# Plot a set of complex numbers with given argument and absolute value bounds

I want to plot the following complex numbers $$z \in \text{(complex numbers)}:\pi/4 < \arg (z) \leq 5 \pi/4,\ 1 \leq |z| < 2$$

I don't know how to graph it so that it would look like 2D without any unecessary details. The closest approach I found is parametric function plotting. I tried to use ContourPlot to graph it, but I just can't seem to do it...

ContourPlot[
Im[F[z[x, y]]],
{3 pi/4 < arg (z) <= 5 pi/4,
1 <= abs (z) < 2},
{x, -.2, .2}, {y, -.2, .2},
PlotRange -> All,
Contours -> Range[-5, 5, .5],
ContourLabels -> True
]


Does anybody know how to graph my set?

With correct Mathematica syntax and increased range of x,y RegionPlot solves your problem:

RegionPlot[ 3 Pi/4 < Arg[x + I y] <= 5 Pi/4 && 1 <= Abs [x + I y] < 2 , {x, - 2,2}, {y, - 2, 2} ]


Thanks to the comment @Bili Debili: Arg returns angle in the range -Pi...Pi, that's why the condition 3 Pi/4 < Arg[x + I y] <= 5 Pi/4 has to be changed

RegionPlot[3 Pi/4 < Abs[Arg[ x + I y]] <= Pi &&1 <= Abs [x + I y] < 2 , {x, - 2, 2}, {y, - 2, 2} ,FrameLabel -> {x, y}]


• Shouldn't the function also go into the third quadrant ?
– VLC
Nov 21, 2020 at 14:07
• That's right, thanks. Arg[…] lies in the range -Pi,Pi Nov 21, 2020 at 14:18
ParametricPlot[
ReIm[r*Exp[I*θ]], {θ, π/4, (5 π)/4}, {r, 1, 2},
MeshFunctions -> {#3 &, #4 &},
Mesh -> {{{π/4, {Thick, Blue, Dashed}}, {(5 π)/
4, {Thick, Blue}}}, {{1, {Thick, Red}}, {2, {Thick, Red,
Dashed}}}}, BoundaryStyle -> None, PlotStyle -> Yellow]


One more way is as follows.

ComplexRegionPlot[ Pi/4 < Arg [z] <= 5 Pi/4 && 1 <= Abs [z] < 2, {z, -2 - 0*I, 2 + 2*I},AspectRatio->Automatic]