Question

z = 1 - raiz quadrada de 3 i​

Answer 1

z=1 -√3i ==> cos(+), sen(-) ==> 4°Q

p=√a²+b²=√1²+√3² = √1+3 = √4 = 2

tgx= b/a= -√3/1= -√3 => X=360-60=300°

z= p(cosx + i.senx)

z= 2(cos300 + i.sen300)