Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
– 1 – i
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 .