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If z satisfies |z-1|<|z+3| then omega=2z...

If `z` satisfies `|z-1|<|z+3|` then `omega=2z+3-i,` (where `i=sqrt-1`) sqtisfies

A

`abs(omega-5-i) lt abs(omega+3+i)`

B

`abs(omega-5) lt abs(omega+3)`

C

`Im(iomega) gt 1`

D

`abs(arg(omega-1)) lt (pi)/(2)`

Text Solution

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The correct Answer is:
B, C, D
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