5

Try ComplexListPlot: a1 = (10/7)*(Cos[Pi/10] + I*Sin[Pi/10]); ComplexListPlot[Table[Callout[a1^k, k], {k, 0, 12}], AxesLabel -> {R, I}, ImageSize -> Large] If you use Callout[N @ a1^k] instead of Callout[a1^k, k] you get:


3

Generate the graphics primitives for Prolog using ListPlot with same plot markers: marker = {{Graphics[{Disk[]}], 0.03}}; ListPlot[Table[{x, x}, {x, 1, 5, 0.1}], Prolog -> {First @ ListPlot[{{4, 5}}, PlotStyle -> Black, PlotMarkers -> marker]}, PlotMarkers -> marker]


2

labels = Style[#, 16] & /@ StringPadRight[#, StringLength@# + 3] & /@ {"Latin America and Caribbean (LAC)", "Europe and Central Asia (ECA)", "Middle East and North Africa (MENA)", "South Asia (SAR)", "Africa (AFR)", "East Asia and Pacific (EAP)", "Organization ...


1

{a, b, c} = {5/2 y + 3 x <= 3 && y < 2/3 && 5/2 y + x <= 2, y < 13/12 - 13/12 x && y < 13/24, y <= 1 - x && y < 2/3}; RegionPlot[{ Callout[a, "a corner", {1/2, 3/5} + {.1, 0.3}, {1/2, 3/5}, CalloutMarker -> "Arrow", Background -> Blue, LabelStyle -> White], ...


1

Something like this? r1 = RegionPlot[ 5/2 y + 3 x <= 3 && y < 2/3 && 5/2 y + x <= 2, {x, 0, 1}, {y, 0, 1}, PlotStyle -> {Opacity[0.2], Yellow}, PlotLabels -> Placed["First region", {.8, .3}]]; r2 = RegionPlot[ y < 13/12 - 13/12 x && y < 13/24, {x, 0, 1}, {y, 0, 1}, PlotStyle -> {...


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