So f (x+2a) = f [(x+a)+a] = [f (x+a)-1]/[f (x+a)+1].

= {[f(x)- 1]/[f(x)+ 1]- 1 }/{[f(x)- 1]/[f(x)+ 1 }

Both numerator and denominator are multiplied by [f (x)+ 1]:

= {[f(x)- 1]-[f(x)+ 1]}/{[f(x)- 1]+[f(x)+ 1]}

=-2/[2 f(x)]=- 1/ f(x)，

That is f(x+2a)=-1/ f(x),

∴f(x+4a)= f[(x+2a)+2a]

=- 1/ f(x+2a)

= f(x)，

Then one period of f(x) is 4a.