Q: How are imaginary exponents defined?

The original question was: How do you do xi (x to the i power), and how on Earth was it developed?  There isn’t really anything to base xi on from previous rules of exponents as it is a completely new idea.

Physicist: Euler, the dude who originally came up with using the imaginary unit, i, as a placeholder for \sqrt{-1} (which, despite having no actual solution, is still something we can talk about), also came up with “Euler’s equation“: e^{i\theta} = \cos{(\theta)} + i \sin{(\theta)}, where “e” is equal to 2.718281828…

Leonhard "Lenny" Euler. A squinty genius.

Euler’s (surprising) equation provides a way to talk about complex exponents (“complex” = “involves i“).  So, Euler not only provided an idea that’s confusing as hell (i), but also a way to deal with it efficiently.

So, for example (using Euler’s equation and some log properties):

\begin{array}{ll}3^i\\=e^{\ln{(3^i)}}\\=e^{i\ln{(3)}}\\=\cos{(\ln{(3)})} + i\sin{(\ln{(3)})}\\\approx 0.455+0.891\,i\end{array}

So why is this formalism used, instead of some other set of rules?  Like everything else in mathematics, it’s a matter of convenience and self consistency.  You’d hope that the usual, old rules would apply in a natural, convenient way.  For example, you’d want x^i x^{-i} = 1.  And that’s exactly what happens:

\begin{array}{ll}x^i x^{-i}\\=\left[e^{\ln{(x^i)}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} + i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} + i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 + \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}

(Here I’ve used: Cos(-x) = Cos(x), Sin(-x) = -Sin(x), i2=-1, and the Pythagorean identity.)

Even more important, this technique is used because it recovers the usual rules for real exponents (exponents that don’t involve i), or at the very least doesn’t mess them up.  It keeps arithmetic nice and self consistent.  After all, when you’re coming up with new math, you want to make sure that you don’t trash what you’ve already got.  Euler’s equation is an “analytic continuation” of the exponential function (e^x) from the real numbers, to the complex ones.  An analytic continuation takes a function defined on a fairly small set, like all the real numbers, and generalizes it to work on a larger set, like all complex numbers (which includes real numbers like “3”, but also includes numbers like “i” and “4-2i”).  It’s not obvious, but it turns out that Euler’s equation is the only “nice” way to define complex exponents.

You’ll find (at least, those people who are so inclined will find) that Euler’s equation, and in particular the method for finding imaginary exponents above, is consistent with all the rules of exponentiation.  Specifically, \left(x^A\right)^B = x^{AB}, x^Ax^B = x^{A+B}, x^Ay^A = (xy)^A, and x^0 = 1.

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29 Responses to Q: How are imaginary exponents defined?

  1. patty says:

    how does e^(i*pi) = -1 ??

  2. Neal says:

    Here’s another way you could think about this, based on “analytic continuation.” For the real numbers, e^x = \sum_{k=0}^\infty x^k/k! . For complex numbers, simply define e^z = \sum_{k=0}^\infty z^k/k!. That gives everything you want!

  3. patty says:

    thanks, another question
    whats wrong with this proof:
    ipi+ipi=0 (this is false)

  4. Peter Davis says:

    The exponential function is multi-valued for non-real solutions. e^(2iπ) = cos(2π) + i sin(2π) = cos(0) + i sin(0) = 1. With this method you could also “prove” 0 = 2iπ = 4iπ = 6iπ = … In reality you can only take the log of both sides of the equation after line 5 when “log” is a single-valued function, or if both arguments are in the same phase (sorry don’t know the technical term)…just as acos(cos(2π))=0 not 2π.

  5. Neal says:

    Nitpick: You mean “The logarithm is multi-valued.” The exponential function is still single-valued, but it is horribly not injective. The technical term is “branch,” but you have the correct idea — you need to choose the values of log with the same phase.

  6. patty says:

    i understand how 2pi(radians) =360=0 (degrees)
    so could the 2pi be in degrees?
    Just cuz i was board, I wanted to find “i” like it was a variable and got weird results

    e^ypi=-1 (find y to see if it is i)
    y= ln(-1) + (-pi)
    y= ln(e^ypi) -pi
    y=ypi-pi; pi=ypi-y; pi= y(pi-1)
    y=pi/(pi-1)=1.4669ish ( what did i do wrong?)

  7. patty says:

    ops my bad, i did the second line wrong so ignore my past comment
    e^ypi=-1 (find y to see if it is i)
    y=ln((-1)^(1/pi)) [i don’t know how to simplify further than this]

  8. Bryn says:

    From your second line:
    πy=Ln(-1)=ln(1)+i(π+2πn)=πi(1+2n) (where n is an integer)
    Then, y=i(1+2n).

