Why does the function $e^{ix}$ have a real part, without using the Euler's formula

Question

I would like to intuitively understand why $e^{ix}$ has a real part, if the the function $e^{ix}$ has an imaginary argument.

I know that
$$e^{ix}=\cos x + i\sin x$$
and I don't need convincing that it is so. I understand how $e^{ix}$ behaves when I rewrite it in the sine/cosine form and that this function can be visualized/illustrated by a circle in complex plane. I also understand how it is derived from Maclaurin series, but it doesn't shed any light on this issue - I see it as "mindless" proof. I struggle to intuitively grasp why this function can even have a real part, when the argument is purely imaginary.

Since this function is also periodic, I also don't understand why it even "falls down" at any point, when $e^x$ does not even have a single stationary point. I assume that it follows from the fact that $i^n$ is periodic, which I comfortably understand, but I can't see how multiplying imaginary number by a real number in an exponent has the same effect.

Side note for context: I am a soon-to-be third semester physics student, in which I am going to have a course on optics, which heavily relies on Euler's formula.

Answer 1

If you can accept that $\frac d{dz}(e^z) = e^z$, here's an intuition I like.

Think of the curve $t\mapsto e^{it}$ in the complex plane. By the chain rule we get $\frac d{dt}e^{it} = ie^{it}$.

Now, remember that multiplying by $i$ geometrically corresponds to rotation by $90^\circ$ (which you can check just by noting that $i\cdot 1 = i$ and $i\cdot i = -1$ and remembering that $\mathbb C$ is $\mathbb R^2$ as real vector space with basis $\{1,i\}$).

We conclude that velocity of our curve is always perpendicular to the position vector. But that means that the curve can't stay on the same line in the complex plane. In particular, the image of the curve can't be contained in the imaginary line $i\mathbb R$.

We can go a step further and even calculate $\frac {d^2}{dt^2} e^{it} = -e^{it}$, so acceleration is always pointed at the origin. If you remember classical physics, this should remind you of uniform circular motion. And that's precisely why the image of our curve is a circle in the complex plane (that's probably circular reasoning, but you asked for intuition anyway).

Answer 2

This is a Question about Intuition , without using Euler (cis) or other trigonometric calculations & formulas.

Here is my Intuitive way to look at it :

(1) Consider $A = e^{ai}$ (where $A$ is Imaginary) & $B = e^{bi}$
Assumption : Let us say that when Exponent is Imaginary , we expect "output" to be Imaginary , with no real Part.
We will see that this Assumption will not work out & will give Contradictions.

Let us multiply the Imaginary values $A$ & $B$ to get $C=e^{(a+b)i}$
According to our assumption , $C$ must also be Imaginary , having no real Part. BUT the Product of 2 Imaginary numbers ($A$ & $B$) must be real ! Contradiction with our Assumption !

(2) We get Same Conclusion (Same Contradiction) when we try with $A \cdot A = A^2 = e^{2ai}$ , which must be real.

(3) More-over , when we consider $D = \sqrt{A} = e^{ai/2}$ , we see that $D$ can not be Imaginary (because $D^2 = A$ must be Imaginary) & $D$ can not be real (because $D^2 = A$ must be Imaginary) : Hence we are Compelled to the Conclusion that $D$ must be Complex ! It must have Imaginary Part as well as real Part !

(4) Lastly , we can check $E = e^{axi}$ where $x$ varies from $1$ to $2$.
We assumed that $x = 1$ will give Imaginary "output". We saw that $x = 2$ will give real "output" earlier.
When we consider intermediate values for $x$ (Eg $1.0001$ , $1.1$ , $1.5$ , $1.9$ , $1.9999$ ) , we will realize that the "output" $E$ can not "Discontinuously" jump from Imaginary to real : It must move through Complex Numbers.

(5) When we see that we have alternating values :
$e^{1ai}$ (Im)
$e^{2ai}$ (Im $\times$ Im = real)
$e^{3ai}$ (Im $\times$ Im $\times$ Im = Im)
$e^{4ai}$ (Im $\times$ Im $\times$ Im $\times$ Im = real)
$e^{5ai}$ (Im $\times$ Im $\times$ Im $\times$ Im $\times$ Im = Im)
$e^{6ai}$ (Im $\times$ Im $\times$ Im $\times$ Im $\times$ Im $\times$ Im = real)
$e^{7ai}$
$e^{8ai}$
$e^{9ai}$
We will realise that this must be a Periodic function !

[[ Initially , when we assumed $A = e^{ai}$ was Imaginary , we made it something like $a=\pi/2$ , which will make all the other calculations give the alternating values , though we require the Euler (cis) formula to more know about that : It is not immediately necessary to get into that. The Intuition given here will not require that calculation ]]

Answer 3

If $e^{i\frac x2}$ were pure imaginary then $$e^{i\frac x2}e^{i\frac x2}=e^{ix}$$ would be real.

Answer 4

Since you are looking for an explanation rather than a proof, the following may be helpful. The elementary trig formula $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$, and a similar formula for $\cos(\alpha+\beta)$, suggest that there is some kind of additivity going on. Indeed, posing Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ as a definition, one obtains the suggestive formula $e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$ typical of the exponential function. This motivational discussion does not require the knowledge of either derivatives or the differential equation $y'=y\,.\quad$ :-)