UPSC NDA Math Model Paper 10

Given complex number, -1 - i
|z|=r=√(−1)2+(−1)2
=√2
Now, −1−i can be represented as point P(−1,−1), which lies in 3 rd quadrant.

Now, θ=−π+tan−1(

Im(z)

Re(z)

)
= -π + π/4
= -3π/4
∴ Polar form: z = √2 (cos(-3π/4) + i sin(-3π/4))
= √2 (cos(3π/4) - i sin(3π/4)) (∵ cos (-θ) = cos θ and sin (-θ) =-sin θ)
Hence, option (2) is correct.