Question

设z^2=(x+yi)^2=5-12i(x,y∈R),求z

Answer 1

（x+yi)^2=x^2-y^2+2xyi=5+12i,

由复数相等的条件得

x^2-y^2=5,①

2xy=12,②

①*6-②*5/2,6x^2-5xy-6y^2=0,

∴x=3y/2,或x=-2y/3.

分别代入②,得y^2=4,或y^2=-9(无实根）,

∴y1=2,y2=-2;

对应的x1=3,x2=-3.