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Question

Suppose z and ω are two complex numbers such that |z|≤1,|ω|≤1, and |z+iω|=|z−iω|=2.Which of the following is true for z and ω?

A
Re(z)=Re(ω)
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B
Im(z)=Im(ω)
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C
Re(z)=Im(ω)
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D
Im(z)=Re(ω)
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Solution

The correct option is D Im(z)=Re(ω)Suppose z=x+iy,ω=α+iβ|z+iω|=2⇒|z|2+|ω|2+iω¯¯¯z−i¯¯¯ωz=4⇒iω¯¯¯z=i¯¯¯ωz=2 ......... (1)Also |z−i¯¯¯ω|=2⇒|z|2+|ω|2+iωz−i¯¯¯ω¯¯¯z=4⇒iωz−i¯¯¯ω¯¯¯z=2Adding (1) and (2), we get⇒i(ω+¯¯¯ω)(z−¯¯¯z)=4⇒i(2iβ)(2x)=4βx=−1....... (3)Subtracting (1) and (2), we geti(ω+¯¯¯ω)(z−¯¯¯z)=0⇒αy=0........ (4)⇒ either α=0 or y=0If y=0, then x2+y2=1⇒x=±1 and z=1 or −1If α=0 the $\alpha^2+\beta^2=1\Rightarrow \beta =\pm 1\Rightarrow \omega =\pm i$So Im(z)=Re(ω)=0

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