

This can be split up though as

$ e^{a \ln(x) + i b \ln(x)} = e^{a \ln(x)} e^{i b \ln(x)} = x^a e^{i b \ln(x)} $

and then you can use euler's formula to take care of the imaginary exponent, so that

$x^{a + ib} = x^a \cos(b \ln(x)) + i x^a \sin(b \ln(x)) $

This formula does not provide much intuition as to what is really happening though. You mentioned that you can understand integer exponentiation as simple repeated multiplication, but I don't think that that is the right way to look at it in complex analysis. I think it is much better to look at exponentiation geometrically.

When exponentiating a real number by a complex value, we found that

$x^{a + ib} = x^a e^{i b \ln(x)} $

so exponentiating $x$ by $a + ib$ gives you the point in the plane with magnitude $x^a$ and angle $b \ln(x)$.

| cite | improve this answer | |
  • $\begingroup$ The geometric explanation of exponentiation is helpful, thanks. I think I am gaining a clearer picture of what is going on. $\endgroup$ – friedo Nov 10 '10 at 22:38
  • $\begingroup$ @friedo, I'm glad I could help. I have thought a lot about this myself, so if you have any more questions I can try to append them to my answer. $\endgroup$ – Eric Haengel Nov 11 '10 at 11:21

The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$.

By definition

$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+i\sin y)$$

which agrees with the real exponential function when $y=0$.

The principal logarithm of $z=x+iy$ is the complex number

$$w=\text{Log }z=\log |z|+i\arg z$$

so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$.

The complex power is

$$z^w=e^{w\text{ Log} z}.$$

In your example $z=x,w=i$ is therefore $x^i=e^{i \text{ Log}x}$.

If $x>0$, $\text{Log }x=\log x$. If $x<0$, $\text{Log }x=\log |x|+i\pi$.


$(-1)^i=e^{i\text{Log }(-1)}=e^{i(i\pi)}=e^{-\pi}$.

$2^i=e^{i\text{Log }(2)}=e^{i\log 2}=\cos (\log 2)+i\sin (\log 2)$.

$(-2)^i=e^{i\text{Log }(-2)}=e^{i(\log 2+i\pi)}=e^{i\log 2}e^{-\pi}=(\cos (\log 2)+i\sin (\log 2))e^{-\pi}.$

| cite | improve this answer | |
  • $\begingroup$ the first equation is missing i at sin(y) $\endgroup$ – MariusM Nov 26 '17 at 21:05
  • $\begingroup$ @MariusM Thanks! Fixed. $\endgroup$ – Américo Tavares Nov 26 '17 at 23:28

It is a way to extend your real variable function into the complex domain in a "nice" (read nice as holomorphic) way. You might ask the question why do we need to even extend in the first place? The need for extension comes from the fact that a $n^{th}$ order polynomial with real co-efficients need not have all its roots as real numbers. So you end up with these complex numbers. Once we find that we have these new numbers, we begin to wonder why should we not try to feed these numbers into functions which are defined on the real variables. But then we find that we have a zillion ways to define these functions for these new complex numbers. A "nice" way to extend these functions is to do in a way that the extended function is holomorphic. That is why you extend $e^{i \theta}$ to the complex plane as $\cos(\theta) + i \sin(\theta)$. Once you do this extension then all the other things related to logarithms and exponentiation follows in a logically coherent way. For instance, we can now define $x^i$ (say $x>0$ or else rewrite $x$ as $-y$ where $y>0$ and write $-1$ as $e^{i \pi}$) as $(e^{\log (x)})^{i} = e^{i \log(x)} = \cos(\log(x)) + i \sin(\log(x))$

This is how I understand it.

| cite | improve this answer | |
  • $\begingroup$ The last definition is valid for a positive real $x$. $\endgroup$ – Américo Tavares Nov 10 '10 at 22:27
  • $\begingroup$ @ Americo: I have changed it. Thanks. $\endgroup$ – user17762 Nov 10 '10 at 22:32

The quantity x^i is well-defined if x is a positive real number. For negative (or complex) values of x, the evaluation is open to multiple interpretations.

You have the following. The natural logarithm of positive real numbers is well-defined, so:

x^i = exp(i*ln(x))

Bear in mind that for real x > 0, all one knows about ln(x) is that it is real. It might be positive, negative, or zero depending on the comparison of x to 1.

Now the function exp can be defined by a power series that converges everywhere in the complex plane. As it turns out, that definition of the exponential function implies the following when the argument is purely imaginary (as here):

x^i = exp(i*ln(x)) = cos(ln(x)) + i*sin(ln(x))

For more see:

[Euler's formula -- Wikipedia]

regards, hm

| cite | improve this answer | |
  • $\begingroup$ How can something "converge everywhere?" That seems like an oxymoron. I still don't understand what an exponent of i "does" $\endgroup$ – friedo Nov 10 '10 at 22:15
  • $\begingroup$ Power series are infinite series. The power series for the exponential function e^z = exp(z) consists of a summation of real coefficients times nonnegative integer powers of z as follows: exp(z) = SUM (z^k)/(k!) for k = 0,1,2,... There are power series, such as the one for ln(x+1), which do not converge for values of |x| > 1. $\endgroup$ – hardmath Nov 10 '10 at 22:38
  • $\begingroup$ "converge everywhere" means the series for exp(x) is, for all values of x (hence "everywhere"), convergent. Not that the series, for a single x, converges to everything. $\endgroup$ – T.. Nov 10 '10 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.