Given, 1 - i can be written in polar form as

      1 - i = r(cos θ + i*sin θ)

=> 1 - i = r*cos θ + i(r*sin θ)

Now, 

r*cos θ = 1 and r*sin θ = -1

Squaring both side and add them, we get

      r² * cos² θ + r² * sin² θ = 1 + 1

=> r² (cos² θ + sin² θ) = 2

=> r² = 2

=> r = √2

Again,

      r*sin θ/r*sin θ = -1/1

=> tan θ = -1

=> θ = -π/4

Now, 

      1 - i = √2{cos (-π/4) + i*sin (-π/4)}

=> 1 - i = √2{cos (π/4) - i*sin (π/4)}