I'm trying to understand why Wolfram plots $y=\sqrt{x^2-4}$ in this way:

I did understand why the blue line is drawn in this way (the real part). What I didn't understand is why the imaginary part is drawn like this.

Note also that a complex function is in the form $f:\mathbb R^2\to \mathbb R^2$ and shouldn't be drawn in a plane.

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    $\begingroup$ Looking at the first graph, it appears that they're imagining it as a function $f:\mathbb{R}\rightarrow\mathbb{C}$, and treating the y-axis as the imaginary axis when referencing the orange line. Notice that the blue line remains at $0$ between $(-2,2)$, indicating that the real part is $0$ for those values. Honestly this is pretty cool and I have not seen it before. $\endgroup$ – Thoth Jan 5 '16 at 13:10

It looks like they are writing $y(x) = f(x) + ig(x)$ where $f$ and $g$ are real-valued functions of a single real variable $x$. You see the plots of the functions $f$ and $g$.

Note that your function appears to be $y:\mathbb R \to \mathbb C$, so you can indeed plot the real and imaginary parts as ordinary functions.


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