UPSC NDA Math Model Paper 9

Given complex number Z = 3 + 3i.
Let rcosθ = 3 and rsinθ = 3
By squaring and adding, we get
r2(cos2θ+sin2θ)=18
∴r=3√2
As we have taken r‌cos‌θ=3
⇒cos‌θ=

3

r

=

3

3√2

=

1

√2

⇒ θ = 45° (Since it is in first quadrant, there won’t be any changes in θ value)
So, on comparing with z = r (cosθ + i sinθ), we can write as 3√2 (cos 45° + i sin 45°)