Q: Why is e to the i pi equal to -1?

Physicist: This equation (e^{i \pi} = -1) was recently voted one of the most famous equations ever.  That isn’t part of the answer, it’s just interesting.

First, you’ll find (by plugging them into a graphing calculator and graphing) that:

1) Sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots

2) Cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots

3) e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots

Where N! = 1 \cdot 2 \cdot 3 \cdots N.

These are called “Taylor expansions” of “Sine”, “Cosine”, and “e to the x”.  If you were to continue the patterns above forever, then you would find that the equality is exact.  There is some very exciting math to back me up on this, but for now just trust.

Know that i^2 = - 1.  This is how you define i.  Therefore, i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i , \ldots

Check it!:

e^{i x} = 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots = 1 + i x + \frac{i^2 x^2}{2!} + \frac{i^3 x^3}{3!} + \frac{i^4 x^4}{4!} + \frac{i^5 x^5}{5!} + \frac{i^6 x^6}{6!} + \frac{i^7 x^7}{7!} + \cdots = 1 + i x - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + i \frac{x^5}{5!} - \frac{x^6}{6!} - i \frac{x^7}{7!} + \cdots

and grouping terms that have “i” in them:

= ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots) + i ( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots )

= Cos(x) + i Sin(x)

Holy crap! e^{ix} = Cos(x) + i Sin(x)!

This is the “Euler Equation”.  Or one of them at least.  Just plug in “x = \pi“.

e^{i \pi} = Cos(\pi) + i Sin(\pi) = (-1) + i(0) = -1.

So the trick is Euler’s equation, which is (surprisingly) true.

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27 Responses to Q: Why is e to the i pi equal to -1?

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  4. patty says:

    whats wrong with this proof:
    ipi+ipi=0 (this is false)

  5. That one guy says:

    @patty. Where did (e to the ipi) squared come from?

  6. patty says:

    look at line 3 to line 4, since (-1) equals (e^ipi), i can rewrite as such, does that help?

  7. Bryn says:

    Logarithms are either multi-valued or use the ‘principle value’ of the argument which is restricted to -π<ϕ≤π. By this I mean that you have Ln(re^(iθ))=ln(r)+i(θ+2πn) with an arbitrary integer n, or ln(re^(iθ))=ln(r)+iθ where θ satisfies -π<θ≤π.

    Because of that, the last line must be either i(π+π+2πn)=i(0+2πm), where n and m are integers, or 0=0. These are both true (for the appropriate values of m and n).

  8. patty says:

    that makes sense, but what if m and n were really high negative #’s( like negative infinity) that we are left with 0^i=1, could this this true?

  9. tau says:

    i^23 = -i, then I can redefine it as (i^4)*^23/4 , so that its value doesn’t change. But since i^4 = 1, then (i^4)*^23/4 will become 1^23/4 which is 1. thus, as we see, we got two answers. How is it possible?

  10. The Physicist The Physicist says:

    Whenever you see a “1/N” in an exponent, you’re doing an Nth root, which drops N different roots. In this case the solutions you get are 1^23/4 = {1, i, -1, -i}.
    The difference comes about because while taking a power produces one answer, but taking a root produces several.

  11. kraguljacm says:

    The exponential function over the complex field has period 2pi.

    So e^ipi*e^ipi = e^(2pi)i = e^0 = 1

    More generally e^(2npi)i = 1 for n = 1, 2, 3, ….

  12. Bob says:

    I had always questioned the equation but never actually knew it had a good proof!Thank you so much for the post!

  13. David says:

    Can a mathematician help a brother out?
    How to plot this function?
    ∫eˣ = f (u ⁿ)

  14. Freya says:

    Patty, your proof really vexed me, too. What is wrong with it is step 3, where you changed 1 into e^0. This only occasionally results in a valid solution.

    To give you a more general example, suppose you had an equation x^2 = 1. Obviously, the solution is ±1, but would it be valid to say that x^2 = x^0 and then 2 = 0? Of course not!

  15. Michael says:

    Why is e^iTau=1 so much prettier than e^iPi=-1? 🙂

  16. fred says:

    Bah, this is clear as mud. Proofs don’t ‘explain’ anything, they prove it assuming a lot of other assumptions about math along the way.

    Where is the simple of how ‘i’ in the exponent creates a negative number?

  17. EMonk says:

    @fred said “Where is the simple of how ‘i’ in the exponent creates a negative number?”

    It doesn’t. The negative number – and incidentally the elimination of i from the result – are entirely the fault of `pi`.

    The `i` in `e ^ (i x)` allows for the development to `e ^ (i x) = Cos(x) + i Sin(x)` as shown in the article above. The reason why this works is the repeating sequence of (non-negative integer) powers of `i`: `1, i, -1, -i, …` which has has two properties of interest: It alternates 1 complex value and 1 non-complex value, and it alternates 2 positive values and 2 negative values.

    If you work through the steps above, it is those alternations that allow the development of `e ^ (i x) = Cos(x) + i Sin(x)`.

    And this is where the negative result comes from: `Cos(x) = -1`

    Or in short: `i` enables the development to `Cos(pi) + i Sin(pi)`, `pi` gives the resultant negative.

  18. MJ Young says:

    Awesome EMonk! Now, why the hell coulndn’t they have added that little extra bit you did to fully clarify!

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  20. OMatthews says:

    Hello. Looking at the Physicist’s proof for e^ix = cos x +i sin x: Could I please ask why it is valid to substitute an imaginary for a real exponent? (i.e. when he goes from e^x to e^ix). I have only yet seen proof that the series used for e^x is valid for real exponents. Thank you

  21. Ruvian says:


    Hello. You can do this because i is just a constant like the number 2. You can substitute by any number you want, being careful not to make any mistake.

  22. Ken says:

    @patty: It’s already been said many ways, but here’s how I’d say it simply:

    In the last step of your “proof” you say that because x^y = x^z, then y = z. This is obviously not true. E.g., i^2 = i^6 but 2 != 6.

  23. Michael says:

    The problem in the proof quoted by Patty is in the definitions. It look right but i is a complex number not a real number. Hence one must first prove those rules for complex numbers and as mentioned earlier in the comments would then discover that they can not be used as quoted.

  24. Garbodor says:

    Logarithms are equal to the imaginary value as the above value squared. Whence albeit imaginary logarithm, factorial. Hope that helps.

  25. Carl Culator says:

    what am i doing here

  26. L says:

    whats wrong with this proof:
    ipi+ipi=0 (this is false)

    I think this:
    (1)^2 = 1, (-1)^2 = 1 then I say 1=-1????

  27. The Physicist The Physicist says:

    In both cases the inverse operation (the log in the first case and the square root in the second) produces multiple results. The problem shows up in the last step where you go from “e^ipi+ipi=e^0” to “ipi+ipi=0”
    (-1)^2=(1)^2, but that doesn’t mean that -1=1. Similarly, e^(2ipi)=e^0 (and both of these are equal to 1), but that doesn’t mean that 2ipi=0.

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