Euler's Identity

[hanhan]

Hi,
I can understand the Euler's Identity.

θ is a real number => jθ is an imaginary number
=> e^ (jθ) also has to be an imaginary number.
But as the formula above:
e^ (jθ) = cosθ + j sinθ
Apparently, it has both real and imaginary components.
Where am I wrong?
Thanks for helps.

 

[misterT]

You should post these to Mathematics and Physics section

EDIT: but don't repost.. moderators will propably move these there.

 

[misterT]

When you take modulus (absolute value) of complex number, you end up with pure real number.

The modulus of complex number is defined that way.

Did I answer the question.. I'm not sure what part of the equations you need help with.

 

[steveB]

θ is a real number => jθ is an imaginary number

True, except for θ=0 when j0 is not clearly imaginary.

=> e^ (jθ) also has to be an imaginary number.

Not true. This is a complex number in general. It can be real, imaginary or complex, depending on the value of θ.


So this is where you are making your mistake. The simplest way to see your mistake is to take the case of θ=0. Here you have exp(0)=1. Obviously, this is real and not imaginary.

 

[hanhan]

Hi,
here is my confusion:
θ is a real number => jθ is an imaginary number => e^ (jθ) also has to be an imaginary number.
That is what I think. It is wrong but I don't know why.
According to Euler formula:
e^ (jθ) = cosθ + j sinθ
Apparently, e^ (jθ) has both real and imaginary components not just imaginary component like what I think.

 

[hanhan]

Thanks steveB.

Not true. This is a complex number in general. It can be real, imaginary or complex, depending on the value of θ.


So this is where you are making your mistake. The simplest way to see your mistake is to take the case of θ=0. Here you have exp(0)=1. Obviously, this is real and not imaginary.

Could you explain why there is a real component in e^ (jθ)? I don't have any intuitive about this?

 

[misterT]

Apparently, e^ (jθ) has both real and imaginary components not just imaginary component like what I think.

Yes.. now you got it.

Numerical example:

2e^ (j3) = 2*cos(3) + 2*j sin(3)

EDIT: The equation is actually called "Eulers Formula".

 

[hanhan]

Yes.. now you got it.

Numerical example:

2e^ (j3) = 2*cos(3) + 2*j sin(3)

EDIT: The equation is actually called "Eulers Formula".

Could you explain why there is a real component in e^ (jθ) without using Eulers Formula or Taylor expansion? Maybe an intuitive way to understand this?

 

[misterT]

Thanks steveB.

Could you explain why there is a real component in e^ (jθ)? I don't have any intuitive about this?

That is not so easy.. it comes backwards from taking the natural logarithm of cosθ + j sinθ:

ln(cosθ + j sinθ)

Mathworld: https://mathworld.wolfram.com/EulerFormula.html
Wikipedia: https://en.wikipedia.org/wiki/Euler's_formula

 

[steveB]

Could you explain why there is a real component in e^ (jθ)? I don't have any intuitive about this?

I agree it's not necessarily easy to see. That's the real advantage of Euler's Identity. The answer for exp(jθ) is not easy to see, but the answer for cosθ+jsinθ is very easy to see.

But, if you want to see it, you can use the Taylor expansion exp(x)=1+x+x^2/2 ...

This is an infinite sum, but if you use x=jθ, you can show the sum is two pieces since exp(jθ)=(1-θ^2/2 ...)+j(θ-θ^3/6 ...), and then we know that the bracketed terms are Taylor expansions of cosθ and sinθ.

If this is too unfamiliar, you can try to find other ways to calculate exp(jθ). But, the bottom line is that the answer is the answer, and the answer is complex in general, even if it's not easy to see directly.

 

[MrAl]

Hi,


To prove this using Taylor Series (plural Series, not singular) we can compute the series for a few different functions and then compare results. The procedure would go like this:

1. Compute the Taylor Series for e^x, call that series Ex.
2. Replace all x in Ex with i*x, reduce, call that series Eix. We now have e^(ix) in series form.
3. Compute the Taylor Series for sin(x). Call that series Sx.
4. Compute the Taylor Series for cos(x). Call that Cx.

Now the identity we want to prove is:
e^(ix)=cos(x)+i*sin(x)

so in terms of our series (plural again) we can state this as:
Eix=Cx+i*Sx

and once the terms of Cx and i*Sx are summed, they come out exactly equal to the terms of Eix using any reasonable number of terms for all series.

But why would we want to do it this way rather than some other way? That's because this method shows the beauty of the whole thing. We start with a complex exponential where the imaginary operator 'i' (denoted as 'j' in electronic work) operates on the exponent, and we convert that into another form where the complex operator operates only on a term as a constant multiplier rather than operating on the variable itself. In other words, we not only got the complex part out of the exponent, we also started with a function of ix and ended up where i is just a constant multiplier of a function (but also summed with another function). So we started with:
f1(i*x)

and ended with this:
f2(x)+i*f3(x)

so overall we managed to accomplish this:
f1(i*x)=f2(x)+i*f3(x)

and not only that but even f2 and f3 are related by Trigonometry! So you start to see why this is such a beautiful idea.

 

[hanhan]

Thanks MrAl,
I knew how to prove this using Taylor expansion but I want to really understand what does e^(jθ) mean. After reading some old threads, I am interested in considering j as a rotate operator. I understand some basic that Ratchit has pointed out. However, I can't understand how we can explain e^(jθ) = cosθ + jsinθ by using the knowledge that j is a rotate operator.

To me, θ is a real number => jθ is an imaginary number
Then I tend to think of e^(jθ) is an imaginary number but that is wrong. Why?
jθ is an imaginary number => e^(jθ) also has to belong to imaginary axis. But it is wrong!!! I don't know why?
I want to understand this by using j as an rotate operator not using Euler's Identity or Taylor expansion.

 

[steveB]

Why do you want to understand it using j as the rotation operator? Is there a basis to think a proof is possible using this idea? If so, can you give us more information about that method, and maybe we can help?

You are asking a strange question. You tend to think exp(jθ) should always be imaginary, but you know that it isn't. The Euler identity and the Taylor expansion both prove that you are right that you are wrong. So, just abandon your wrong idea and keep the right idea.
Asking why is like asking why 2+2=4. We know it is from experience and we can prove it mathematically. If someone comes along and insists that they think 2+2 should be 5, but they know it isn't and ask why it isn't, what would you tell them?

Anyway, j is a rotation operator when you multiply by j. Hence, the right hand side of the Euler identity can be described as the addition of a real number cosθ vectorially added to the real number sinθ rotated by 90 degrees in the complex plane (hence making that part imaginary). Complex numbers are like vectors, and must be added in that way. But, it's not clear how you use that to prove the Euler identity.

 

[hanhan]

Hi,
I have just found a great explanation here: https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

 

[steveB]

That is a nice explanation of complex exponentials. Does it answer your question? If so, what part answered it?

 

[MrAl]

Hi again,


Here is a very nice 3d geometrical interpretation showing what e^(i*x) and it's rotation i*e^(i*x) means...

https://hightechavenue.blogspot.com/2013/07/analog-avenue-helix-rotations.html

Note that 'x' is the vertical axis and that would be normal for 't' in relativity theory.

But i also already mentioned that 'i' times something makes everything imaginary real, and everything real imaginary, and when we go from real to imaginary or imaginary to real that's a rotation in the complex plane, and you can see the helixes starting out 90 degrees out of phase with each other (90 degrees in the complex plane). So when we multiply by 'i' we are really rotating the helix by 90 degrees. By that simple rule then when we multiply by 'i' a second time, we effectively rotate the helix another 90 degrees so we should see a new helix that starts out 180 degrees ahead of the first. One more multiplication by 'i' and we move ahead another 90 degrees yet, and one last multiplication and we are a full 360 degrees ahead which is often interpreted as no rotation. This follows the pattern i explained previously where we have four possibilities but that's all.

 

[hanhan]

That is a nice explanation of complex exponentials. Does it answer your question? If so, what part answered it?

Hi,
Yes. My problem is that I don't understand what does e^(jθ) really mean by using j operator.
In the topic he said:
e^i = lim(1 + 100% i /n) ^n ( when n approaches infinity )
( and here is a related topic about exponential growth https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ )
Then by using the growth = ( 1 + return )^x in the topic https://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/ )
I can understand what does (1 + 100% i /n) ^n mean.
Finally this part in the topic https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ helped me.

