2. Complex function exp (jωt) as a model of the real function cos (ωt)

Rotating Fourier Series

Chapter 2. Complex function exp (jωt) as a model of the real function cos (ωt)

Chapter 2.1 Introduction
The purpose of this article is to treat sine/cosine functionsc as rotating vectors. For example, vector (1,0) rotating with speed ω0=1/sec,  corresponds to the real function 1*cos(ω0t)=1cos(1t). I remind you that the angular velocity ω0=1 (more precisely ω0=1/sec) is assigned to the period T=2πsec≈6.28sec. This approach will greatly facilitate the intuive understanding of the formulas related to the Fourier Series, Fourier and Laplace Transforms.
Note on ω0 pulsation
Why not ω alone? Because this is an article about Fourier Series and ω0 is the pulsation of the fundamental harmonic, while ω=n*ω0 refers to the n-th harmonic of the periodic  function f(t).

Chapter 2.2 Euler Equation
I assume you know the complex numbers and the complex function exp(jω0t), wich are used in many fields. For me, electrician are rotating vectors. They show the phase shift ϕ between the current and voltage better than in ordinary time graphs. If you are not sure, I recomend the article “Complex Numbers” from the top bar.

Fig. 2-1
Euler Equation
Treat α generally as good x, that is, as a real number related to the equation! And jα is an imaginary number on the Im z axis. Only from this equation results its interpretation with a circle and a right triangle. Here you can cleaarly see that α is the angle in radians.
For any real number α, the complex value z=exp(jα) will be somewhere on the circle with radius A=1. At first it seems as obvious as the definitions of sine and cosine. But what is the relationship between the complex exponential function exp (jα) and trigonometry? Mr. Euler in his day could only use complex, addition, subtraction, multiplication and division. What about the function exp (jω0t)? How to calculate it using only the above-mentioned four actions? Euler already knew, however, that the real function exp (x) is the sum of infinitely many decreasing polynomials, such as the real number 1 is the infinite sum of the series 1/2 + 1/4 + 1/8 + 1/16 + …
He treated the complex exponential function exp (jx) in a similar way. Ie. He calculated its value as the sum of polynomials using only four operations on complex numbers. And he probably was surprised when the point z = exp (jx) began to spin around a circle with radius A = 1. Only then did he associate x with the angle α. And he expected the function to go to infinity somewhere. As for any honest real exponential function! Substitute in the Euler formula in order α = 0, 0.25π, 0.5π, 0.75π … 2π. You will see the point z = exp (jα) spin in a circle!

Chapter 2.3 Video Support
The main advantage of the article is animation. It is much more imaginative than a simple chart.

Fig. 2-2
Description of the buttons and video indicators
-Start clicking starts the animation and changes to the Stop button
-Clock the current experiment time, also shown as a yellow bar
-Simulation time differently – experiment duration, here 13 sec.
-Full screen clicking enlarges the screen, re-clicking reduces etc …
-Stop Clicking stops the simulation and changes to a Restart button
-Restart as the name suggests.

Chapter 2.4 Study of the various harmonic motions f(t) as their two-dimensional versions F (jω0t)
This chapter is just a pretext to get acquainted with the complex function Start*exp(jω0t). I emphasize that the parameters Start and jω0t are complex numbers! Now that we are familiar with video handling, we will examine harmonic motions for different initial states of the rotating vector Start and pulsation ω0. You will see that the time function x(t) is the projection of the rotating vector Start*exp (jω0t) onto the real axis Re z. So x(t)=Re z {Start * exp (jω0t)}.
We will examine the 5 harmonic motions.
1.1cos (1t) as Start*exp(jω0t) for Start=1 and ω0=1/sec
2.1sin (1t) as Start*exp(jω0t) for Start=1*exp(-jπ/2)=1*exp(-j90°=-j and ω0=1/sec
3.1cos (1t-π/4) as Start*exp(jω0t) for Start=1*exp(-jπ/4)=1*exp(-j45 °)≈0.707-j0.707 and ω0=1/sec
4.1cos (2t) as Start*exp(jω0t) for Start=1 and ω0=2/sec
5. 0.5cos (1t) as Start*exp (jω0t) for Start=0.5 and ω0=1/sec

