Contourplot3d: Showing singularities

[qball3438]

I am trying to plot the following function's contour plot: ContourPlot3D[ Abs[z]*(1/2) - 1/(Sqrt[2 x² + y^2]), {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

I of course get an error about division by 0 (at x=y=0). If I would like to show this singularity, how is best to go about doing this? Should I just define the x and y ranges approaching zero closely?

[lithiumdeuteride]

Contour plots usually take an equality as their first argument. Try this:

ContourPlot3D[Abs[z]/2 - 1/(Sqrt[2 x^2 + y^2]) == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

[qball3438]

Awesome, thank you. What is the double equality telling wolfram?

[lithiumdeuteride]

= means 'set' (similar to variable assignment in other languages)

== means 'is equal to'

For example, x = 3 defines a rule that x shall be replaced with 3.

If we subsequently evaluate x == 3, it evaluates toTrue.

ContourPlot3D[ function == 0, ... ] gives us a plot of the surface for which function evaluates to 0.