Problem:
Let f(z)=z+bz+a and g(z)=f(f(z)), where a and b are complex numbers. Suppose that ∣a∣=1 and g(g(z))=z for all z for which g(g(z)) is defined. What is the difference between the largest and smallest possible values of ∣b∣ ?
Therefore either B=C=0 (and A=D ) or A+D=0. In the former case b=−1,f(z)=z−1z+a, and g(z)=1+a(1+a)z=z, as required, unless a=−1. (Note that a=−1 in this case would imply f(z)=1, which contradicts g(g(z))=z.)
In the latter case 1+2a+b2=0, so ∣b∣2=∣2a+1∣. Because ∣a∣=1, the triangle inequality yields
1=∣2∣a∣−1∣≤∣2a+1∣≤2∣a∣+1=3
so 1≤∣b∣≤3. The minimum ∣b∣=1 is attained when a=−1 and b=1 (or as above, when b=−1 ). The maximum ∣b∣=3 is attained when a=1 and b=±3i. The required difference is 3−1.
Note: The conditions imply that a lies on the unit circle in the complex plane, so 2a+1 lies on a circle of radius 2 centered at 1 . The steps above are reversible, so if b2=−1−2a, then g(g(z))=z (unless a=b=−1 ). Therefore b2 can be anywhere on the circle of radius 2 centered at -1 , and ∣b∣ can take on any value between 1 and 3.