Chapter 2.4.1 f (t)=1*cos (1t) as F(jω0t)=1*exp (j1t) i.e. as Start*exp(jω0t) for Start=1 and ω0=1/sec
The most important conclusion:
The rotating vector F(jω0t)=1*exp(j1t) is a two-dimensional version of the function f(t)=x(t)=1*cos(1t). Here ω0=1/sec.
Later it will turn out that almost every periodic function f(t) on the period T corresponding to the pulsation ω=2π/T
has its two-dimensional version as the sum F (jt)=c1*exp(j1ω0t)+c2*exp(j2ω0t)+c3*exp(j3ω0t)…
c1, c2, c3 …, cn are complex numbers or vectors an+jbn as initial states (fort=0) of spinning vectors cn*exp (jnω0t).
In the two-dimensional version of F(jω0t) of the function f(t), some features are more visible and intuitive than in the one-dimensional.

Fig. 2-3
Fig. 2-3a*
x(t)=1cos(1t)
It is a one-dimensional motion only along the real axis Re z on the complex plane Re z, Im z.
Click the “Start” button or the drawing to see the 2 T periods of the harmonic motion.
We know the amplitude A=1 and the time of 2 periods, i.e. 2T≈12 sec≈12.56 sec≈4πsec shown on the video stopwatch, i.e. T=2πsec-> ω0=2/T=1/sec.
The absolute values ​​of speed are the highest in the middle, and the smallest, that is, zero at the ends “at turns”.
*On the occasion. I’ll teach you Polish. Rys. 2-3a it’s Fig.2-3a 
Rys. 2-3b
Complex function 1exp (jω0t) as a rotating vector with length 1 and initial state (1,0)
Note:
Since the Start point=z=(1,0)=1+0j
The complex function has the value Start*exp(j1t)=(1+0j)*exp(j1t)=1*exp(j1t) and it is a rotating vector of length 1 and the initial state (1,0).
It can be seen that the animation from Rys. 2-3a is a projection of the animation from Fig. 2-3b onto the real axis Re z
Or otherwise
Re 1exp (j1t) = 1cos (1t)
The projection of the rotating vector onto the real axis Re z, moves as in Rys. 2-3a.
Some people almost equate exp(jω0t) with cos (ω0t). It is not exact, but it is accurate and pictorial. Just like the famous phrase “I am for and even against”. We know what’s going on, although the logic is wrong.
You can see the initial state of the circulating vector Start=(1+0j)=+1 in Rys 2-3b before the animation.
Remember. I will not repeat it a second time!
Each Start vector is an initial state and “stands still”. Here Start=z=1+0j, but it could be, for example, Start=z=2-3j.
When Start is multiplied by exp (jω0t), the vector Start*exp(jω0t) starts to spin at ω0 angular speed. You will find out about it in the next animations as well.
Rys. 2-3c
Time plot x(t)=1cos (1t) (ω0 = 1 / sec)
-The horizontal axis (“x”) is the time t in seconds.
Characteristic points 0sec 0.5πsec≈1.57sec πsec≈3.14sec 0.75πsec≈2.36sec 2πsec≈6.28 sec
– the vertical axis (“y”) is x(t) in units
Try to stop the simulation “around” at these times and compare Rys. 2-3a, Rys. 2-3b and Rys. 2-3c.

