Supremum of all y-coordinates of the Mandelbrot set Let $M\subset \mathbb R^2$ be the Mandelbrot set. What is $\sup\{ y : (x,y) \in M \}$? Is this known?
To be more descriptive: What is the supremum of all y-coordinates of all black points in the following picture:

Picture File:Mandel zoom 00 mandelbrot set.jpg by Wolfgang Beyer licensed under CC-BY-SA 3.0
 A: Robert Munafo's Mu-ency site calls this the northernmost point of the Mandelbrot set, giving the coordinate as  $-0.207107867093967+1.122757063632597i$. A quick search on Google and OEIS turns up no references.
A: I made a supremum image page of 31 images leading to the conjectured Supremum point.
$f(x)=x^2+C$.  If we take C=Robert Munafo's point, we can see that perhaps this point, is the point where $f^{14}(0)=-f^{1}(0),\;\;f^{15}(0)=f^{2}(0)$, where $C\approx -0.207107867093967+1.122757063632597i$ which leads to $f^{n}(0)$ repeating with a period of 13, after the preperiod.  Then the point C is one of the zeros of the polynomial with 2^13 terms.  It is the solution nearest that point.  We numerically estimate the zero iterating with Newton's method, since the polynomial is too large to work with.  The result is printed to 60 decimal digits.  Because this solution eventually repeats, by definition the point never escapes to infinity so it is a member of the Mandelbrot set.  Such preperiodic points are called Misiurewicz points, and are algebraic numbers.
$$C \approx 
-0.207107867093967732893764544285894983866865721506089742782655+
1.12275706363259748461604158116265882079904682664638092967742i$$
Here is an image vertically centered on that point, $C \pm 10^{-28}i$, with a small green box at the center of the image.  From appearances, the point C might be the "top" of the Mandelbrot...  If so, than the nothernmost point is that Misiurewicz point.  But I have no idea how to prove it.  We do know that as we zoom into the Mandelbrot at a Misiurewicz point, that in the limit, zooming in is self-similar rather than increasingly chaotic.  For the point in question, zooming in about $10^6$ seems to be self similar, so that an image zoomed in by $10^{-22}$ or by $10^{-28}$ or by $10^{-34}$ would all look nearly identical to this image. Also there is apparently no rotational component in the self similarity; I think no rotation would be required if this Misiurewicz is the top and it would be nice to show that there is no rotation;  see the supremum image page I made.
There is Wolf Jung's Mandel program, available at http://www.mndynamics.com/ which has some nice tutorials on Misiurewicz points.

