We have, z = – 1 – i
Let – 1 = r cosθ …(i)
and – 1 = r sinθ …(ii)
Squaring and adding (i) and (ii), we get
r2(cos2θ+sin2θ) = 1 + 1 ⇒ r2 = 2 ⇒ r = √2
Substituting the value of r in (i) and (ii), we get
√2 cos θ = −1,√2 sin θ = - 1
⇒ cos θ = −

1

√2

, sin θ = −

1

√2

⇒ cos θ = - cos (

π

4

) = - sin (

π

4

)
Here, cosθ < 0 and sinθ < 0.
∴ θ lies in the third quadrant.
∴ θ = - (π−

π

4

) = - (

4π−π

4

) =

−3π

4

∴ The required polar form is z = √2[cos(

−3π

4

)+isin(

−3π

4

)]