  9. Neal says:

    Okay, first of all you have to remember that the only reason “ln” makes sense for real numbers is that the exponential function is one-to-one. If you have two real numbers a and b, e^a = e^b if and only if a = b.

    In the complex numbers, you no longer have this nice property. Instead, e^z = e^w if and only if Re(z) = Re(w) and Im(z) – Im(w) = 2k\pi for some integer k. Since you can find many different z and w with e^z = e^w, you cannot(!) define a logarithm function! Therefore, it makes no sense to take ln of both sides of your second equation.

    To solve e^[y\pi] = -1 for y, instead you need to let y be a complex number u + iv. Then write -1 = e^[y\pi] = e^[u\pi]e^[iv\pi]. Conclude that u = 0, so that -1 = e^[iv\pi]. What can v possibly be?

  10. patty says:

    Ok that makes sense, thanks for taking your time, my math knowledge is at a low level but I am still intellectually curious. I have one more weird thing that I am confused on
    535.49165^i=1 ; I understand that 2k\pi for some integer k implies here so “i” is not 0, but it acts like it. My real question is that lets say integer k was a really high negative # (like -infinity), would we then assume that 0^i=1?

  11. Neal says:

    Your computation is correct, although it misses the detail that the square root (you raised both sides to the 1/2 power) is, like the logarithm, not well-defined, so you implicitly made a choice about which square root to use. But your choice was consistent, so there was no problem.

    Anyway, yes, e^[i2\pi] = 1. You may therefore conclude that (e^[2\pi])^i = 1.

    What’s really going on behind the scenes here is not that i is (or acts like) 0, it’s that 2\pi acts like 0. Remember from the post that e^[it] = cos(t) + isin(t). What is e^[i2\pi]? e^[i4\pi]? e^[i*-500\pi]?

    This answers your question! As angles, 2k\pi = 0 regardless of what k is — it can be as high negative as we want. Geometrically, it doesn’t matter how many times you’ve walked around a circle, you’ll always end up back where you started. Thus, for any high negative number, (small)^i = 1.

    [I may post later with more details … this is actually pretty interesting 🙂 ]

  12. patty says:

    e^(-inf*pi*i)=indeterminate [we don’t know where in the circle it stops or ever stops], but only real (non-“i”) values are 1 and -1
    therefore, lets just assume that:
    e^(-inf*pi*i)= positive or negative 1
    0^i= positive or negative 1

    [this is so interesting that I am procrastinating my work and not even feeling guilty 🙂 ]

  13. Neal says:

    Okay, let’s back up. It’s not clear what “0^i” means. In fact, to define complex exponents, you have to use the exponential function and the logarithm: z^w = e^[wlog(z)]. What does “log(z)” mean? It means “pick one of the preimages of z under the exponential map.”

    Fact: the exponential map has an “essential singularity” at infinity. This means that as you approach infinity from any direction, you can get close to any value you want.

    What does this have to do with 0^i? If you try to write down what the symbols mean, 0^i = e^[ilog(0)].

    Problem: log(0) doesn’t make sense because the exponential function is always nonzero. Nothing maps to zero. So instead of log(0), you have to take log(some sequence that tends to zero). Depending on your choice of sequence and choice of branch of log, when you write down the limit, you can get 0^i = anything you want.

    For example, if you take the sequence it, t\to 0, since log(it) = t + ik\pi/2 (where k is some integer), (it)^i = e^[ilog(it)] = e^[i(t + ik\pi/2)] = e^[it]e^[-k\pi/2] limits to e^[-k\pi/2], not zero, as t goes to 0. So here we see that 0^i = e^[-k\pi/2].

  14. patty says:

    “0^i = e^[-k\pi/2]”
    so 0^i can equal 1??? that is weird

  15. Neal says:

    Yep! That’s why it doesn’t make any sense to talk about “0^i” 🙂 It’s like “0/0” — it can be anything you want!

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  17. John Gabriel says:

    A complex number is ill-defined. To say that i^2=-1 implies that sqrt(-1) is a real magnitude (not modulus), but the process of finding “square roots” applies only to real positive magnitudes. In reality, all complex numbers are in fact equivalent to two-dimensional vectors, that is, a + bi = (a,b). Multiplication and division are non-standard. Complex numbers cannot be ordered because true division is not possible using complex numbers. This is not because of the *Well Ordering Principle”, rather the Well Ordering Principle is an indirect result of the division process. Think about it as follows: division (a form of measure) arose from the concept of ratio and ratio is the comparison of two magnitudes or well defined numbers. The key word here is *comparison*, not well-ordering principle.

