(z-5)/(z'-5)=(z-5)/(z-5)',令(z-5)=r(cosA+isinA),r>0,則(z-5)'=r(cosA-isinA),
(z-5)/(z'-5)=(cosA+isinA)/(cosA-isinA)=cos(2A)+isin(2A),故a=cos(2a),z=
(rcosA+5)+irsinA,|z|^2=9=(rcosA+5)^2+(rsinA)^2=r^2+10rcosA+16=0,
cosA=[(-r^2-16)/(10r)]-1,a=cos(2A)=2(cosA)^2-1=2[(r^2+16)/(10r)]^2-1
=(1/50)(r+(16)/r)^2-1>=(1/50)[2(root(r(16/r))]^2-1=64/50-1=7/25