Complex Function Iteration 2
Let $ f(z)= \frac{z+a}{z+b} $ and $ g(z)=f(f(z)) $, where $ a $ and $ b $ are complex numbers. Suppose that $ \left| a \right| = 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 $ \left| b \right| $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
After a bit of algebra, we obtain:
\[h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S},\]where $ P=(a+1)^2+a(b+1)^2 $, $ Q=a(b+1)(b^2+2a+1) $, $ R=(b+1)(b^2+2a+1) $, and $ S=a(b+1)^2+(a+b^2)^2 $. In order for $ h(z)=z $, we must have $ R=0 $, $ Q=0 $, and $ P=S $. The first implies $ b=-1 $ or $ b^2+2a+1=0 $. The second implies $ a=0 $, $ b=-1 $, or $ b^2+2a+1=0 $. The third implies $ b=\pm1 $ or $ b^2+2a+1=0 $.
Since $ |a|=1\neq 0 $, in order to satisfy all 3 conditions we must have either $ b=1 $ or $ b^2+2a+1=0 $. In the first case $ |b|=1 $. For the latter case, note that $ |b^2+1|=|-2a|=2 $, so $ 2=|b^2+1|\leq |b^2|+1 $ and hence $ 1\leq|b|^2\Rightarrow1\leq |b| $. On the other hand, $ 2=|b^2+1|\geq|b^2|-1 $, so $ |b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3} $.
Thus, $ 1\leq |b|\leq \sqrt{3} $. Hence, in any case the maximum value for $ |b| $ is $ \sqrt{3} $ while the minimum is $ 1 $ (which can be achieved in the instance where $ |a|=1,|b|=\sqrt{3} $ or $ |a|=1,|b|=1 $ respectively). The answer is then $ \boxed{\sqrt{3}-1} $.
Freeflows