Convert the given complex number in polar form 3+i

The complex number is \( \sqrt{3}+i\)

Let rcosθ = \( \sqrt{3}\) and rsinθ = 1

Squaring and adding

= \( r^2cos^2θ + r^2sin^2θ = (\sqrt{3})^2+1^2\)

= \( r^2(cos^2θ + sin^2θ) =3+1\)

= \( r^2=4\)

= \( r=\sqrt{4}=2\)  [Conventionally, r > 0]

= 2cosθ = \( \sqrt{3}\) and 2sinθ = 1

= \( cosθ = \dfrac{\sqrt{3}}{2} and \;sinθ = \dfrac{1}{2}\)

\( θ =\dfrac{\pi}{6}\)  [As θ lies in the I quadrant]

= \( \sqrt{3}+i= rcosθ + irsinθ \)

= \(2 cos\dfrac{\pi}{6}+ i2sin\dfrac{\pi}{6}\)

\( =2(cos\dfrac{\pi}{6}+isin\dfrac{\pi}{6})\)