若复数(a-i)^2在复数平面内对应的点在y轴负半轴上,则实数a的值
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若复数(a-i)^2在复数平面内对应的点在y轴负半轴上,则实数a的值
若复数(a-i)^2在复数平面内对应的点在y轴负半轴上,则实数a的值
若复数(a-i)^2在复数平面内对应的点在y轴负半轴上,则实数a的值
(a-i)^2=a^2-2ai+i^2
=(a^2-1)-2ai
此数在y轴负半轴上
则有:a^2-1=0 (1)
-2a0
所以 a=1
(a-i)平方=a^2-2ai i^2 i^2=-1
a^2-2ai-1 在Y负半轴 a^2-1=0
a=±1 -2a应<0 所以a=1
The answer is a=1. Imagine the point $A$ moves along the x-axis, from the positive infinity to the negative infinity. Then the vector a-i moves in the lower half-plane from right to left. In order to ...
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The answer is a=1. Imagine the point $A$ moves along the x-axis, from the positive infinity to the negative infinity. Then the vector a-i moves in the lower half-plane from right to left. In order to make its square being on the negative y-axis, one has to take a-i to be of angle -\pi/4, that is, a=1. This completes the solution.
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