Fourier Series Animation
Chapter 4 Complex Fourier Series Animation
Chapter 4.1 Animation tools
The tools come from an excellent article  https://betterexplained.com/articles/an-interactive-guide-to-the-fourier-transform .
It facilitates the Complex Fourier Series with the real part understanding. See Fig. 3-12 Chapter 3.4.
It’s near to the classical Complex Fourier Series now. See Fig. 3-21 Chapter 3.5.

Chapter 4.2  f(t)=1 constant function 

Fig. 4-1 f(t)=c0=1 function animation
Note that you return from animation to the article by “Windows arrow clicking”.
Click Start animation and write in something in the appropriate windows then. I.e. Write “1 0 0 0” in the Cycles window and tick the other small windows. The most simple time function will be executed.
The effect are:
yellow stationary point on the circle with the R=1
– blue moving point as a f(t)=1 time function.
It’s possible to program  for example  3 rotating vectors c1, c2, c3 with ,, 3ω pulsation + c0 constant. The period T=3 sec will be measured in the chapter 4.3, i.e. ω=2.1/sec. So, you can test undermentioned Complex Fourier Function 

Fig. 4-2
Complex Fourier Function when “1 0 0 0” is written in the window
Cycles “1 0 0 0” means c0=1 and c1=c2=c3=0. There are any rotations. Or other words. c0=1 and it doesn’t rotate.  The  other c1,c2,c3  vector lenghts are null. (they don’t exist). So, the blue line in the Fig. 4-1 is a f(t)=1constant time function.

Chapter 4.3 f(t)=1*exp(j1ωt) function
Click  Start animation

Fig. 4-3 f(t)=1*exp(jωt)-animation
See Fig. 4-2 and write in “0 1 0 0” in the Cycle window. It means that c1=1 and c0=c2=c3=0. So the z=1*exp(jωt) complex function is programmed. You see the rotating blue point. It’s position z=1*exp(jωt) is a just a complex function!
The point projection on the Re z moves acc. to the f(t)=1*cos(ωt) time function.
The most important conclusion is the association between the rotating point z=1*exp(jωt) complex function, and the f(t)=1*cos(ωt) time function.  Four phases point movement i.e. ωt=0º,90º,180º,270º are yellow points on the circle and appropriate yellow vertical lines on the right f(t) time function. These f(t) values “1, 0, -1, 0” are shown in the Time window too.
The animation will be paused when clicked in the any place.
Try to stop in these ωt=0º,90º,180º,270º phases. It isn’t easy but you will be a champion after some attempts I hope!
Measure 10 rotations time. It should be circa t=30 sec. I.e T=3sec period and ω=2π/3ek≅2.1/sec.
The next click starts animation again.

Chapter  4.4 f(t)=1*exp(j1ωt-30º)  function
The function will be 30º delayed when compared to Fig. 4-3.
Click Start animation

Fig. 4-4 f(t)=1*exp(jωt-30°)  animation
Write in “0 1:-30 0 0” Cycles window.
The z=1*exp(jωt-30°)  function and the appropriate f(t)=1*cos(jωt-30°) time function is executed now. More strictly–>30°=π/6.
Note that:
f(t)=1*exp(jωt-30°) as a complex function or  rotating blue point
– f(t)=1*cos(ωt-30º) as a time funcion
contains this same information!
For example it may be i(t) current function.
It is very useful for electicians. The blue point position as an arrow is easy to current phase and amplitude imagine.
The
30º phase is a i current delay regarding the sinusoidal input voltage.

Chapter  4.5 f(t)=c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt) function
Let’s explore the sum f(t)=c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt)–>see Chapter 3.3.1 :
c1=1-0.2j
c2=0.6-0.4j
c3=0.4-0.4j
Will be the same chart as Fig. 3-6b Chapter 3?
Unfortunately, the Cycles window accepts c1, c2 i c3 complex coefficients as a module-phase version.
So, we need to change the numbers versions:
c1=1-0.2i–>r=1.02 φ=-11.301º
c2=0.6-0.4j–>r=0.721 φ=-33.7º
c3=0.4-0.4j–>r=0.567 φ=-45º
I propose to use the sure Wolfram Alfa program. For example what is a  c1=1-0.2i?
Note that imaginary number is here and not as in our article! Click WolframAlfa. 

Fig. 4-5 c1=1-0.2i  for c1 –>r=1.02 φ=-11.301º  convertion
Convert c2 and c3 too and use converted numbers in the experiment.
Click  Start animation


Fig. 4-6 f(t)=1.02*exp(1ωt-11.301º)+0.721*exp(2ωt-33.7º)+0.567*(3ωt-45°)  function animation
Write in window Cycles “0  1.02:-11.301  0.721:-33.7 0.567:-45”
First
in Cycles is c0=0
Second is c1=1-0.2i  as  r=1.02 φ=-11.301º
Third is c2=0.6-0.4j as r=0.721 φ=-33.7º
Fourth is c3=0.4-0.4j as r=0.567 φ=-45º
Notes:
You see c3 in the Cycle  window as only 0.567 and you don’t see -45º. The window is too small!
-The upper Fig. 4-6 part shows harmonics and the sum Total and Parts are ticked)
-The lower Fig. 4-6 part shows the sum only (Parts isn’t ticked)
The lower right chart is similar to the Fig. 3-6b Chapter 3. But they aren’t identical. Why? There are different scales.
The job for You
-Note the association between vectors and appropriate harmonics.
-Try to stop the process in the phases 0º, 90º, 180º i 270º when window parts isn’t ticked.
It isn’t easy but possible. 
Are the f(t) values and vector lengths roughly equal?
More strictly-are they proportional? (the rotating vector and f(t) process have different scales).

Conclusion
Complex Fourier Series is a infinity sum of the rotating vectors cn*exp(jnωt).
Their rotations nω are bigger and theirs vectors cn are smaller when n–>∞.
The projection on the Re z of this sum moves exactly as f(t) time function. see–>Fig. 3-13 Chapter 3.

But classical Complex Fourier Series are oppositely rotating vectors–>see Fig. 3-20 and Fig 3-21 Chapter 3.4.1