設z=m+ni
|z|≤1
m²+n²≤1
z+z共軛|z|=a+bi
m+ni+(m-ni)√(m²+n²)=a+bi
[1+√(m²+n²)]m+[1-√(m²+n²)]ni=a+bi
a=[1+√(m²+n²)]m
b=[1-√(m²+n²)]n
ab=(1-m²-n²)mn
由均值不等式得mn≤(m²+n²)/2
ab≤[1-(m²+n²)](m²+n²)/2=(-1/2)[(m²+n²)²-1/2]²+1/8
當m=n=1/4時,ab有最大值(ab)max=1/8