Proving $e^{i\pi} = -1$ without proving $e^{ix} = \cos x + i\sin x$

Question

Just as the title asks: I'm looking for a proof of Euler's identity without first proving the general case of $e^{ix} = \cos x + i\sin x$.

Answer 1

I outlined in a previous question a proof that $$\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n = \cos x + i\sin x$$

That outline (about halfway through the answer) can be altered just to prove it for $x=\pi$ pretty directly. Basically:

$$\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}=1+g(n)$$

where $ng(n)\to 0$, so we can prove that:$$\left(\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}\right)^n\to 1$$

You can prove that $ng(n)\to 0$ geometrically. You essentially only need to prove that:

$$\left|\,\cos(\pi/n) - 1\right| = O(1/n^2)$$ and $$\left|\,\sin(\pi/n)-\pi/n\right| = O(1/n^2)$$

The first is easy to prove. The distance from $(\cos x,\sin x)$ to $(1,0)$ is $\sqrt{2-2\cos x}$ and $x$ is the length of the arc of the circle between those two points. So $$0\leq 1-\cos\pi/n\leq \frac{\pi^2}{2}\frac{1}{n^2}$$

The second requires more care. We can easily see $0\leq \sin\pi/n \leq \pi/n$ since $\sin \pi/n$ is the distance of $(\cos\pi/n,\sin\pi/n)$ to the real line, and $\pi/n$ is the length of a longer path to the real line.

Now, take the points $P=(1,0)$ and $Q=(\cos 2\pi/n,\sin 2\pi/n)$. Draw the tangents to the unit circles at these points and find their intersection, $R$. Then show that the length of the path $PRQ$ is $2\tan \pi/n$. This is a path "outside" the circle, so it must be greater than the path of length $2\pi/n$ along the circle between $P$ and $Q$, so we get inequality: $$\tan\pi/n \geq \pi/n$$ This gives us the inequality:

$$\frac{\pi}{n}\cos \pi/n \leq \sin\pi/n \leq\frac{\pi}n$$

Which means that $$|\pi/n-\sin\pi/n|\leq \frac{\pi}{n}(1-\cos\pi/n) = O(1/n^3)$$

Answer 2

In this answer, it is shown that $$ e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n $$ is a point whose absolute value is $1$ and whose angle is $x$. No mention is made of sines and cosines until the end when converting from polar coordinates. Since $-1$ has absolute value $1$ and angle $\pi$, this would indicate that $$ e^{i\pi}=-1 $$ without first computing $e^{ix}=\cos(x)+i\sin(x)$.

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Some have complained that the proof above, since it proves that $e^{ix}$ is the point with absolute value $1$ and angle $x$, actually proves $e^{ix}=\cos(x)+i\sin(x)$. Howver, since I've never seen a proof of $e^{i\pi}=-1$ that doesn't first, in essence, prove that $e^{ix}=\cos(x)+i\sin(x)$, this all seems like an exercise in obfuscation anyway. Here is a more carefully hidden version of the same argument.

