Question

Find the modulus and argument of the complex number z= -1 - i√3

Answer 1

z=-1-i√3

let r cos∅=-1 and r sin∅=-√3

on squaring and adding , we get

(r cos∅)²+(r sin∅)²=(-1)²+(-√3)²

r²(cos²∅+sin²∅)=1+3

r²=√4

r=2

modulus=2

∴2cos∅=-1 and 2sin∅=-√3

cos∅=-1/2 and sin∅=-√3/2

since both the value of sin∅ and cos∅ are negative and sin∅ and cos∅ are negative in III Quadrant,

Argument = -(π-π/3) = -2π/3

thus , the modulus and argument of the complex number -1-√3i are 2 and -2π/3 respectively

i hope this will help u

Step-by-step explanation:

Answer 2

Given:-

• z = -1 - i√3

To find:-

• Find the modulus and argument of the complex number..?

Solutions:-

• z = -1 - i√3

• Let rCos∅ = -1 and rSin∅ = -√3

On squaring and adding, we obtain

=> (rCos∅)² + (rSin∅)² = (-1)² =(-√3)²

=> r²(Cos²∅ + sin²∅) = 1 + 3

=> r² = 4 [Cos²∅ + Sin²∅ = 1]

=> r = √4 = 2 [Convertionally, r > 0]

Therefore,

Modulus = 2

=> 2Cos∅ = -1 and 2Sin∅ = -√3

=> Cos∅ = -1/2 and Sin∅ = -√3/2

Since both the value of Sin∅ and Cos∅ are negative and Sin∅ and Cos∅ are negative in (iii). quadrant,

Argument = -(π - π/3) = -2π/3

Thus, the modules and argument of the complex numbers -1 - √3i and 2 and -2π/3 respectively.