Question

Find the modules and argument of -1-i√3​

Answer 1

Answer:

2, -2π/3

Step-by-step explanation:

Modulus = √[(-1)^2 + (√3)^2]

= √(1 + 3)

= 2

cos theta= -1/2

sin theta = -√3/2

Therefore, Theta = -(π-π/3) = -2π/3

Answer 2

Step-by-step explanation:

modulus : lzl=√(-1)^2 +(√3)^2=2

let ¢be acute angle by tan¢ =| I'm(z) /re(z))

tan¢= l -√3/-1 l or √3/1 or√3

¢ = π/3

z lies in 3rd quadrant

arg(z) = -(π -¢}= -(π -π/3) =2π/3

SO that it is the answer I hope it will help u.