    Complex theory is not required to prove any mathematical result since the trigonometric properties of sine and cosine are in fact the real reasons certain theorems are true. In some cases it is easier to use complex theory for such proofs but the same results can be proved without complex theory.

    Euler’s identity e^(i * pi) = -1 (which is easy to show if one assumes i^2=-1) is nonsense. To wit, if one assumes this identity is true, then ln -1 = i * pi (where ln is the natural logarithm) which is obviously false. For starters, the exponential function is *always* positive given a well-defined number. Providing i*pi as input to the function is garbage in and what results is garbage out, that is, e^(i * pi) = -1.

    Note that e^a=b ln b=a

    Complex theory is the study of non-number (*) or partial-number concepts with properties used in the study of number theory and algebra.

    Although sqrt(-1) is not a number, its role in complex theory is to help formulate properties or prove results in number theory.

    Students are able to understand such an explanation rather than many hand-waving arguments that are unsound.

    (*) It is just as easy to say that a polynomial of degree n has k real roots as opposed to saying a polynomial has n complex roots where a complex root is in fact not a number.

    Group theorists fancy thinking of a group binary operation as a “squaring” operation if the operation is applied to the same element (**). However, this is errant thinking since binary operations are not the same, even though two groups may have the same fundamental properties.

    Complex “numbers” are not only *not* numbers, they are *not* magnitudes also.

    Some fallacious arguments involving complex numbers:

    1. -1=(sqrt(-1))^2
    2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
    3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
    4. sqrt((-1)*(-1))=1

    Academics will claim that step 3 is incorrect because sqrt(a)*sqrt(b)=sqrt(a*b) iff a and b are both greater than 0. Idiot academics (includes most mathematics professors) do not realize that when they claim sqrt(-4) = 2i, they have just contradicted this fact, that is,

    sqrt(a)*sqrt(b) implies sqrt(a*b) where a, b > 0
    sqrt(a*b) implies sqrt(a)*sqrt(b) where a, b > 0

    Since sqrt(-4) = sqrt(4 x -1) is not equal to sqrt(4)*sqrt(-1) !!!

    I claim that step 1 is also incorrect because the binary operation of squaring is valid only when performed on well-defined elements, that is, numbers.

    In the set of complex numbers, the binary operation of *multiplication* has a dual nature: with real numbers a permutation or multiplicative principle is immediately evident, whereas with complex numbers this may or may not be the case. This chaotic organization is yet another reason why complex number theory is based on ill-defined concepts. In group theory, a cyclic group of order 4 can be represented using complex objects 1, i, -1, -i. A C4 group with complex objects is fundamentally similar to a C4 rotation group (0, tau/4, tau/2, tau) where the group operation is anticlockwise rotation by tau/4 and tau=2(pi) .

    Is it human nature or just that modern academics are too stupid to realize these facts? How does one tell a student that the concept sqrt(-1) is a magnitude when there is no way of measuring its dimensions, and yet has learned throughout his studies (till that stage) that magnitudes can be measured – indeed, a magnitude becomes a number once it’s measurable. Just as well, inferential suspension of knowledge is a core part of the human learning process, for otherwise it would be impossible for any student to move beyond this ridiculously ill-defined concept!

    Student: What does sqrt(-1) mean?
    Teacher: Hmm. It’s a number and these are the operations that are valid…

    No wonder it’s called *complex* theory – mathematicians are themselves confused. A better name would be Pseudo-number theory or the use of ill-defined concepts in arriving at some useful theory regarding well-defined numbers.

    (**) Claiming that i^2=-1 is equivalent to saying (stone)^2 = stick or (any object)^2= caterpillar because i is ill-defined. Indeed, i^2=-1 is exactly the same as claiming that [(0,1)]^2=-1 which is absurd.

    An example of how an educator should explain complex theory:

    Suppose that a quintic polynomial has only one pair of non-solutions (DO NOT call these roots because they are NOT numbers!), that is, two non-solutions containing i, then we can assume that three real roots exist.

    The previous paragraph still fails the requirements for being well-defined, but it is more accurate than most of the rubbish you have heard muttered by your professor of mathematics.

    Even better, an educator might say, a quintic polynomial has n real roots rather than arrive at conclusions about its solution based on non-attributes.

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  19. Ron says:

    “1. -1=(sqrt(-1))^2
    2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
    3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
    4. sqrt((-1)*(-1))=1

    Academics will claim that step 3 is incorrect because sqrt(a)*sqrt(b)=sqrt(a*b) iff a and b are both greater than 0.”

