复数题z1+z2=根号2,z1z
复数题z1+z2=根号2,z1z2=1，z1的22次方-z2的22次方的值为
因为z1z2=1
所以，z2=1/z1
则，z1+z2=z1+(1/z1)=√2
===> z1^2-√2z1+1=0
===> z1=(√2±√2i)/2
所以，z1=(√2+√2i)/2，z2=(√2-√2i)/2
或者，z1=(√2-√2i)/2，z2=(√2+√2i)/2
①当z1=(√2+√2i)/2，z2=(√2-√2i)/2时
z1=cos(π/4)+iain(π/4)，z2=cos(π/4)-isin(π/4)=cos(-π/4)+isin(π/4)
所以，z1^22-z2^22=[cos(π/4...全部
复数题z1+z2=根号2,z1z2=1，z1的22次方-z2的22次方的值为
因为z1z2=1
所以，z2=1/z1
则，z1+z2=z1+(1/z1)=√2
===> z1^2-√2z1+1=0
===> z1=(√2±√2i)/2
所以，z1=(√2+√2i)/2，z2=(√2-√2i)/2
或者，z1=(√2-√2i)/2，z2=(√2+√2i)/2
①当z1=(√2+√2i)/2，z2=(√2-√2i)/2时
z1=cos(π/4)+iain(π/4)，z2=cos(π/4)-isin(π/4)=cos(-π/4)+isin(π/4)
所以，z1^22-z2^22=[cos(π/4)+iain(π/4)]^22-[cos(-π/4)+isin(-π/4)]^22
=[cos(22π/4)+isin(22π/4)]-[cos(-22π/4)+isin(-22π/4)]
=2isin(22π/4)=2isin(22/4-4π)=2isin(3π/2)
=-2i
②当z1=(√2-√2i)/2，z2=(√2+√2i)/2时
z1=cos(π/4)-iain(π/4)=cos(-π/4)+isin(π/4)，z2=cos(π/4)+isin(π/4)
所以，z1^22-z2^22=[cos(-π/4)+iain(-π/4)]^22-[cos(π/4)+isin(π/4)]^22
=[cos(-22π/4)+isin(-22π/4)]-[cos(22π/4)+isin(22π/4)]
=-2isin(22π/4)=-2isin(22/4-4π)=-2isin(3π/2)
=2i
综上：z1^22-z2^22=±2i。
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