Help Understanding Complex Roots I was reading a graphical explanation of complex roots, and between Figures 7 and 8 I became confused. 



The roots appear in the imaginary plane, but I don't understand why the original function must be inverted before the graphical representation fits. I realize that inversion is necessary for the graphical representation of imaginary roots to fit, but the lack of explanation bothers me. I would like to understand this inverstion better than "we do it because it fits the data." 
 A: Comment: In general, a nonzero quadratic has the form $y = A(x-a)^2+b^2$ where $A,a,b$ are real constants, but I set $A=1$ for the sake of clarity.
The graph is given to extend $y=f(x) = (x-a)^2+b^2$ to $y=Re(f(z)) = Re((z-a)^2+b^2)$ where $z = x+it$ is a complex variable. The graph given indicates all points in $(t,x,y)$ space such that the equation $y=Re(f(z))$ is satisfied. In particular,
$$ y = Re((x+it-a)^2+b^2) = (x-a)^2-t^2+b^2  $$
Considering $x=a$ which is a vertical plane in the pictures,
$$ y = -t^2+b^2 $$
This is a parabola which opens downward in the $x=a$ plane. It has $y=0$ when $t = \pm b$. This is why the graph opens downward. On the other hand, the plane $t=0$ (otherwise known as the $xy$-plane) we have $y = (x-a)^2+b^2$ which is a parabola with vertex $(a,b^2)$ which opens upward with no $x$-intercept.
The function is not inverted, it is extended to a complex domain, in this larger context the three-dimensional graph looks as pictured. However, beware the real picture casts $y = Re(f(x+it))$ as a surface, it is one-equation in three variables. This gives two parameters at most points on its solution set. The picture you post focuses on two interesting curves on this surface (it has parametrization $X(u,v) = (u,v,Re(f(u+iv)))$ )
Also, I was tempted to say it was the graph of $y = f(x+it)$, but that is more difficult to graph directly... think about it.
