Calculating with complex numbers - Printable Version +- QB64 Phoenix Edition (https://qb64phoenix.com/forum) +-- Forum: Chatting and Socializing (https://qb64phoenix.com/forum/forumdisplay.php?fid=11) +--- Forum: General Discussion (https://qb64phoenix.com/forum/forumdisplay.php?fid=2) +--- Thread: Calculating with complex numbers (/showthread.php?tid=586) |
Calculating with complex numbers - Kernelpanic - 07-05-2022 This is from the QBasic Tech Reference book, and is intended to demonstrate math with complex numbers. I myself can't see through it anymore, but anyone who deals with mathematics can perhaps use it. Code: (Select All) 'Loesung quadratischer Gleichungen S. 231, QBasic Referenz - 5. Juli 2022 RE: Calculating with complex numbers - BSpinoza - 07-06-2022 @Kernelpanic: Your program only solves quadratic equations with real solutions (roots). If you try to solve a quadratic equation with complex roots, you will get an error. example: x^4 - 4 x +7 = 0 This equation has two complex roots: x_1 = 2 - sqr(3i) x_2 = 2 + sqr(3i) I have attached a program, which solves the roots of a polynomial up to degree 10. It gives in the most cases all roots of a polynomial and uses the Lin-Bairstow-Method. RE: Calculating with complex numbers - euklides - 07-06-2022 Very good program, Bruno ! RE: Calculating with complex numbers - Kernelpanic - 07-06-2022 (07-06-2022, 07:07 AM)BSpinoza Wrote: @Kernelpanic: Excellent program, but unfortunately I have no idea anymore of the matter. I had to first look up again what a polynomial is. I misunderstood something in the description, one should be able to solve all quadratic equations with it. Rechnet man nicht mit reellen , sondern mit komplexen Zahlen, ist die Wurzel von -1 als i (imaginäreZahl) definiert. Die Darstellung einer komplexen Zahl ist dann Realteil + Imaginärteil. In diesem Kalkül können alle quadratischen Gleichungen gelöst werden. RE: Calculating with complex numbers - bplus - 07-06-2022 Yah, Solves quadratics, no calculating complex numbers except if root of the quadratic given in standard form with a, b, c as real numbers because again no complex calculation of b^2-4*a*c eg b = -i ??? RE: Calculating with complex numbers - triggered - 07-07-2022 "Yah, Solves quadratics, no calculating complex numbers except if root of the quadratic given in standard form with a, b, c as real numbers because again no complex calculation of b^2-4*a*c eg b = -i ???" Bplus is the only guy who can slowly turn a run-on sentence into a run-on question. |