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°)