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I was thinking today whether this can be a backward proof for $i^2= -1$ or not, where i is the imaginary unit!

we know:
$$e^{ix}= \cos x + i \sin x; \hspace{1cm} (Euler's formula)$$ where $i$ is the imaginary unit and the argument $x$ given in radians.

so,
$$e^{i \pi}= \cos \pi + i \sin \pi= -1$$

$$e^{i \pi/2}= \cos \pi/2 + i \sin \pi/2= i$$

Now, we can write
$$i=e^{i \pi/2}= e^{i (\pi - \pi/2)} = e^{i \pi}/e^{i \pi/2}=-1/i $$ ;
$$i=-1/i $$

SO,
$$i^2=-1 $$

Thanks

S. Amani
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    What is your definition for that i which you mean to prove that i^2 = -1? – dxiv Oct 13 '16 at 04:45
  • I said that i is the imaginary unit. – S. Amani Oct 13 '16 at 04:49
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    What is your definition for the imaginary unit, if not $i^2 = -1$? – dxiv Oct 13 '16 at 04:50
  • The imaginary number i is defined solely by the property that its square is −1 [Wikipedia] – S. Amani Oct 13 '16 at 04:55
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    OK, so the imaginary unit i is defined by i^2=-1. Then why do you try (or need) to prove that i^2=-1? You just said that that's its definition. – dxiv Oct 13 '16 at 04:58
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    You can not assume any formula involving i without first defining what i is. I highly doubt that any texts prove anything about i before defining it. If you have a very specific question about a very specific text, then please edit your question and fill-in all the details. Otherwise, you are just running circles around the definition of i. – dxiv Oct 13 '16 at 05:12
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    Do you honestly not see that your argument is completely circular? I have never seen a text where $i^2=-1$ was introduced after eulers formula. But that doesn't matter. Euler's formula is gibberish unless, e, sin, cos, i, and exponentiation have all been defined first. And then we have to prove eulers formula. All that requires we know $i^2=-1$ by definition first. We can't prove something if what we use to prove it is already based on assuming it is true. Surely you see that, don't you? – fleablood Oct 13 '16 at 06:13
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    This is posible but you have to decide what the definition a complex number is to begin with as well as what the "unit" is. And then I think Euler's formula must be given as an axiomatic definition. I honestly think this will get you nowhere. – fleablood Oct 13 '16 at 06:24

3 Answers3

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In your proof, you used both the property that $e^a/e^b=e^{a-b}$ and $e^{ix}=\cos x + i\sin x.$ However the proofs of both of those properties require you to use $i^2 = -1.$ So it is circular! However, if you were given those two properties, then the proof would be valid. But it is much more likely that it is already given that $i^2 = -1$ as a definition of $i$ !

Teoc
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Your argument is interesting, but I think at best it is incomplete. If you don't start with $i^2 = -1$, then you are claiming:

If there exists a number $i$ such that $e^{ix}= \cos x + i \sin x $, then this number satisfies $i^2 = -1$.

You still need to prove the existence of such a number.

Ovi
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  • @TheGreatDuck The OP didn't prove that B is true (or anything else for that matter), because he simply didn't define what A was. – dxiv Oct 13 '16 at 04:47
  • @TheGreatDuck I don't see the If there exists a number i such that... anywhere in the question. Rather, it is just a straight assertion we know... which implies that i would have been defined elsewhere prior, but that definition is never spelled out. – dxiv Oct 13 '16 at 04:54
  • @TheGreatDuck Sorry, not sure what your point is, and I'll leave it at that. The OP got confused while trying to (circularly) prove a definition. Ovi pointed out that circularity. – dxiv Oct 13 '16 at 05:34
  • First you have to define what $e^z $ means for a complex number. Actually even before that you have to define what a complex number is. I think this is doomed. – fleablood Oct 13 '16 at 06:18
  • You can't define anything on the complex numbers unless you define what a complex number is. And you can't do that without multiplication. You can't define e^z on complex numbers unless you know how to take derivatives and you can't have the essential du/dx=-idv/dy if we don't already have i^2=-1. And without that we can not define e^z. – fleablood Oct 13 '16 at 07:26
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First you have two define what a complex number is. We may define it as a set of values with a two digit component i.e. as a set of {(a,b)} but if we don't somehow include as part of the definition that (a,b)x (c,d)=(ac-db,ad+bd) we can not conclude anything as other definitions and conditions yield equal legitimate systems (for example $\mathbb R^2$-- something must distinguish that $\mathbb C $ is somehow something different than $\mathbb R^2$ and that difference can only be one of definition.)

Now the only reason we can claim Euler's Formula, is because of this definition. We don't have any intrinsic idea what $e^{z=(a,b)} $ could mean. But we do know that if we want $e^z $ to have any of the properties it does in the reals we must have $\frac {d e^{z}}{dz}=e^z $.

And from there we are stuck if we don't have a definition for what multiplication of complex numbers is in the first place.

If we have it by definition that (a,b)(c,d)=(ac-bd,ad+bc) then we know $i=(0,1)^2=(-1,0) =-1$ so there is nothing to prove.

From there we can determine if $f (z)= u (z)+iv (z) $ and $f '(z) $ exists as a limit in both the real and imaginary unit that $du/dx=dv/dy$ and $du/dy=-idv/dx $ and that if $d e^z /dz = e^z$ that only be possible if $e^{(x,y)}=e^x (\sin y + i \cos y) $. And that's the only reason Euler's function works-- because we first defined $i^2 =-1$

Now, we could do things backwards. We could define $e^{(x,y)}=(e^x,0) ((1,0)\sin y +(0,1)\cos y $ and from that as an axiom, prove $(0,1)^2=(-1,0) $ as you did but that is arbitrary and backwards.

fleablood
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