Free42 and cube root of complex number

02242021, 02:47 PM
Post: #8




RE: Free42 and cube root of complex number
(02242021 01:35 PM)Thomas Okken Wrote:(02242021 08:59 AM)Werner Wrote: Please, leave it as it is  Free42 would be one of the only ones to get it right ;) Hi, Thomas Okken Can you elaborate what is decided ? If Free42 do support 0, then sqrt(40i) = sqrt(4*exp(pi*i)) = 2*exp(pi/2*i) = 2i But, user need to know "0" is actually inputed. 0 should displayed as "0", not "0" 40*i should displayed as "4 i0", not "4 i0" If not, (80i)^(1/3) = (8+0i)^(1/3) = 1 + sqrt(3)*i 

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Messages In This Thread 
Free42 and cube root of complex number  Ajaja  02232021, 11:05 PM
RE: Free42 and cube root of complex number  Thomas Okken  02242021, 02:02 AM
RE: Free42 and cube root of complex number  Albert Chan  02242021, 10:26 AM
RE: Free42 and cube root of complex number  Paul Dale  02242021, 02:52 AM
RE: Free42 and cube root of complex number  JF Garnier  02242021, 08:28 AM
RE: Free42 and cube root of complex number  Werner  02242021, 08:59 AM
RE: Free42 and cube root of complex number  Thomas Okken  02242021, 01:35 PM
RE: Free42 and cube root of complex number  Albert Chan  02242021 02:47 PM
RE: Free42 and cube root of complex number  Thomas Okken  02242021, 02:53 PM
RE: Free42 and cube root of complex number  Werner  02252021, 06:08 AM
RE: Free42 and cube root of complex number  Albert Chan  02252021, 04:40 PM

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