User blog comment:Sbiis Saibian/Googology101 - Part II/@comment-10429372-20141024190431/@comment-5982810-20141025182327

@LP9

Now to try an answer your question. In the case of 10^z we may write it as: e^(z*ln10) Let z be the complex number x+i*y. In this case we have e^(xln10+i*yln10). The magnitude of this complex number will be e^(xln10) or just 10^x (10 to the power of the real part). What is the magnitude of x+y*i? Well we just use the Pythagorean theorem here and say that it's sqrt(x^2+y^2). So what if we want z and 10^z to be the same distance to zero (but not necessarily headed in the same direction). Then we just need the magnitudes to be equal. So we need 10^x = sqrt(x^2+y^2) This is equivalent to saying Solving for y we obtain y = sqrt(100^x-x^2). What this means is that for real "x" we can compute an imaginary part "y" that will have the property that 10^z will have the same absolute value as z. All fixed points of 10^z = z must be points on the curve of y = sqrt(100^x-x^2) or y = -sqrt(100^x-x^2). Hence Vel's graph.
 * 10^z | = | z |

Why are there some solutions? Because as x-->inf we have that y-->inf. Remember that "y" represents the angle portion of 10^(x+i*y). So every time the angle goes around about 2*pi radians the angle matches up with the angle of x+i*y, and the result is a fixed point. But the y component is sqrt(100^x-x^2) which is approximately 10^x, so this means the angles rotational speed increases exponentially with respect to x. The result is that fixed points will get denser and denser as we fan out (as also seen in Vel's graph).

That's what I was able to gather from my analysis.