Intuition for $e^{ix}$ lies on the unit circle.

Question

I do know that for any $x$, the complex number $e^{ix}$ have to lie on the unit circle because if we plot the points for $\cos x + i\sin x$ for each $x$, we will eventually form a circle.

Are there any more intuitive approach on why $e^{ix}$ lies on the unit circle?

Answer 1

Here's a perspective I like: consider the function $f:\Bbb R \to \Bbb C$ given by $f(t) = e^{it}$, and note that we can think of $\Bbb C$ as a two dimensional space (that is, $\Bbb C$ is isometric $\Bbb R^2$). Now, we note that if we apply the usual rules of calculus, we have $$ \frac {df}{dt} = i\,e^{it} $$ Now, let's split this function into its real and imaginary parts. That is, say that $$ f(t) = f_R(t) + i f_I(t) $$ where $f_R(t),f_I(t) \in \Bbb R$. Equivalently, we can think of $f(t) = (f_R(t),f_I(t))$ as tracing out a trajectory in $\Bbb R^2$. Along these lines, we have $$ f'(t) = if(t) = i[f_R(t) + i f_I(t)] = -f_I(t) + if_R(t) $$ Equivalently, we see that when we follow the trajectory of $f(t)$, its tangent vector points in the direction of $(-f_I(t),f_R(t))$.

Note: we are not assuming Euler's formula, which would tell us that $f_R(t) = \cos(t)$ and $f_I(t) = \sin(t)$.

Now, we use a handy fact:

For $\alpha: \Bbb R \to \Bbb R^2$, we have $$ \frac{d}{dt}\|\alpha(t)\|^2 = \frac{d}{dt} \langle \alpha(t),\alpha(t) \rangle = 2 \langle \alpha'(t), \alpha(t) \rangle $$ where $\langle \cdot,\cdot \rangle$ denotes the usual "dot product".

With that, we may now conclude that $$ \frac {d}{dt} |f(t)|^2 = \langle (f_R(t),f_I(t)), (-f_I(t),if_R(t)) \rangle = 0 $$ That is: because $f'(t)$ points in a direction orthogonal to the position $f(t)$ (relative to the origin), the magnitude of $f(t)$ (i.e. its distance from the origin) remains constant.

Now, since $|f(0)| = |e^0| = 1$, we can conclude that for any $t$, $|f(t)| = 1$. That is, $f(t)$ traces a trajectory on the unit circle. That is, for any $t$, $e^{it}$ is on the unit circle, which is what we wanted to show.

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It is notable that from here, we may see that $|f'(t)| = \|\alpha'(t)\| = 1$, which tells us that the unit circle is traced one length unit per unit length in $t$, which is enough for us to deduce Euler's formula in its full form.

Answer 2

I don't think there is any intuition - it just is.

Once $e^{ix} = \cos(x)+i\sin(x)$ has been established it is obvious, but until then, no.

In my opinion, any of the answers here that attempt to produce an intuitive reason require more advanced reasoning than deriving Euler's (or d'Alembert's) formula.

Answer 3

It depends upon what you mean by the symbol $e^{ix}$ for all real $x$. One way to introduce this symbol is to use the definition $$e^{z} = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}, z\in\mathbb{C}\tag{1}$$ and in particular if $z = ix, x \in\mathbb{R}$ then $$f(x) = e^{ix} = \lim_{n \to \infty}\left(1 + \frac{ix}{n}\right)^{n}\tag{2}$$ We can now prove easily that $|f(x)| = 1$ for all $x \in \mathbb{R}$. It is in fact better to show equivalently that $|f(x)|^{2} = 1$ for all $x \in \mathbb{R}$. This we do next.

Clearly $$|f(x)|^{2} = \lim_{n \to \infty}\left|1 + \frac{ix}{n}\right|^{2n} = \lim_{n \to \infty}\left(1 + \frac{x^{2}}{n^{2}}\right)^{n}\tag{3}$$ Now as $n \to \infty$ it is clear that after some value of $n$ we have $|x/n| < 1$ and hence by Bernoulli's inequality we have $$1 - n\cdot\frac{x^{4}}{n^{4}}\leq \left(1 - \frac{x^{4}}{n^{4}}\right)^{n} \leq 1$$ and hence by Squeeze theorem we have $$\lim_{n \to \infty}\left(1 - \frac{x^{4}}{n^{4}}\right)^{n} = 1\tag{4}$$ Similarly we can show that $$\lim_{n \to \infty}\left(1 - \frac{x^{2}}{n^{2}}\right)^{n} = 1\tag{5}$$ Dividing equation $(4)$ by equation $(5)$ we get $$\lim_{n \to \infty}\left(1 + \frac{x^{2}}{n^{2}}\right)^{n} = 1$$ and thus by equation $(3)$ we conclude that $|f(x)|^{2} = 1$ for all real $x$. It follows that the complex number $f(x) = e^{ix}$ lies on the unit circle $|z| = 1$ for all real $x$.

Note further that the argument of $f(x)$ can be handled even more easily. Clearly we can see that $$\operatorname{arg}f(x) = \lim_{n \to \infty}n\arctan(x/n) = x$$ and thus it is clear that $f(x) = e^{ix} = \cos x + i\sin x$. See this answer for another proof of $f(x) = \cos x + i\sin x$.

Answer 4

You can just calculate it directly, using the power laws and the fact that $\overline{\mathrm e^z}=\mathrm e^{\overline z}$: $$\left|\mathrm e^{\mathrm ix}\right|^2 = \mathrm e^{\mathrm ix}\,\overline{\mathrm e^{\mathrm ix}} = \mathrm e^{\mathrm ix}\mathrm e^{-\mathrm ix} = \mathrm e^{\mathrm ix-ix} = \mathrm e^0 = 1$$

On why $\overline{\mathrm e^z}=\mathrm e^{\overline z}$:

There are several definitions of $\mathrm e^x$, so the details differ, but the common theme is the same: $z$ is the only complex quantity in the definition.

1. The series definition: $$\mathrm e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$ Note that $k!$ is a real number for each $k$, $\overline{x+y}=\overline x+\overline y$ and $\overline{x\cdot y}=\overline x\cdot\overline y$. Since $z$ is the only complex quantity in the definition, the relation holds for the partial sums: $$\overline{\sum_{k=0}^n \frac{z^k}{k!}}=\sum_{k=0}^n \frac{\overline z^k}{k!}$$ And since $\overline z$ is continuous, the same holds for the limit.

2. The limit definition $$\mathrm e^z = \lim_{n\to\infty}\left(1+\frac{z}{n}\right)^n$$ Again, $z$ is the only quantity that is not real, and for each $n$, the expression under the limit is constructed only from sums and products (note that $\frac{z}{n}=\frac{1}{n}\cdot z$ where $\frac1n$ is real, and that $x^n=x\cdot x\cdots x$), therefore $$\overline{\left(1+\frac{z}{n}\right)^n}=\left(1+\frac{\overline z}{n}\right)^n.$$ And again, the same holds for the limit because $\overline z$ is continuous.