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Question

Let z and ω two complex number such that |z|1,|ω|1 and |ziω|=|zi¯¯¯ω|=2, then z equals to

A
1 or i
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B
i or i
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C
1 or 1
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D
i or 1
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Solution

The correct option is A i or i
Let z=x+iy and ω=a+ib
|ziω|=2 and |ziω|=2
|x+iyi(a+ib)|=2;|x+iyi(a+ib)|=2
|(x+b)+i(ya)|=2;|(xb)+i(ya)|=2
(x+b)2+(ya)2=2;(xb)2+(ya)2=2
(xb)2+(ya)2=4=(x+b)2+(ya)2
(xb)2=(x+b)2
(xb)=(x+b) or xb=(x+b)
b=0 or x=0
if b = 0
ω=a not a complete number.
so x = 0
z=iy
z=±p(|z||6|)

1097572_1174856_ans_63551cdb8799458da490e14c4cb85e51.png

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