复数题目
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解:
令z=cosa+sina i
w=(cosa+sina i)^2-i+1
=(cosa)^2+2cosasina i-(sina)^2-i+1
=cos(2a)+1+[sin(2a)-1]i
|w|=√[(cos(2a)+1)^2+(sin(2a)-1)^2]
=√[(2cos(2a)-2sin(2a)+3]
=√[2√2cos(2a+π/4)+3]
-1≤cos(2a+π/4)≤1
3-2√2≤2√2cos(2a+π/4)+3≤3+2√2
3-2√2≤|w|≤3+2√2
若z属于C,|z|=1,复数w=z^2-i+1,则|w|的取值范围是( [3-2√2,3+2√2] )
令z=cosa+sina i
w=(cosa+sina i)^2-i+1
=(cosa)^2+2cosasina i-(sina)^2-i+1
=cos(2a)+1+[sin(2a)-1]i
|w|=√[(cos(2a)+1)^2+(sin(2a)-1)^2]
=√[(2cos(2a)-2sin(2a)+3]
=√[2√2cos(2a+π/4)+3]
-1≤cos(2a+π/4)≤1
3-2√2≤2√2cos(2a+π/4)+3≤3+2√2
3-2√2≤|w|≤3+2√2
若z属于C,|z|=1,复数w=z^2-i+1,则|w|的取值范围是( [3-2√2,3+2√2] )
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