根号a+bi
问题描述：
根号a+bi
求解根号下a+bi
答
设(x+yi)^2 = a+bi所以x^2-y^2+2xyi = a+bix^2-y^2 = a2xy = b设m = x^2,n = -y^2所以m+n =a,4mn = -b^2,即mn = -b^2/4这时要分情况讨论（根据判别式是否>0）,下面只说其中一种情况：假设a,b不全为0,设a^2+b^2 = c^2...