On the left, \(f(z)\), with a movable point \(a\).
The center shows \(f(z)-f(a)\). This shifts \(f(z)\) so that there is a root at \(a\).
In this center plot, if \(f\) is differentiable, we can see the the function appears to look like a linear function around \(a\):
It has a single, evenly distributed cycle of colors around the root at \(a\)
On the right is a plot of \(f'(z)\), with the value of \(f'(a)\) a marked on the graph.
We can interpret the value of \(f'(a)\) as a scaling and rotation from \(1\), indicated by the angle.
As you drag point \(a\), notice that in the center graph the angle formed from the original postive real axis color (red to white boundary in original coloring, yellow stripe in alternative coloring) to right is the same measure as the angle of \(f'(a)\).
This shows that the amount of 'twist' at point \(a\) is the same as \(f'(a)\).
However the direction of twist is reversed, because this is a domain color plot, rather than a range color plot. In a range color plot each point \(z\) stays the same color, but is transformed and moved to a new location \(f(z)\), so this shows the transformation directly. In a domain color plot, each point \(f(z)\) is left in its original location \(z\) and simply given a new color. Because the points are re-colored, rather than moved, points that are the same color don't correspond, and the transformation appears to be in the opposite direction.
When you drag point \(a\) on the left graph, the remaining graphs will update to show the derivative at the new \(a\).
Please be patient, this update may take some time.
Enter your function here:
\(f(z)= \) →
\(r \in\) [0,]
\(r\) scale