Kitonum

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17 years, 235 days

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These are replies submitted by Kitonum

@vv  Thanks for this elaboration. It is clear that these inequalities  (a>2  and  so on) were found by hand.

Compare with solution in Mathematica:

@Carl Love  I started from the desired result for OP and came to the following interpretation: a set is given, each element of which is the product of two factors. One of these factors is fixed (in the question these are  a, b, c ). Find the set of remaining factors.

@Carl Love OP probably just misused the term "common factor". I have shown how to get the desired result in another obvious interpretation, which OP obviously had in mind.

@nm  I wrote "In each specific example ..."

@acer  You wrote  "And how about requesting their spacing evenly by arclength rather than by value of the independent parameter? "

See the post  https://www.mapleprimes.com/posts/213452-Uniform-Point-Plot  with a procedure for this. 

@Carl Love  Sorry, but it seems to me that for a beginner, your method will be difficult to understand.

Another way for the same:

plots:-display(plot(sin(x), x= -Pi..Pi), plot(sin(x),x= -Pi..Pi, style=point, symbol= box, numpoints= 16, adaptive= false));


We can write very short for more symbols:

plot(sin(x), x= -Pi..Pi, style=pointline, symbol= box, numpoints= 30, adaptive= false);

                  

@ecterrab  OP wants to plot the equation  |z+1|*|z-1|=1  (he mistakenly calls it a function), not independent graphs of the left and right sides of this equation.

@abdulganiy  Use the plots:-textplot  command.

@JAMET You have to do it yourself. It's quite easy if you first study the solutions I've shown.

@AHSAN  See my updated answer.

@janhardo I think it is useful to know the physical meaning of this curvilinear integral: if  2*x+y^2  is the linear density at the point  (x,y) of the segment, then the value of the integral is the mass of this segment. 

@Preben Alsholm  Thank you for the improvement.

@ActiveUser  See the corrected answer.

You have already built it. cos(Pi/4)  is a constant. And what is the question?

@Scot Gould You're right. Using  add  instead of  sum  avoids many problems. 

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