Let – 3 = r cos θ and 0 = r sin θ ……….(i)

Squaring both sides, we get

9= r² cos² θ and 0 = r² sin² θ

Adding both the equations, we get

9 + 0 = r² cos² θ + r² sin² θ

⇒ 9 = r² (cos² θ + sin² θ)

⇒ 9 = r² or r² = 9 [∵ sin² θ + cos² θ = 1]

⇒ r = √9

⇒ r = 3 (conventionally, r>0)

Substituting r = √2 in (i), we get

– 3 = 3 cos θ and 0 = 3 sin θ

⇒ cos θ = – 1 and sin θ = 0

⇒ θ = cos^-1 (-1) and θ = sin^-1 0

∵ We know that the complex number – 3 lies on the real axis (x-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

∴ θ = cos^-1 cos π and θ = sin^-1 sin 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 z = 3 (cos π + i sin π)