What is Imaginary Growth?

Combining x- and y- coordinates into a complex number is tricky, but manageable. But what does an imaginary exponent mean?

Let's step back a bit. When I see "3^4" I think of it like this:

3 is the end result of growing instantly (using e) at a rate of ln(3). 3 = e^ln(3)
3^4 is the same as growing to 3, but then growing for 4x as long. So 3^4 = e^(ln(3) * 4) = 81
Instead of seeing numbers on their own, you can think of them as something e had to "grow to". Real numbers, like 3, give an interest rate of ln(3) = 1.1, and that's what e "collects" as it's going along, growing continuosly.

Regular growth is simple -- it keeps "pushing" a number in the same, real direction it was going. 3 × 3 pushes in the original direction, making it 3 times larger (9).



Imaginary growth is different -- the "interest" we earn is in a different direction! It's like a jet engine that was strapped on sideways -- instead of going forward, we start pushing at 90 degrees.

The neat thing about a constant orthogonal (perpendicular) push is that it doesn't speed you up or slow you down -- it rotates you! Taking any number and multiplying by i will not change its magnitude, just the direction it points.

Intuitively, here's how I see continuous imaginary growth rate: "When I grow, don't push me forward or back in the direction I'm already going. Rotate me instead."

 

[hanhan]

Hi again,


Here is a very nice 3d geometrical interpretation showing what e^(i*x) and it's rotation i*e^(i*x) means...

https://hightechavenue.blogspot.com/2013/07/analog-avenue-helix-rotations.html

Note that 'x' is the vertical axis and that would be normal for 't' in relativity theory.

But i also already mentioned that 'i' times something makes everything imaginary real, and everything real imaginary, and when we go from real to imaginary or imaginary to real that's a rotation in the complex plane, and you can see the helixes starting out 90 degrees out of phase with each other (90 degrees in the complex plane). So when we multiply by 'i' we are really rotating the helix by 90 degrees. By that simple rule then when we multiply by 'i' a second time, we effectively rotate the helix another 90 degrees so we should see a new helix that starts out 180 degrees ahead of the first. One more multiplication by 'i' and we move ahead another 90 degrees yet, and one last multiplication and we are a full 360 degrees ahead which is often interpreted as no rotation. This follows the pattern i explained previously where we have four possibilities but that's all.

Hi,
I understand how to get i*e^(i*x) from e^(i*x). But I don't understand what does e^(i*x) mean. How can you know that e^(i*x) has the Helix?
( I understand e^2, e^x, e^n but not e^j2, e^ jn,...)

 

[steveB]

Hmmm. You say you don't understand exp(jθ), but you seem to describe it well, as if you do understand it. I think this makes it harder for us to help because it's not entirely clear what you feel you are missing.

Here is my best guess at what might help. Like you say, multiplying by j is a 90 degree rotation. Well, j=exp(j∏/2); hence, we see that exp(jθ) causes a rotation by the angle θ. So, when you say that you don't know what exp(j2) means, just consider this number as the number 1 times the number exp(j2). The number 1 is a vector that points a distance of 1 unit on the real axis and the multiplication by exp(j2) is like an instruction to rotate the vector by 2 radians. Or, the number exp(j2) is just a complex number with magnitude 1 and angle 2 radians.

It really doesn't need to be any more complicated than this. Complex numbers can be represented in polar form as A*exp(jθ) or in rectangular form as x+jy. The value "A" is the magnitude and θ is the angle.The relations x=Acosθ and y=Asinθ, as well as A=sqrt(x^2+y^2) and θ=atan(y/x) allow you to convert between the two representations. After that, just work with complex numbers and complex functions to become comfortable and familiar with them.

 

[hanhan]

Thanks SteveB,
I am relearning about the system of numbers in the book: https://www.tuks.nl/pdf/Reference_Material/Steinmetz/CP%20Steimetz%20-%20Engineering%20Mathematics%20-%20A%20Series%20of%20Lectures%20Delivered%20at%20Union%20College%20-%201911.pdf
That is interesting. I will read it carefully and ask later.