Chapter 2.4.2 f(t)=1*sin(1t) jako F(jt)=Start*exp(jω0t)  dla Start=-1j and ω0=1/sek

Fig. 2-4
Rys. 2-4a
x(t)=x(t)=1sin(1t) 
This is the projection of the circulating vector from Rys. 2-4b onto the real axis Re z.
Rys.2-4b
Complex function -1j*exp (j1t) for ω0=1/sec, otherwise exp(-jπ/2)* exp (j1t) because -1j =exp(-jπ/2)
Note:
Since the Start point=0-1j=-1j
The complex function has the value Start*exp(j1t)=-1j*exp (j1t).
You can see the initial state of the circulating vector Start=-1j in Rys.2-4b before the simulation. Compare with Rys. 2-3b. You can clearly see the truth, known for centuries, that sin(ω0t) lags by π/2=90º with respect to cos (ω0t).
Rys.2-4c
Time plot x(t)=1sin (1t)

Chapter 2.4.3  f(t)=1*cos(1t-π/4)  as F(jt)=Start*exp(jω0t)  for Start=1*exp(-jπ/4)=0.707-j0.707 and ω0=1/sek

Fig. 2-5
Rys. 2-5a
x (t)=1cos(1t-π / 4) i.e. x (t)=Acos (ω0t-ϕ) for A=1 ω0=1/sec and ϕ=-π/4=-45º
This is the projection of the circulating vector from Rys. 2-4b onto the real axis Re z.
Rys 2-5b
Complex function 1*exp(j1t-π/4) as 1*cos (1t-π / 4).
Notice that you can see the delay ϕ=-π/4=-45º beautifully here. Check with pythagoras that amplitude A = 1 and that ϕ=-π/4=-45º.
We can write a circulating vector in different ways.
Start*exp(j1t)=exp(-jπ/4)*exp(j1t)=(1/√2-j1/√2)*exp(j1t)≈(0.707-j0.707)*exp (j1t).
The initial state of the circulating vector Start=exp(-jπ/4) is shown in Rys. 2-5b.
Rys. 2-5c
Time plot x(t)=1cos(1t-π/4)

Chapter 2.4.4 f(t)=1*cos(2t)  as F(jt)=Start*exp(jω0t)  for Start=1and ω0=2/sek
We increased the speed to ω0=2/sec in relation to Rys. 2-3

Fig. 2-6
The course is 2 times faster than in Fig. 2-3. Besides, the initial state Start is the same, i.e. Start=(1+0j)=+1
Compare with the corresponding animations in Fig. 2-3.
Rys 2-6a
x (t) = 1cos (2t).
Rys 2-6b
Complex function 1exp (j2t) as 1cos (2t)
Rys. 2-6c
Time plot x(t)=1cos (2t)

Chaptwer 2.4.5 f(t)=0.5cos(1t) as F(jω0t)=Start*exp(jω0t)  for Start=0.5 and ω0=1/sek

Fig. 2-7
Rys. 2-7a
x(t)=0.5cos(1t).
2 times smaller A=0.5 amplitude compare with Fig. 2-4!
Rys 2-7b
Complex function 0.5exp (j1t) as 0.5cos (1t)
Rys. 2-7c
Time plot x(t)=0.5cos (1t).

Chapter 2.5   Functions Start*exp(jω0t)
Compare once again the previously discussed complex functions for different parameters Start and ω0.
1- Start=1 and ω0=1/sek
2- Start=1*exp(-jπ/2)=-j1 and ω0=1/sek
3- Start=1*exp(-jπ/4) and ω0=1/sek
4- Start=1 and ω0=2/sek
5- Start=0.5 and ω0=1/sek

Fig. 2-8
5 versions Start*exp(jω0t)
These functions are shown in Fig. 2-4b … Fig. 2-7b as rotating vectors. Their ends indicate the points z whose coordinates are precisely the complex functions Start*exp(jω0t).
The “beetroot” vector is the initial state of each function. This is the Start parameter that occurs before exp(jω0t). Carefully analyze each of the 5 waveforms taking into account the Start and ω parameters. Compare, for example, Rys. 2-8b and Rys. 2-8a. Here it is better to see that sin(1t)->-j1exp (j1t) is delayed by 90° in relation to cos(1t)->1exp (j1t).

