Let – 1 = r cos θ and – 1 = r sin θ ……….(i)

Squaring both sides, we get

1= r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

1 + 1 = r² cos² θ + r² sin² θ

⇒ 2 = r² (cos² θ + sin² θ)

⇒ 2 = r² or r² = 2 [∵ sin² θ + cos² θ = 1]

⇒ r = √2

⇒ r = √2 (conventionally, r>0)

Substituting r = √2 in (i), we get

– 1 = √2 cos θ and – 1 = √2 sin θ

∵ We know that the complex number –1 – i lies in the third quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .