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Express it in the polar form: (i-1) / (cospi/3) + (isin pi/3).Also Find the arguement and modulus.

$Let z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}} = \frac{i - 1}{\frac{1}{2} + \frac{\sqrt{3}}{2} i} = \frac{2 (i - 1)}{1 + \sqrt{3} i} Rationalize the numerator and denominator with the conjugate of denominator = \frac{2 (i - 1)}{1 + \sqrt{3} i} \times \frac{1 - \sqrt{3} i}{1 - \sqrt{3} i} = \frac{2 (i + \sqrt{3} - 1 + \sqrt{3} i)}{1 + 3} = \frac{\sqrt{3} - 1 + (\sqrt{3} + 1) i}{2} = \frac{\sqrt{3} - 1}{2} + i \frac{\sqrt{3} + 1}{2} ∴ z = \frac{\sqrt{3} - 1}{2} + i \frac{\sqrt{3} + 1}{2} This in form z=x+iy where x= \frac{\sqrt{3} - 1}{2} and y= \frac{\sqrt{3} + 1}{2} r = modulus of z= \sqrt{x^{2} + y^{2}} = \sqrt{{(\frac{\sqrt{3} - 1}{2})}^{2} + {(\frac{\sqrt{3} + 1}{2})}^{2}} = \sqrt{\frac{3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3}}{4}} = \sqrt{2} in the given comples number both x and y are positive Hence θ = \tan^{- 1} | \frac{y}{x} | \Rightarrow θ = \tan^{- 1} | \frac{\frac{\sqrt{3} + 1}{2}}{\frac{\sqrt{3} - 1}{2}} | = \tan^{- 1} (\frac{\sqrt{3} + 1}{\sqrt{3} - 1}) = \tan^{- 1} (\frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}) \Rightarrow θ = \tan^{- 1} (\frac{\tan \frac{π}{4} + \tan \frac{π}{6}}{\tan \frac{π}{4} - \tan \frac{π}{6}}) = \tan^{- 1} (\frac{\tan \frac{π}{4} + \tan \frac{π}{6}}{1 - \tan \frac{π}{4} \tan \frac{π}{6}}) = \tan^{- 1} (\tan (\frac{π}{4} + \frac{π}{6})) \Rightarrow θ = \frac{π}{4} + \frac{π}{6} = \frac{5 π}{12} t he polar form of z=r(cos θ + i \sin θ) \Rightarrow z = \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12}) ∴ The polar form of z= \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}} is \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$

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