1. **State the problem:** Find the modulus and argument of the complex number $$\frac{1 + \sin\theta + i \cos\theta}{1 + \sin\theta - i \cos\theta}$$ and show that for $$\theta = \frac{\pi}{3}$$, the expression equals $$i$$. 2. **Recall formulas:** - The modulus of a complex number $$z = x + iy$$ is $$|z| = \sqrt{x^2 + y^2}$$. - The argument $$\arg(z)$$ is the angle $$\phi$$ such that $$z = |z|(\cos\phi + i\sin\phi)$$. - For a quotient $$\frac{z_1}{z_2}$$, modulus is $$\frac{|z_1|}{|z_2|}$$ and argument is $$\arg(z_1) - \arg(z_2)$$. 3. **Calculate numerator and denominator moduli:** - Numerator: $$z_1 = 1 + \sin\theta + i \cos\theta$$ $$|z_1| = \sqrt{(1 + \sin\theta)^2 + (\cos\theta)^2} = \sqrt{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}$$ Using $$\sin^2\theta + \cos^2\theta = 1$$, $$|z_1| = \sqrt{1 + 2\sin\theta + 1} = \sqrt{2 + 2\sin\theta} = \sqrt{2(1 + \sin\theta)}$$. - Denominator: $$z_2 = 1 + \sin\theta - i \cos\theta$$ $$|z_2| = \sqrt{(1 + \sin\theta)^2 + (-\cos\theta)^2} = \sqrt{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta} = \sqrt{2(1 + \sin\theta)}$$. 4. **Modulus of the quotient:** $$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{\sqrt{2(1 + \sin\theta)}}{\sqrt{2(1 + \sin\theta)}} = 1$$. 5. **Calculate arguments:** - $$\arg(z_1) = \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)$$ - $$\arg(z_2) = \tan^{-1}\left(\frac{-\cos\theta}{1 + \sin\theta}\right) = -\tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)$$ 6. **Argument of the quotient:** $$\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) = \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right) - \left(-\tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)\right) = 2 \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right)$$. 7. **Simplify the argument expression:** Using the identity $$\tan\frac{x}{2} = \frac{\sin x}{1 + \cos x}$$, set $$x = \frac{\pi}{2} - \theta$$: $$\tan\left(\frac{\pi}{2} - \theta\right) = \frac{\cos\theta}{1 + \sin\theta}$$ Therefore, $$\arg\left(\frac{z_1}{z_2}\right) = 2 \tan^{-1}\left(\tan\left(\frac{\pi}{2} - \theta\right)\right) = 2 \left(\frac{\pi}{2} - \theta\right) = \pi - 2\theta$$. 8. **Evaluate at $$\theta = \frac{\pi}{3}$$:** - Modulus is 1. - Argument is $$\pi - 2 \times \frac{\pi}{3} = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$$. 9. **Express the quotient at $$\theta = \frac{\pi}{3}$$:** $$\frac{z_1}{z_2} = 1 \times (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$ This is not equal to $$i$$, so check the problem statement carefully. 10. **Check the problem's second part:** The problem states: $$\left[ \frac{1 + \sin(\frac{\pi}{3}) + i \cos(\frac{\pi}{3})}{1 + \sin(\frac{\pi}{3}) - i \cos(\frac{\pi}{3})} \right] = i$$ Calculate numerator and denominator explicitly: - $$\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$$ - $$\cos(\frac{\pi}{3}) = \frac{1}{2}$$ Numerator: $$1 + \frac{\sqrt{3}}{2} + i \times \frac{1}{2} = \left(1 + \frac{\sqrt{3}}{2}\right) + i \frac{1}{2}$$ Denominator: $$1 + \frac{\sqrt{3}}{2} - i \times \frac{1}{2} = \left(1 + \frac{\sqrt{3}}{2}\right) - i \frac{1}{2}$$ Divide numerator by denominator: $$\frac{\left(1 + \frac{\sqrt{3}}{2}\right) + i \frac{1}{2}}{\left(1 + \frac{\sqrt{3}}{2}\right) - i \frac{1}{2}} = \frac{a + ib}{a - ib}$$ where $$a = 1 + \frac{\sqrt{3}}{2}$$ and $$b = \frac{1}{2}$$. Multiply numerator and denominator by conjugate of denominator: $$= \frac{(a + ib)(a + ib)}{a^2 + b^2} = \frac{a^2 + 2iab - b^2}{a^2 + b^2} = \frac{a^2 - b^2}{a^2 + b^2} + i \frac{2ab}{a^2 + b^2}$$ Calculate: - $$a^2 = \left(1 + \frac{\sqrt{3}}{2}\right)^2 = 1 + 2 \times 1 \times \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 + \sqrt{3} + \frac{3}{4} = \frac{7}{4} + \sqrt{3}$$ - $$b^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ So, - $$a^2 - b^2 = \left(\frac{7}{4} + \sqrt{3}\right) - \frac{1}{4} = \frac{6}{4} + \sqrt{3} = \frac{3}{2} + \sqrt{3}$$ - $$a^2 + b^2 = \left(\frac{7}{4} + \sqrt{3}\right) + \frac{1}{4} = 2 + \sqrt{3}$$ - $$2ab = 2 \times \left(1 + \frac{\sqrt{3}}{2}\right) \times \frac{1}{2} = \left(1 + \frac{\sqrt{3}}{2}\right) = a$$ Therefore, $$\frac{z_1}{z_2} = \frac{\frac{3}{2} + \sqrt{3}}{2 + \sqrt{3}} + i \frac{1 + \frac{\sqrt{3}}{2}}{2 + \sqrt{3}}$$ Simplify numerator and denominator by rationalizing or approximate numerically: - Numerically, $$\frac{3/2 + 1.732}{2 + 1.732} = \frac{3.232}{3.732} \approx 0.866$$ - $$\frac{1 + 0.866}{3.732} = \frac{1.866}{3.732} \approx 0.5$$ So the quotient is approximately $$0.866 + 0.5 i$$ which equals $$\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$$, not $$i$$. 11. **Re-examine the problem:** The problem likely intends $$Z = 1 + \sin\theta + i \cos\theta$$ and the expression is $$\frac{Z}{\overline{Z}}$$ where $$\overline{Z} = 1 + \sin\theta - i \cos\theta$$. Since $$\frac{Z}{\overline{Z}}$$ has modulus 1 and argument $$2 \arg(Z)$$, then for $$\theta = \frac{\pi}{3}$$, Calculate $$\arg(Z) = \tan^{-1}\left(\frac{\cos\theta}{1 + \sin\theta}\right) = \tan^{-1}\left(\frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{0.5}{1.866}\right) \approx \tan^{-1}(0.268) = \frac{\pi}{12}$$. Therefore, $$\arg\left(\frac{Z}{\overline{Z}}\right) = 2 \times \frac{\pi}{12} = \frac{\pi}{6}$$ This contradicts the problem's claim that the expression equals $$i$$ (which has argument $$\frac{\pi}{2}$$). 12. **Conclusion:** - The modulus of the expression is 1. - The argument is $$\pi - 2\theta$$. - For $$\theta = \frac{\pi}{3}$$, the expression equals $$\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$$, not $$i$$. Hence, the problem's second part appears incorrect or requires re-interpretation. **Final answers:** - Modulus: $$1$$ - Argument: $$\pi - 2\theta$$ - At $$\theta = \frac{\pi}{3}$$, value is $$\cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$.