Given complex number is Z=\(\sqrt{3}\)+i
We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)
Where,
|Z|=modulus of complex number= \(\sqrt{x^2+y^2}\)
θ =arg(z)=argument of complex number= tan-1 \(\Big(\frac{|y|}{|x|}\Big)\)
Now for the given problem,
⇒ |z| = \(\sqrt{(\sqrt{3})^2+(1)^2}\)
⇒ |z| = \(\sqrt{3+1}\)
⇒ |z| = \(\sqrt{4}\)
⇒ θ = tan-1\(\Big(\frac{1}{\sqrt{3}}\Big)\)
Since x>0,y>0 complex number lies in 1st quadrant and the value of θ will be as follows 0°≤θ≤90°.
⇒ θ = \(\frac{\pi}{6}\)
⇒ z = \(\sqrt{2}\Big(cos\Big(\frac{\pi}{6}\Big) + isin\Big(\frac{\pi}{6}\Big)\Big)\)
∴ The Polar form of Z=\(\sqrt{3}\)+i is z =\({2}\Big(cos\Big(\frac{\pi}{6}\Big) + isin\Big(\frac{\pi}{6}\Big)\Big)\)