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⇒|(cosθ1−sinθ2)+i(sinθ1+cosθ2)|=2⇒cos2θ1+sin2θ2−2sinθ2cosθ1+sin2θ1+cos2θ2+2sinθ1cosθ2=4⇒2(sinθ1cosθ2−sinθ2cosθ1)=2⇒sin(θ1−θ2)=1⇒θ1−θ2=π2,−3π2 …(1)∵θ1,θ2∈(−π,π]⇒θ1±θ2∈(−2π,2π]
|z−i¯¯¯¯w|=2⇒|(cosθ1−sinθ2)+i(sinθ1−cosθ2)|=2
⇒cos2θ1+sin2θ2−2sinθ2cosθ1+sin2θ1+cos2θ2−2sinθ1cosθ2=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.