Chapter 2.6 Complex function F jt)=(a-jb)*exp (jω0t) as a two-dimensional version of f(t)=a*cos(ω0t)+b*sin (ω0t)
Chapter 2.6.1 General Description
Dessert at the end. So how to build a trajectory F(jω0t) for any sinusoidal function
f (t)=a*cos (ω0t)+b*sin(ω0t)=c*cos(ω0t-ϕ)
From Chapter 2.4.1 we know that the complex function 1*exp (jω0t) corresponds to the time function 1*exp (ω0t).
Similarly, according to Chapter 2.4.2 of the complex function -1j*exp (jω0t) corresponds to 1*sin (ω0t).
We will write it like this:
1 * exp (jω0t) <==> 1 * cos (ω0t)
-1j * exp (jω0t) <==> 1 * sin (ω0t)
This is the case with amplitudes a=1 for cosine and b = 1 for sine.
It works for any amplitudes a and b:
a * exp (jω0t) <==> a * cos (ω0t)
-jb * exp (jω0t) <==> b * sin (ω0t)
Instead of writing that the function corresponds to something, you can more closely, like the following

Fig. 2-9
Re (a-jb)*exp(jω0t)=a*cos(ω0t)+b*sin(ω0t)
The vector (a-jb) is spinning at speed ω0.
The projection of the rotating vector (a-jb)*exp(jω0t) onto the real axis is:
real part (a-jb) * exp (jω0t)
i.e. Re (a-jb)*exp(jω0t)
that is  a*cos(ω0t)+b*sin(ω0t).
This is a generalization of the simplest case from Rys. 2-3 where a=1 and b=0.
a
Complex function (a-jb)*exp(jω0t) as vector rotating at speed ω0 (a-jb)
It has 2 components: –
–real (cosine) a*exp (jω0t)
-imaginary (sinus) -jb*exp (jω0t)
What you see in Rys. 2-9a is the initial state of the rotating vector, i.e. for t = 0.
I emphasize that “lonely” a and b are real numbers!
b
The projection of the rotating a*exp(jω0t) vector onto the real axis is a*cos(ω0t)
c
The projection of the rotating -jb*exp(jω0t)  vector onto the imaginaryl axis is b*sin(ω0t)
d
The projection of the sum of the rotating vectors, i.e. the red vector, onto the real axis is
a*cos(ω0t)+b*sin(ω0t)=c*cos(ω0t-ϕ)
It’s not obvious, but that’s trigonometry! Instead of proof*, let’s substitute specific values, e.g. a=0.75 and b=1.25 and check the animation.
* This can be done by a high school math class student.

Chapter 2.6.2 Concrete example F (jω0t) = (0.75-j1.25) * exp (jω0t) as a two-dimensional version of
f (t) = 0.75 * cos (ω0t) + 1.25 * sin (ω0t)

Fig. 2-10
Rotating vector (0.75-j1.25)*exp(1jt) as f(t)=0.75*cos(1t)+1.25*sin(1t)=1.458*cos(1t-59.04°)
Rys. 2-10a
Rotating vector 0.75 * exp (1jt) and its projection onto the Re z axis as a function of time 0.75*cos(1t)
Rys. 2-10b
Rotating vector -j1.25*exp(1jt) and its projection on the Re z axis as a function of time 1.25*sin(1t)
Rys. 2-10c
Sum of the 2 left vectors as a spinning vector (0.75-j1.25)*exp(1jt). Its projection on the Re z axis is a function of time f(t)=0.75*cos(1t)+1.25*sin(1t)=1.458*cos(1t-59.04°) and its timing diagram is shown in the animation below.

Fig. 2-11
f(t)=1.458*cos(1t-59.04°)=0.75*cos(1t)+1.25*sin(1t)
The angular shift of 59.04 °≈1.03 radians is shown in the diagram.