令z1=a+bi z2=c+di
|z1+z2|=|a+c+(b+d)i|=√[(a+c)²+(b+d)²]
|z1-z2|=|a-c+(b-d)i|=√[(a-c)²+(b-d)²]
∴(a+c)²+(b+d)²=(a-c)²+(b-d)²
即2ac+2bd=-2ac-2bd
ac+bd=0
(z1/z2)²=[(a+bi)/(c+di)]²=[(a+bi)(c-di)/(c²+d²)]²
=[(ac+bd+bci-adi)/(c²+d²)]²
=[(bc-ad)i/(c²+d²)]²
=-(bc-ad)²/(c²+d²)²