Z^2=3-4i 求复数Z
人气：295 ℃ 时间：2020-04-15 14:21:06
解答
z=a+bia,b∈R则a²+2abi-b²=3-4i所以a²-b²=32ab=-4b=-2/aa²-4/a²=3a^4-3a²-4=0(a²-4)(a²+1)=0所以a²=4a=±2b=-2/a所以z=2-iz=-2+i
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