Z^2=3-4i 求复数Z

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Z^2=3-4i求复数ZZ^2=3-4i求复数ZZ^2=3-4i求复数Zz=a+bia,b∈R则a²+2abi-b²=3-4i所以a²-b²=32ab=-4b=

Z^2=3-4i 求复数Z
Z^2=3-4i 求复数Z

Z^2=3-4i 求复数Z
z=a+bi
a,b∈R
则a²+2abi-b²=3-4i
所以a²-b²=3
2ab=-4
b=-2/a
a²-4/a²=3
a^4-3a²-4=0
(a²-4)(a²+1)=0
所以a²=4
a=±2
b=-2/a
所以z=2-i
z=-2+i

Z=6-8i

设z=a+bi
Z^2=3-4i
(a+bi)(a+bi)=3-4i
(a^2-b^2)+2abi=3-4i
a^2-b^2=3
2ab=-4
ab=-2
a=-2/b
代入：a^2-b^2=3
4/b^2-b^2=3
b=±1
a=±2
∴z=2-i或z=-2+i

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