= √3i + 1
Let Z = √3i + 1 = r(cosθ + isinθ)
Now , separating real and complex part , we get
1 = rcosθ ……….eq.1
√3 = rsinθ …………eq.2
Squaring and adding eq.1 and eq.2, we get
4 = r2
Since r is always a positive no., therefore,
r = 2,
Hence its modulus is 2.
Now, dividing eq.2 by eq.1 , we get,
\(\frac{rsin\theta}{rcos\theta}=\frac{\sqrt3}{1}\)
tanθ = √3
Since cosθ = 1/2, sinθ = √3/2 and tanθ = √3
therefore the θ lies in first quadrant.
Tanθ = √3, therefore θ = π/3
Representing the complex no. in its polar form will be
Z = 2{cos(π/3)+i sin(π/3)}