高二数学，复数

(1)z+2πi=x+(y+2π)i
(z+2πi)*=a^x(cos(y+2π)+isin(y+2π))=a^x(cosy+isiny)=z*
(2)设z1=x1+y1i,z2=x2+y2i,则z1+z2=(x1+x2)+(y1+y2)i
(z1+z2)*=a^(x1+x2)*(cos(y1+y2)+isin(y1+y2))
z1*=a^x1*(cosx1+isiny1),z2*=a^x2*(cosx2+isiny2)
z1*×z2*=a^(x1+x2)*(cosy1cosy2+icosy1siny2+isiny1cosy2-siny1siny2)
=a^(x1+x2)*[(cosy1cosy2-siny1siny2)+i(siny1cosy2+cosy1siny2)]
=a^(x1+x2)*[cos(y1+y2)+isin(y1+y2)]=(z1+z2)*
超级简单,你就是懒得写我跟你讲.