Complex solutions to simple equations
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From: dragonman on 9 Nov 2009 05:55 On Nov 8, 11:10 pm, David Bailey <d...(a)removedbailey.co.uk> wrote: > dragonman wrote: > > I want the solutions to x^n=1 to appear in the form r(cos theta +isin > > theta) and then to graph them on an Argand diagram. Any advice given > > would be much appreciated. > > Didn't your teacher tell you how to do this, or give you a reading list? > > This board is not here to answer homework questions - what would be the > point? > > David Baileyhttp://www.dbaileyconsultancy.co.uk For goodness sake I am a teacher. Feel free to google me. Its Mathematica's representation of solutions I wanted. Feel free to apologise any time. Many thanks to the helpful posters above.
From: David Park on 10 Nov 2009 05:57 I generated a solution using Presentations. It should appear in a few days on Peter Linday's site: http://blackbook.mcs.st-and.ac.uk/~Peter/djmpark/html/ at the University of St Andrews School of Mathematics and Statistics: http://www.mcs.st-and.ac.uk/ There are both Mathematica notebooks and PDF files there for solutions to various MathGroup questions. The solution uses the ComplexPolar representation of complex numbers that is in Presentations. There are also routines for expanding ComplexPolar expressions and for converting to and from Cartesian form. This is often used in the introductory part of a complex analysis course and then dropped for the exponential representation. The polar notation is however quite useful in calculations, for displaying complex numbers and in Presentations for specifying iterators in polar plots. And, of course, it is also useful for didactic purposes. David Park djmpark(a)comcast.net http://home.comcast.net/~djmpark/ From: dragonman [mailto:morrisneedleman(a)gmail.com] I want the solutions to x^n=1 to appear in the form r(cos theta +isin theta) and then to graph them on an Argand diagram. Any advice given would be much appreciated.
From: AES on 10 Nov 2009 06:01 In article <hd8sf9$5vo$1(a)smc.vnet.net>, dragonman <morrisneedleman(a)gmail.com> wrote: > > For goodness sake I am a teacher. Feel free to google me. Its > Mathematica's representation of solutions I wanted. Feel free to > apologise any time. > Many thanks to the helpful posters above. A very nice reply! -- and one I'd hope may have some impact on this group. Out of curiosity, did your personal background or vocabulary (pre-Mathematica) include some conceptual understanding of any of the terms local vs global variables functional vs procedural programming mapping as they are used in Mathematica?
From: dragonman on 10 Nov 2009 06:02 On Nov 9, 9:55 pm, dragonman <morrisneedle...(a)gmail.com> wrote: > On Nov 8, 11:10 pm, David Bailey <d...(a)removedbailey.co.uk> wrote: > > > dragonman wrote: > > > I want the solutions to x^n=1 to appear in the form r(cos theta +is in > > > theta) and then to graph them on an Argand diagram. Any advice given > > > would be much appreciated. > > > Didn't your teacher tell you how to do this, or give you a reading list ? > > > This board is not here to answer homework questions - what would be the > > point? > > > David Baileyhttp://www.dbaileyconsultancy.co.uk > > For goodness sake I am a teacher. Feel free to google me. Its > Mathematica's representation of solutions I wanted. Feel free to > apologise any time. > Many thanks to the helpful posters above. I am a teacher.
From: DrMajorBob on 10 Nov 2009 06:09
RootPlotStyle cuts off some points at the edges of the plot, so I'll suggest a small improvement or two. RootPlotStyle[poly_, z_, ptsize_] := Module[{pts = {Re[z], Im[z]} /. NSolve[poly == 0, z], limits}, limits = (ptsize {-1, 1} + Through[{Min, Max}@#]) & /@ Transpose(a)pts; ListPlot[pts, AspectRatio -> Automatic, PlotStyle -> {PointSize[ptsize], Hue[0.8521]}, PlotRange -> limits] ] /; PolynomialQ[poly, z] n = 7; poly = z^n - 1; sol = Solve[poly == 0, z] /. (a_ -> b_) :> a -> ComplexExpand[b]; z /. sol RootPlotStyle[poly, z, .06] and n = 24; poly = z^n - 1; sol = Solve[poly == 0, z] /. (a_ -> b_) :> a -> ComplexExpand[b]; Grid(a)Partition[z /. sol // Simplify, 4] RootPlotStyle[poly, z, .06] Bobby On Mon, 09 Nov 2009 04:45:22 -0600, ynb <wkfkh056(a)yahoo.co.jp> wrote: > On 11=E6=9C=887=E6=97=A5, =E5=8D=88=E5=BE=8C8:48, dragonman > <morrisneedle.. .@gmail.com> wrote: >> I want the solutions to x^n=1 to appear in the form r(cos theta +isin >> theta) and then to graph them on an Argand diagram. Any advice given >> would be much appreciated. > > RootPlotStyle[poly_, z_] := > ListPlot[{Re[z], Im[z]} /. > NSolve[poly == 0, z], AspectRatio -> > Automatic, PlotStyle -> > {PointSize[0.06], Hue[0.8521]}] /; > PolynomialQ[poly, z] > > n = 7; > poly = z^n - 1; > sol = Solve[poly == 0, z] /. > (a_ -> b_) :> a -> ComplexExpand[b]; > z /. sol > RootPlotStyle[poly, z] > > n = 24; > poly = z^n - 1; > sol = Solve[poly == 0, z] /. > (a_ -> b_) :> a -> ComplexExpand[b]; > TableForm[z /. sol] > RootPlotStyle[poly, z] > -- DrMajorBob(a)yahoo.com |