# Plotting solutions of systems in complex plane

I'm interested in plotting solutions of systems of polynomials in the complex plane, in the following way.

Let $f_1,\ldots f_s$ be complex polynomials in variables $x_1,\ldots x_n$. I know how to find the solutions using a code like this

eqs = {x^2 + y^3 + 1 == 0, x + y^2 - 1 == 0}
pts = Solve[eqs, {x, y}]


But I'm having some trouble to plot the values of $x,y$ n the complex plane. I tried some naive thing like

ListPlot[pts]


but nothing works. I'm lost, any help is welcome, thanks.

• Get them using {x, y} /. % Apr 8, 2015 at 3:19
• I forgot a little detail in my answer, sorry. It's corrected. Apr 8, 2015 at 3:21
• A math problem: If you have ${x,y}$ as a point and $x$ and $y$ are complex, how does one show them on complex plane? Apr 8, 2015 at 3:30
• $x$ is one point and $y$ is another, their are not considered as coordinates. Apr 8, 2015 at 3:33
• Nov 21, 2015 at 19:34

You have

eqs = {x^2 + y^3 + 1 == 0, x + y^2 - 1 == 0};
pts = Solve[eqs, {x, y}];


To get the data points and the plot:

data = {x, y} /. pts // Flatten;
ListPlot[{Re[#], Im[#]} & /@ data]

• I think it's perfect! Apr 8, 2015 at 3:45
• Do you mind just explain a little (or recommend some reading) what #,&,/ and @ stands for? Apr 8, 2015 at 3:47
• Thanks! /@ is shorthand for Map. It basically maps {Re[#], Im[#]} over all elements in data one by one. This page is really good to start! Apr 8, 2015 at 3:53

Starting from version 10.1 you can use ReIm:

eqs = {x^2 + y^3 + 1 == 0, x + y^2 - 1 == 0};
pts = {x, y} /. Solve[eqs, {x, y}];

ListPlot[ReIm[pts]]


• Or even ListPlot@ReIm@Values@Solve[eqs, {x, y}]. Nov 21, 2015 at 14:35
• @MichaelE2, hi, please let me know what do you think about this question : mathematica.stackexchange.com/questions/206901/… . Do you think it’s some how related?
– S.S.
Sep 26, 2019 at 15:10
• I have Mathematica 9.
– S.S.
Sep 26, 2019 at 15:12