Question

find the modulus and argument of -2+2i√3​

Answer 1

Step-by-step explanation:

modulus z = √(x^2 + y^2)

x= -2, y= 2√3

z = √(4 + 4*3) = 4

argument theta = tan-1 (y/x)

= tan-1 (2√3/-2)

= tan-1 (-√3) = -60°

theta = -60°

Answer 2

NOTE: |x| is modulus function.

let the equation be

z=-2+2i√3

modulus of z

|z|=√(x²+y²)

where x and y are the coefficients in the general equation of

z=x+iy

In the given equation

x=-2

y=2√3

Plugging the values

|z|=√((-2)²+(2√3)²)

=√4+12

=√16

=4

Now,

argument of z is given by

arg(z)=Ф

Ф=π-tan⁻¹(|y|/|x|)        (since its in 2nd quadrant)

Ф=π-tan⁻¹(2√3/2)

=π-tan⁻¹(√3)

=π-π/3

=2π/3

Hope this helps.