1.已知z=x+yi,且z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z=3,求|z|的最大值,和复数z的实部与虚部和最小值2.若x,y满足x^2+(y-1)^2=1,不等式x+y+m≥0恒成立,求m的范围
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![1.已知z=x+yi,且z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z=3,求|z|的最大值,和复数z的实部与虚部和最小值2.若x,y满足x^2+(y-1)^2=1,不等式x+y+m≥0恒成立,求m的范围](/uploads/image/z/8502732-36-2.jpg?t=1.%E5%B7%B2%E7%9F%A5z%3Dx%2Byi%2C%E4%B8%94z%E4%B9%98%E4%BB%A5%E5%85%B1%E8%BD%AD%E5%A4%8D%E6%95%B0z%2B%EF%BC%881-2i%EF%BC%89z%2B%EF%BC%881%2B2i%EF%BC%89%E5%85%B1%E8%BD%AD%E5%A4%8D%E6%95%B0z%3D3%2C%E6%B1%82%7Cz%7C%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E5%92%8C%E5%A4%8D%E6%95%B0z%E7%9A%84%E5%AE%9E%E9%83%A8%E4%B8%8E%E8%99%9A%E9%83%A8%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC2.%E8%8B%A5x%2Cy%E6%BB%A1%E8%B6%B3x%5E2%2B%28y-1%29%5E2%3D1%2C%E4%B8%8D%E7%AD%89%E5%BC%8Fx%2By%2Bm%E2%89%A50%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82m%E7%9A%84%E8%8C%83%E5%9B%B4)
1.已知z=x+yi,且z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z=3,求|z|的最大值,和复数z的实部与虚部和最小值2.若x,y满足x^2+(y-1)^2=1,不等式x+y+m≥0恒成立,求m的范围
1.已知z=x+yi,且z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z=3,求|z|的最大值,和复数z的实部与虚部和
最小值
2.若x,y满足x^2+(y-1)^2=1,不等式x+y+m≥0恒成立,求m的范围
1.已知z=x+yi,且z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z=3,求|z|的最大值,和复数z的实部与虚部和最小值2.若x,y满足x^2+(y-1)^2=1,不等式x+y+m≥0恒成立,求m的范围
z乘以共轭复数z+(1-2i)z+(1+2i)共轭复数z
=x²+y²+(1-2i)(x+yi)+(1+2i)(x-yi)
=x²+y²+x+2y-(2x-y)i+x+2y+(2x-y)i
=x²+y²+2x+4y
=3
∴x²+y²+2x+4y=3
(x+1)²+(y+2)²=8
z=x+yi在以(-1,-2)为圆心,2√2为半径的圆上
|z|=√(x²+y²)
表示为圆上到原点的距离
∵原点在圆内部
∴圆心到原点距离
=√5
∴|z|最大值=√5+2√2
设z=x+y
y=-x+z
当y=-x+z与圆相切时
z有最大值,最小值
最小值=-7(此时x=-3,y=-4)
(2)
x^2+(y-1)^2=1
以(0,1)为圆心,半径=1的圆
令圆x^2+(y-1)^2=1与直线x+y+m=0相切
即圆心(0,1)到直线x+y+m=0的距离为1
|1+m|/√2=1,m=-1+√2,m=-1-√2,
结合图像可知,当m≥-1+√2时,
x+y+m≥0恒成立