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Question

If z and w be two complex numbers such that |z|1,|w|1 and |z+iw|=|zi¯¯¯¯w|=2, then

A
number of solutions for z is 1
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B
number of solutions for z is 2
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C
number of solutions for w is 1
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D
number of solutions for w is 2
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Solution

The correct option is D number of solutions for w is 2
|z+iw||z|+|iw|
2|z|+|i||w|
|z|+|w|2
But from given condition, |z|+|w|2
|z|=|w|=1

Let z=cosθ1+isinθ1 and w=cosθ2+isinθ2 (|z|=|w|=1)|z+iw|=2|(cosθ1sinθ2)+i(sinθ1+cosθ2)|=2cos2θ1+sin2θ22sinθ2cosθ1+sin2θ1+cos2θ2+2sinθ1cosθ2=42(sinθ1cosθ2sinθ2cosθ1)=2sin(θ1θ2)=1θ1θ2=π2,3π2 (1)θ1,θ2(π,π]θ1±θ2(2π,2π]

|zi¯¯¯¯w|=2|(cosθ1sinθ2)+i(sinθ1cosθ2)|=2
cos2θ1+sin2θ22sinθ2cosθ1+sin2θ1+cos2θ22sinθ1cosθ2=4
2(sinθ1cosθ2+sinθ2cosθ1)=2
sin(θ1+θ2)=1
θ1+θ2=π2,3π2 (2)

Solving equation (1) and (2)
(θ1,θ2)=(π,π2),(0,π2),(0,3π2),(π,π2)
θ1,θ2(π,π]
(θ1,θ2)=(π,π2),(0,π2)

Case-1:
θ1=π,θ2=π2
z=1,w=i

Case-2:
θ1=0,θ2=π2
z=1,w=i

Hence, there are 2 possibilities for z and w.

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