z*z的共轭复数+2i*z的共轭复数=3+ai,求z,用a表示
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解决时间 2021-11-30 09:02
- 提问者网友：佞臣
- 2021-11-29 12:28
z*z的共轭复数+2i*z的共轭复数=3+ai,求z,用a表示
最佳答案
- 五星知识达人网友：北方的南先生
- 2021-11-29 13:19
设z=x+yi
z*z=(x+yi)²=x²-y²+2xyi
z*z 的共轭复数为 x²-y²-2xyi
z 的共轭复数为x-yi 代入
x²-y²-2xyi+2i(x-yi)=x²-y²-2xyi+2xi+2y
=x²-y²+2y+2x(1-y)i
=3+ai
则有
x²-y²+2y=3
2x(1-y)=a
x =
√(4+2√(4+a²))/2
-√(4+2√(4+a²))/2
√(4-2√(4+a²))/2
-√(4-2√(4+a²))/2
y =
(a+2*(4+2*(4+a^2)^(1/2))^(1/2)-1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4+2*(4+a^2)^(1/2))^(1/2)+1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a+2*(4-2*(4+a^2)^(1/2))^(1/2)-1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4-2*(4+a^2)^(1/2))^(1/2)+1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
z*z=(x+yi)²=x²-y²+2xyi
z*z 的共轭复数为 x²-y²-2xyi
z 的共轭复数为x-yi 代入
x²-y²-2xyi+2i(x-yi)=x²-y²-2xyi+2xi+2y
=x²-y²+2y+2x(1-y)i
=3+ai
则有
x²-y²+2y=3
2x(1-y)=a
x =
√(4+2√(4+a²))/2
-√(4+2√(4+a²))/2
√(4-2√(4+a²))/2
-√(4-2√(4+a²))/2
y =
(a+2*(4+2*(4+a^2)^(1/2))^(1/2)-1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4+2*(4+a^2)^(1/2))^(1/2)+1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a+2*(4-2*(4+a^2)^(1/2))^(1/2)-1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4-2*(4+a^2)^(1/2))^(1/2)+1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
全部回答
- 1楼网友：玩世
- 2021-11-29 14:01
y =
(a+2*(4+2*(4+a^2)^(1/2))^(1/2)-1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4+2*(4+a^2)^(1/2))^(1/2)+1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a+2*(4-2*(4+a^2)^(1/2))^(1/2)-1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4-2*(4+a^2)^(1/2))^(1/2)+1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
(a+2*(4+2*(4+a^2)^(1/2))^(1/2)-1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4+2*(4+a^2)^(1/2))^(1/2)+1/4*(4+2*(4+a^2)^(1/2))^(3/2))/a
(a+2*(4-2*(4+a^2)^(1/2))^(1/2)-1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
(a-2*(4-2*(4+a^2)^(1/2))^(1/2)+1/4*(4-2*(4+a^2)^(1/2))^(3/2))/a
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