I was solving some exercises in complex analysis in preparation for a qualifying exam, and I came across a problem which asked me to prove that if $x, y \in \mathbb{R}$ then

$$ e^{ix} = e^{iy} \iff x - y = 2 \pi n \quad \text{for some} \ n \in \mathbb{Z} $$

At first I thought it was quite easy and what I thought is that it was kind of obvious if I used Euler's formula and equated real parts and imaginary parts, so that

$$ e^{ix} = e^{iy} \iff \cos{x} = \cos{y} \quad \text{and} \quad \sin{x} = \sin{y} $$

and thought of this in terms of the corresponding points in the unit circle. But that doesn't seem very rigorous to me. So I would like to ask the following.

How can this be proved using another definition for the trigonometric functions like the series definition or the definition as solutions of the differential equation $y'' = -y$ with some initial conditions? And what properties of the trigonometric functions have to be used to prove this?

Note I thought of doing the following but still can't complete the argument.

$$ e^{ix} = e^{iy} \iff e^{i(x-y)} = 1 \iff \cos{(x-y)} = 1 \quad \text{and} \quad \sin{(x-y)} = 0 $$

and now I would like to conclude that this happens if and only if $x-y = 2 \pi n$ but don't know how to justify this.

Any help would be very appreciated. Thank you.

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    $\begingroup$ The sine function is zero precisely at integer multiples of $\pi$, so that covers the sine part... $\endgroup$ – J. M. is a poor mathematician Sep 4 '11 at 16:49
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    $\begingroup$ Expanding on that: since both sine and cosine are $2\pi$-periodic, it suffices to consider their values within a restricted domain. Just note where the cosine becomes $1$ and the sine becomes $0$ in that restricted domain, and then... $\endgroup$ – J. M. is a poor mathematician Sep 4 '11 at 16:54
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    $\begingroup$ You don't even need to worry about the sine... $\cos(x-y) = 1$ precisely when $x-y = 2\pi n$. (And $\sin(x-y) = 0$ follows because $\cos^2 \theta + \sin^2 \theta = 1$.) $\endgroup$ – ShreevatsaR Sep 4 '11 at 16:57
  • $\begingroup$ @Shree: I think you've got what Adrián needs; maybe you can write an answer? $\endgroup$ – J. M. is a poor mathematician Sep 4 '11 at 16:59
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    $\begingroup$ @J. M. I just found a treatment of this in Lars Ahlfors' Complex Analysis book. In section 3 of chapter 2 (titled The exponential and trigonometric functions) he defines the exponential and in turn the sine and cosine using the exponential, and proves some of the usual formulas. Then he treats the periodicity question and proves that the exponential $e^{iz}$ is periodic, calls the period $2\pi$ and proves that it has the usual values at the points $0, \pi, 2\pi$ etc. This along with ShreevatsaR's answer completely solve my question. Thank you also for your suggestions. $\endgroup$ – Adrián Barquero Sep 4 '11 at 20:21

As you observed, $$ e^{ix} = e^{iy} \iff e^{i(x-y)} = 1 \iff \cos{(x-y)} = 1$$ (The condition $\sin{(x-y)} = 0$ is redundant because it follows from $\cos{(x-y)} = 1$ using $\cos^2\theta + \sin^2\theta = 1$.)

Now if $\cos\theta = 1$, then $\theta = 2n\pi$ pretty much by definition. One way of seeing this is that $\cos$ is periodic with period $2\pi$, and the only $\theta$ in $[0,2\pi)$ for which $\cos\theta = 1$ is $\theta=0$. Said differently, if $2\pi n$ is the multiple of $2\pi$ that is closest to $\theta$, then writing $\theta = 2\pi n + \phi$ where $-\pi < \phi \leq \pi$, we have $1 = \cos(\theta) = \cos(2\pi n + \phi) = \cos(\phi)$, so $\phi = 0$.

Alternatively, $\cos\theta$ is the projection on the x-axis of a point that has rotated by angle $\theta$ from its initial point $(1,0)$, so if $\cos\theta = 1$ (and therefore $\sin\theta = 0$), then the point has come back to $(1,0)$, so it must have made a number of full turns: if it made $n$ turns, then we say $\theta = 2\pi n$.


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