Answer:
Option B
Explanation:
We have
z=√2√1+√3i⇒z=√2+2√3i
z=√(√3+i)2⇒z=±(√3+i)
z lie in third quadrant
∴ z=−√3−i
⇒|z|=√(−√3)2+(−1)2=√3+1=√4=2 and
tanθ=|−1−√3|=1√3⇒θ=π6
Since θ lie in 3rd quadrant
∴ arg(z)=−(π−π6)=−5π6
Hence, z=2(cos(−5π6)+isin(−5π6))