复数z=a+bi(a,br)满足z的平方=4+3i,求z

复数z=a+bi(a,br)满足z的平方=4+3i,求z
复数z=a+bi(a,br)满足z的平方=4+3i,求z

复数z=a+bi(a,br)满足z的平方=4+3i,求z
(a+bi)²=4+3i
a²-b²+2abi=4+3i
a²-b²=4
2ab=3,
2得b=3/(2a)
将b代入1:a²-9/(4a²)=4
4(a²)²-16a²-9=0
(2a²-9)(2a²+1)=0
得a²=9/2
故a=±3/√2,
b=3/(2a)=±1/√2
故z=±(3+i)/√2