Let z and ω  be two complex numbers such that |z|≤1, |ω|≤1  and |z+iω|=|z−iω¯|=2 , then Z equals

We have 2=|z+iω|≤|z|+|ω| ∴|z|+|ω|≥2But given that |z|≤1  and  |ω|≤1⇒|z|+|ω|≤2From (1) and (2) |z|=|ω|=1Also|z+iω|=|z−iω¯|  ⇒|z+iω|2=|z−iω¯|2  (z+iω)(z¯−iω¯)=(z¯+iω)(z−iω¯)⇒zz¯+iωz¯−izω¯+ωω¯=zz¯−izω¯+iωz+ωω¯⇒ωz¯−ω¯z+ω¯z¯−ωz=0⇒(ω+ω¯)(z¯−z)=0z=z¯  or  ω=−ω¯⇒Im(z)=0⇒Re(ω)=0Also |z|=1 , |ω|=1⇒z=1  or  −1  and  ω=i   or  −i