@Carl Love 

f:= z-> z^3*cos(1/(z-2)):
coeff(series(f(2+1/z),z),z);

or

f:= z-> z^3*cos(1/(z-2)):
coeff(series(f(1/(z-2)),z=2),z-2);

@Alfred_F I also don't like how the solution is formulated. This often happens to me, but usually (this case too!) it is easy to reformulate (and fill the gaps) and obtain a real solution.

@Alfred_F  Note that a faster version is:

Q5:=n -> irem(add(convert(n,base,10)),5):
for n to 100000 do
if   Q5(n)<>0 then next 
elif Q5(n+1)=0 then print(nmin=n); break else n++ fi od:

BTW, Copilot AI did it well:

@Rouben Rostamian  

SMTLIB:-Satisfy(1+9+a+b = 1968 - (1900 + 10* a + b)) assuming a::nonnegint,b::nonnegint,a<=9,b<=9;

             {a = 4, b = 7}

I agree. Maybe a new field "accepted answer" could be introduced. But in the case of a single answer, the OP may even now click for "best answer".

@Alfred_F  After executing the code of my answer, f1+f2+f3 gives a sum of squares.
Note however that f1,f2,f3 are not unique; changing the order of the indeterminates, one obtains distinct representations. There are infinitely many representations!

@nm Actually such a command exists:

restart;
f:=419*x^2 + 116*x*y - 426*x*z + 78*y^2 - 142*y*z + 133*z^2 - 1604*x - 682*y + 1086*z + 2306:
SMTLIB['Satisfy'](f=0) assuming real;

                    {x = 7, y = 11, z = 13}

@Carl Love Yes, thank you, corrected.

@janhardo  What about T=1? (Not to mention T=0.)

It will be never possible to study such problems just looking at a graph.

For example, it can be shown that any polynomial of degree n is expressible as the sum of n+1 periodic functions.
E.g. f(x) = x is a sum of two periodic (but discontinuous) functions. Could someone comment on the periods of these functions by looking at a line? (Actually these periods are arbitrary but incommensurable.)

@Alfred_F If a spherical bullet impacts a plane, the "impression" is a circle (actually a disk) only if the (linear) trajectory is orthogonal to that plane. So, what is the significance of r1, r2, r3?

P.S. It seems that unlike your other problems, this one if very vague.

@Preben Alsholm But in my post,  f(complex) is a "function of a complex number" after evaluation.
Othewise, the simple example of a constant function
f := x -> a;

can be used.

For your example:

'f'(2+3*I).b - b.'f'(2+3*I);

returns <0,0;0,0>.

@Preben Alsholm Actually A, B commute (.) if at least one of them is a complex number i.e. C is in the center of (Expressions, .).
The strange fact is that any function of a complex number commutes (.) with any expression:
f(2+3*I).b - b.f(2+3*I) = 0. 

This reminds me of the following fact:

If the matrix M is defined by

m[i,j] = ( d(1,i) + d(1,j) - d(i,j) ) / 2 

then the set {P_i} can be embedded in R^k  iff  rank(M) <= k.
 

@Andiguys Your C_1 is
{0 <= 0.000023339978559*p1 - 0.85357635872}

Probably you want to replace the previous plot2 with:

plot2 := plot([solve(rhs(C_1[])), t, t = -0.2e9 .. 0.1e9])

I don't understand why you complicate things for such a simple problem.