[BUG ?] Who can explain to me the CAS options "Use i" and "complex"? - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: [BUG ?] Who can explain to me the CAS options "Use i" and "complex"? (/thread-10431.html) [BUG ?] Who can explain to me the CAS options "Use i" and "complex"? - dg1969 - 04-02-2018 11:21 AM Hi, Here is an equation for which czeros returns a bad solution when none of the "use i" or "complex" options are checked... (Physical prime Firmware 13441). $$(3+2i)z+(1-i)\bar{z}=1$$ Code: cZeros((3+2*i)*z+(1-i)*CONJ(z)=1,z) give me the wrong answer $$\frac{1}{4+i}$$ But with "Use i" or with "Complex" checked it give the correct answer: $$\frac{2-i}{11}$$ Code: csolve((3+2*i)*z+(1-i)*CONJ(z)-1=0,z) always return the good answer with or without options checked... Thank's RE: Who can explain to me the CAS options "Use i" and "complex"? - dg1969 - 04-02-2018 03:00 PM More precisely my question is: Why this wrong answer rather an answer like "no real solution" ? If "Use i" and "Complex" are not checked that assumes (that's what I imagine) that $$z$$ is real. And of course if $$z\in \mathbb{R}$$ then $$\bar{z}=z$$... The "wrong solution": $$\frac{1}{4+i}$$ is in fact the only complex solution that satisfy this other equation:: $$(3+2i)z+(1-i)z-1=0$$ In my mind if the options "Use i" and "Complex" ( I assume $$z \in \mathbb{R}$$ )are not checked then with my initial equation the cZeros() function should answer "No solution" or else the system should display an alert indicating that the options need to be modified for a complex response to be provided. For me there is a problem. Because I was looking for a complex solution. The calculator gives me a bad one and I may not realize it ... RE: [BUG ?] Who can explain to me the CAS options "Use i" and "complex&... - Arno K - 04-02-2018 03:25 PM First of all, czeros is not the function to use when you solve an equation, nevertheless it should not provide a wrong answer. as far as I remember the complex home setting allows complex output from real input (solve x^2=-1). When you enter "assume (z,complex)" first or go to CAS settings and check complex there you will get the correct result czeros( (3+2i)*x+(1-i)*conj(z)=1) will work. Arno RE: [BUG ?] Who can explain to me the CAS options "Use i" and "complex&... - parisse - 04-03-2018 08:13 AM Complex mode is now temporarily turned on in cZeros in source code. RE: [BUG ?] Who can explain to me the CAS options "Use i" and "complex&... - dg1969 - 04-03-2018 04:17 PM (04-03-2018 08:13 AM)parisse Wrote:  Complex mode is now temporarily turned on in cZeros in source code. Thank you M. Parisse ! :o) RE: [BUG ?] Who can explain to me the CAS options "Use i" and "complex&... - dg1969 - 04-03-2018 04:23 PM (04-02-2018 03:25 PM)Arno K Wrote:  First of all, czeros is not the function to use when you solve an equation, Arno I'm not sure I understand correctly. What is the use of this function czeros ? In what situation should I use it? I take this thread to ask for some help. I understand that the "Complex" option means that variables can be complex. What is the difference with "Use i"? RE: [BUG ?] Who can explain to me the CAS options "Use i" and "complex&... - Arno K - 04-03-2018 07:31 PM (04-03-2018 04:23 PM)dg1969 Wrote:  Arno I'm not sure I understand correctly. What is the use of this function czeros ? In what situation should I use it? I take this thread to ask for some help. I understand that the "Complex" option means that variables can be complex. What is the difference with "Use i"? czeros computes zeros of the given expression, that is something like 'x^2+x' is enough to be entered, you don't need to enter '...=0', solve is used for equations 'x^2+x=-3'. If you check use i in CAS-settings, polynomials will be factored down to linear terms, without that 'factors (x^2+1)' results in {x^2+1,1}, with use i the output is {x+i,1,x-i,1}. Concerning Complex in CAS settings, I think that is always switched on, regardless to the mark there. In Home settings you can decide if you want complex results show up as '1+3*i' or as '(1,3)', with that checkmark there you can enter sqrt(-1) and the result is i, wizhout the same input is commented as Error. Arno