If `arg(z_(1))=(2pi)/3 and arg(z_(2))=pi/2`, then find the principal value of `arg(z_(1)z_(2))` and also find the quardrant of ` z_(1)z_(2)`.

Explanation

`arg(z_(1)z_(2)) =arg(z_(1) + arg(z_(2)) +2kpi`
`= (2pi)/3 + pi/2 + 2kpi = ( 7pi)/ 6 + 2k pi`
To bring the argument in the principal range k =-1
`arg(z_(1)z_(2))= (7pi)/6 -2pi = (5pi)/6` which shows that ` z_(1)z_(2)` lies in `III^("rd")`quadrant.