Trying to explain to a non-mathematician the meaning of complex numbers, I came across a nice geometric intuition for "seeing" the complex zeros of a function. Suppose $f:\mathbb{C}\longrightarrow\mathbb{C}$ is a function. Then \begin{equation} f(x+iy) = A(x,y) + iB(x,y) \end{equation} The problem is that we cannot graph $f$ because we have a 3-dimensional constrained visual understanding. But if we consider the locus $\Gamma$ of points on the plane such that $B(x,y)=0$, we can actually plot a graph with the information about the zeros of $f$. On $\Gamma$ the function takes only real values so we can plot $f$ as a height map over $\Gamma$ in $3D$ space.

For example, if $f(x+iy)=(x+iy)^2$, we have that \begin{equation} f(x+iy) = x^2-y^2 + 2ixy \end{equation} and setting $2xy=0$ gives us the union of the x-axis and the y-axis. Plotting $f$ over this set gives the union of two parametric curves: \begin{equation} (t,0,t^2)\qquad\text{and}\qquad (0,t,-t^2) \end{equation} If we now consider the function $f(x+iy)=(x+iy)^2 + c$ we can see that what the parameter $c$ does is to shift these parametric curves up and down along the $z$-axis. The intersection of these curves with the $z=0$ plane shows how the zeros of the function move around in a very visual way.

In short, setting the imaginary part of a complex function equal to zero defines the locus of points on $\mathbb{C}$ or on the plane where the function takes only real values, letting us see that part of the function as a height map.

Question: does this geometric locus have a name? Is it used in some way somewhere?

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    $\begingroup$ [+1] Funny! Unfortunately not bijective. Never seen it before... $\endgroup$ – Jean Marie Feb 11 '17 at 16:50
  • $\begingroup$ With $f(z)=z^3$ it's even nicer. What is it that's not bijective? $\endgroup$ – marco trevi Feb 11 '17 at 16:56
  • $\begingroup$ You might want to check out the domain coloring method. It uses hue to indicate the argument of the complex value and brightness for the magnitude. For example, here is the plot of the function you discussed. The zeroes are very clear from this diagram. $\endgroup$ – Agnishom Chattopadhyay Feb 11 '17 at 17:13
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    $\begingroup$ By "not bijectiv"e I meant that from a given reprentation you cannot retrieve a unique function $Z=f(z)$. $\endgroup$ – Jean Marie Feb 11 '17 at 17:24
  • $\begingroup$ ah, that's really unfortunate $\endgroup$ – marco trevi Feb 11 '17 at 17:25

Deeparaj Bhat gave me the following idea:

Because all such values which essentially have no imaginary part are real, we can consider them to be in $\mathbb{R}$. Hence, what we are looking for is the pre-image of $\mathbb{R}$ and can be denoted $f^{-1}(\mathbb{R})$ with a slight abuse of notation.

Then, you could think of your plot as the plot of $|f(.)|$ restricted to $f^{-1}(\mathbb{R})$

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  • $\begingroup$ Yes, but I guess I'm plotting just $f$, not $|f|$ $\endgroup$ – marco trevi Feb 12 '17 at 9:07
  • $\begingroup$ @marcotrevi sure, but they are the same thing when the map is restricted to this domain. $\endgroup$ – Agnishom Chattopadhyay Feb 12 '17 at 9:22
  • $\begingroup$ no, it's not true- take $f(x)=x^3$. Then $|f|$ is always non negative while $f(-2)=-8$... $\endgroup$ – marco trevi Feb 12 '17 at 11:11

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