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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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    [+1] Funny! Unfortunately not bijective. Never seen it before... – 2017-02-11
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    With $f(z)=z^3$ it's even nicer. What is it that's not bijective? – 2017-02-11
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    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](http://imgur.com/a/ZVDza) is the plot of the function you discussed. The zeroes are very clear from this diagram. – 2017-02-11
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    By "not bijectiv"e I meant that from a given reprentation you cannot retrieve a unique function $Z=f(z)$. – 2017-02-11
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    ah, that's really unfortunate – 2017-02-11
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    @AgnishomChattopadhyay yes, I am aware of the domain coloring method but I find that this way of seeing the real part of $f$ gives more insight...to me, at least – 2017-02-11

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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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    Yes, but I guess I'm plotting just $f$, not $|f|$ – 2017-02-12
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    @marcotrevi sure, but they are the same thing when the map is restricted to this domain. – 2017-02-12
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    no, it's not true- take $f(x)=x^3$. Then $|f|$ is always non negative while $f(-2)=-8$... – 2017-02-12