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

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.

• Every complex number has a real part, but I guess it's not what you are asking here. Please clarify. Note also that even though $e^{ix}=\cos x+i\sin x$, when $x$ is not real the real part of $e^{ix}$ is not $\cos x$. Commented Jul 13, 2023 at 10:39
• Can you understand that the function $f(z) = z^2$ is completely real when the argument is purely imaginary? $f(i) = -1$ is real! Commented Jul 13, 2023 at 10:49
• @Trebor is having a good point. This is usual in maths. Using his example, it would be analog why don't we have negative numbers for $x<0 ; f(x)=x^2$. You mave also look up the definition and plot in real numbers into the polynomials, as Trebon hinted at. Commented Jul 13, 2023 at 10:59
• I think $e^{i0}=1$ should be easy however you define the exponential. Thus for $x=0$, certainly $e^{ix}$ has nonzero real part. Commented Jul 13, 2023 at 11:22
• @Cesareo, no offense, but you should read the whole question. Commented Jul 13, 2023 at 12:18

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).

• this is an excellent answer for a physicist like the OP, +1 Commented Jul 13, 2023 at 12:29
• It's a good answer for someone who understands complex numbers and the exponential function but I doubt that @Henry05 will understand it. To me, his question is like saying: "I understand the proof that the square root of two is irrational but why is that true, intuitively?". Commented Jul 13, 2023 at 12:43
• I agree with that assessment , @JohnDouma , which was why I avoided rotation & trigonometry & calculus in my Answer. I concentrated only on why that function can not be only Imaginary & why it has to be real and/or Complex. This is a "Soft Question" hence there can be no Single Correct Answer !
– Prem
Commented Jul 13, 2023 at 14:29
• I have to respectfully disagree with you. $(e^x)' = e^x$ is high school mathematics, chain rule is high school mathematics, velocity, acceleration and uniform circular motion is high school physics, complex numbers - high school mathematics, powers of $i$ high school mathematics. The only university level mathematics is tiny bit of linear algebra to explain that multiplication by $i$ corresponds to rotation by $90^\circ$, which honestly, you can believe is true solely from the fact that you know how to calculate powers of $i$. Commented Jul 13, 2023 at 14:38
• @John Douma, yes, knowing a thing and understanding a thing are indeed different, which is why I wrote my answer that should be intuitive enough for physicists. You still didn't point out what exactly is problem with my answer and why you think OP wouldn't be able to understand it, which I'd find very helpful, and otherwise, this exchange pointless. If it's "because OP doesn't know analytic function theory", I find that absurd since clearly every mathematician saw complex exponential function before learning any deep results from complex analysis. Commented Jul 13, 2023 at 15:10

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 ]]

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

• In Case you did not see it , Just to inform you : That is Point 1 & Point 2 ( & maybe Point 3 ) in My Elaborate Answer : No worry though !
– Prem
Commented Jul 15, 2023 at 11:03
• I falsely red flag. Sorry. Commented Jul 15, 2023 at 14:56
• I am not getting what you are saying ? Is there something wrong with my Comment ?
– Prem
Commented Jul 15, 2023 at 16:33
• No. There is nothing wrong with your comment. Thanks for commenting. It is not easy writing with telephone. I am used to keyboard. Commented Jul 15, 2023 at 20:01

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 $$y'=y\,.\quad$$ :-)