Given complex number is z = sin120° – icos120°
⇒ z = \(\frac{\sqrt{3}}{2} - i\Big(\frac{-1}{2}\Big)\)
⇒ z = \(\frac{\sqrt{3}}{2} + i\Big(\frac{1}{2}\Big)\)
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| = \(\sqrt{\Big(\frac{\sqrt{3}}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\)
⇒ |z| = \(\sqrt{\frac{3}{4}+\frac{1}{4}}\)
⇒ |z| = \(\sqrt{1}\)
⇒ |z|=1
Since x>0,y>0 complex number lies in 1st quadrant and the value of θ will be as follows 0°≤θ≤90°.
⇒ θ = tan-1\(\frac{1}{\sqrt{3}}\)
⇒ θ = \(\frac{\pi}{6}\)
⇒ z = \(1\Big(cos\Big(\frac{\pi}{6}\Big) + isin\Big(\frac{\pi}{6}\Big)\Big)\)
∴ The Polar form of z = sin120° – icos120° is z = \(1\Big(cos\Big(\frac{\pi}{6}\Big) + isin\Big(\frac{\pi}{6}\Big)\Big)\)