Using this formulation of $e^{i\pi}$: $$ e^{i\pi}=\lim_{n\to\infty}\left(1+\frac{i\pi}{n}\right)^n\tag{2} $$ For a complex number $z$, let $|z|$ be its magnitude and $\arg(z)$ be its angle. If it is not already known, only a small amount of algebra and trigonometry is needed to show that $$ \begin{align} |wz|&=|w|\cdot|z|\tag{3a}\\ \arg(wz)&=\arg(w)+\arg(z)\tag{3b} \end{align} $$ Induction then shows that \begin{align} |z^n|&=|z|^n\tag{4a}\\ \arg(z^n)&=n\arg(z)\tag{4b} \end{align} Let us take a closer look at $1+\dfrac{i\pi}{n}$. $$ \begin{align} \left|\,1+\frac{i\pi}{n}\,\right|&=\sqrt{1+\frac{\pi^2}{n^2}}\tag{5a}\\ \tan\left(\arg\left(1+\frac{i\pi}{n}\right)\right)&=\frac{\pi}{n}\tag{5b} \end{align} $$ Using $(4a)$, $(5a)$, and $(2)$, we get $$ \begin{align} |e^{i\pi}| &=\lim_{n\to\infty}\left|\,1+\frac{i\pi}{n}\,\right|^n\\ &=\lim_{n\to\infty}\left(1+\frac{\pi^2}{n^2}\right)^{n/2}\\ &=\lim_{n\to\infty}\left(1+\frac{\pi^2}{n^2}\right)^{\frac{n^2}{2n}}\\ &=\lim_{n\to\infty}e^{\frac{\pi^2}{2n}}\\[12pt] &=1\tag{6} \end{align} $$ It can be shown that when $x$ is measured in radians $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{7} $$ Using $(4b)$, $(5b)$, and $(7)$, we get $$ \begin{align} \arg(e^{i\pi}) &=\lim_{n\to\infty}n\arg\left(1+\frac{i\pi}{n}\right)\\ &=\lim_{n\to\infty}n\arg\left(1+\frac{i\pi}{n}\right) \frac{\tan\left(\arg\left(1+\frac{i\pi}{n}\right)\right)}{\arg\left(1+\frac{i\pi} {n}\right)}\\ &=\lim_{n\to\infty}n\frac{\pi}{n}\\ &=\pi\tag{8} \end{align} $$ Using $(6)$ and $(8)$, we get that $e^{i\pi}$ has magnitude $1$ and angle $\pi$. That is, $$ e^{i\pi}=-1 $$

Answer 3

Here’s another argument that has, I think, some gaps. Consider the map $p\colon\mathbb R\to\mathbb C$, by $p_t=e^{it}=\lim_n(1+it)^n$. It’s additive, in the sense that $p_{s+t}=p_sp_t$, and it maps into the unit circle, since $\overline{p_t}=p_{-t}=1/p_t$, so that $p_t\overline{p_t}=1$. So we have a map from the real numbers to the circle group. What’s the speed (not velocity) of the moving point $p_t$? Clearly (?) uniform, because of the additivity, and you can calculate the derivative at $t=0$ as $$ \lim_{h\to0}\lim_{n\to\infty}\frac{(1+ih)^n-1}{h}\,, $$ clearly (?) equal to $i$, so that the speed is $|i|=1$. (Of course you can remove this question-mark just by appealing to to what we know about the derivative of the exponential function.) Moving around the circle at uniform unit speed, has to get halfway around in time $\pi$.

Answer 4

We shall use that $\exp(z_1+z_2)=\exp(z_1)\cdot\exp(z_2)$ for arbitrary $z_1$, $z_2\in{\mathbb C}$, and that $\bigl|e^{i\tau}\bigr|=1$ for real $t$. The labels $\sin$ and $\cos$ only appear as abbreviations for certain expressions, and no knowledge about these functions is implied.

Note that the series $$\sin t:={\rm Im}(e^{it})=t-{t^3\over 6}+\ldots, \quad \cos t:={\rm Re}(e^{it})=1-{t^2\over 2}+\ldots\tag{1}$$ are alternating for $|t|<1$. It follows that $$\sin 1>{5\over6}>{1\over\sqrt{2}}\ .$$ As $\sin 0=0$ and $t\mapsto e^{it}$ is continuous there is a $\tau\in\ ]0,1[\ $ with $$\sin\tau ={1\over\sqrt{2}}\ .$$ Since $\tau<1$ it follows from $(1)$ that $\cos\tau>0$, and $\bigl|e^{i\tau}\bigr|=1$ implies $\cos\tau=\sqrt{1-\sin^2\tau}={1\over\sqrt{2}}$.

Therefore $e^{i\tau}={1\over\sqrt{2}}(1+i)$, and putting $\pi:=4\tau$ we get $$e^{i\pi}=\left(e^{i\tau}\right)^4={1\over 4}(1+i)^4=-1\ .$$

Answer 5

Just take the log of both sides:

$$\ln e^{i\pi}=\ln(-1)\Rightarrow i\pi\ln e=i\pi\Rightarrow i\pi=i\pi$$