    Actually John…. The flaw is in step 4 where it is conveniently forgotten by you that the sqrt(1) (from step 3: sqrt((-1)*(-1))) also equals -1. I read through your websites for your “new calculus,” and I am just not impressed with people who claim every professor is an idiot and every mathematician in history is an idiot, to include people like Newton, Leibniz and Cauchy. following that up with ridiculous conspiracy theories is just icing on the cake.

  20. James Hall says:

    I think you are mistaken and Gabriel is correct. You would do well to check your logic. The flaw is really in Step 3 and not step 4 as you incorrectly assumed.

    To be able to claim that sqrt(a)*sqrt(b)=sqrt((a)*(b)), it must be known beforehand that both a and b are real numbers.

    Proposition 4 would be correct if nothing preceded it because (-1)*(-1) is a positive value and hence a square root is possible.

  21. Diane Porter says:

    If e to the iπ equals -1, then any positive number to any imaginary number must be equal to a real, negative number.

    So what is 4 to the i?

  22. The Wonderer says:

    Why not also have a negative number to the power i is a real number?

  23. Freya says:

    @Diane Porter:

    The example was given earlier in the article.

    4^i = e^ln(4^i)

    = e^i ln(4)

    = cos(ln(4)) + i sin(ln(4))

    = 0.183456975 + 0.98302774 i

    Pi is a special case. Since cos π = -1 and sin π = 0:
    e^iπ = cos π + i sin π
    = -1 + i(0)
    = -1.

    Another special is e^(iπ/2) = cos π/2 + i sin π/2 = 0 + i(1) = i. Yet another is e^(3iπ/2) = -i (try this yourself to understand why).

  24. Freya says:

    e^ix has no imaginary part only when sin(x) = 0, which doesn’t leave many possibilities.

  25. Xerenarcy says:

    @James Hall
    nothing wrong with this without any assumptions about a and b:
    sqrt(a) * sqrt(b) = sqrt(a * b)

    whether you know some property about a or b is irrelevant. sqrt is exponentiation by a half, and the usual exponentiation rules (per the post) still apply:

    (a^0.5) * (b^0.5) = (a * b)^0.5

    or are you saying that exponentiation does not distribute amongst the multiplicands?

  26. Xerenarcy says:


    1. -1=(sqrt(-1))^2
    2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
    3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
    4. sqrt((-1)*(-1))=1

    sqrt(a) = +b or -b
    such that (+b)^2 = a and (-b)^2 = a
    since the result of sqrt is squared, and both values behave identically, this implies #1 is correct.

    since #1 is necessarily true numerically and algebraically, and since sqrt is dual-valued…
    [from #1] -1 = sqrt(-1) * sqrt(-1)

    stepping aside briefly, sqrt(-1), will have two solutions alternating sign. before you say this is illogical, consider that we have negative and positive infinity and that these are distinct, despite an infinity in your result being technically incorrect (the correct term is an undefined value).

    since sqrt(-1) = some +b or -b, we have 4 possible solutions to #2

    -1 = (-b) * (-b)
    -1 = (-b) * (b)
    -1 = (b) * (-b)
    -1 = (b) * (b)

    which reduces to two solutions by equivalence:
    -1 = (-b) * (-b) = (b) * (b) = b^2
    -1 = (-b) * (b) = (b) * (-b) = -(b^2)

    so at best to your argument’s credit this is half wrong and half right. no one wins, but a solution does exist irrespective of what b is, so moving on…

    once more, the fault here is assuming sqrt yields only one valid solution.


    using sqrt’s dual-valued nature, it is more correct to say:
    sqrt((-1)*(-1)) = +(sqrt(-1)*sqrt(-1)) or -(sqrt(-1)*sqrt(-1))

    sqrt(1) = +(-1) or -(-1) = 1 or -1

    no fancy tricks here, just high school math.

    but to appease all… ill address it directly…

    sqrt(-1)*sqrt(-1) = sqrt((-1)*(-1)) = sqrt(1) = 1

    sqrt(1^2) = 1 or -1
    1^2 = 1
    sqrt(1) = 1 or -1

    sqrt(-1)*sqrt(-1) = sqrt(1) = 1 or -1

    i can soundly say that this does not disprove i or roots of negative numbers. what it does demonstrate is that using only part of the available solutions creates logical paradoxes.

  27. MJ Young says:

    Imaginary exponents only rotate a value, here pi, in an “orthogonal” or 90 degree direction from the number line, they do not increase its value; opposed to regular exponents which move a number left or right along the number line.

  28. Tai Mao says:

    I would love. To look in the future and see children learning about the number i and how it was deemed an imaginary number. Thinking about the number i make sense to me just like negative numbers. However, at one point the idea of a number less than zero was quite absurd. I call i a number, and I don’t view it as a “imaginary” number.

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