Explanation
$$\sin\dfrac{6\pi}{5}+i(1+\cos\dfrac{6\pi}{5})$$$$=2\sin\dfrac{3\pi}{5} \cos\dfrac{3\pi}{5} +2i\cos^2\dfrac{3\pi}{5}$$$$=-2\cos\dfrac{3\pi}{5}(-\sin\dfrac{3\pi}{5}-i\cos\dfrac{3\pi}{5})$$ $$=-2\cos\dfrac{3\pi}{5}(\cos\dfrac{9\pi}{10}+i\sin\dfrac{9\pi}{10})$$So argument is $$\dfrac{9\pi}{10}$$
$$x=9^{\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$$$$x=9^{\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}}$$$$x=9^{\dfrac{1}{2}}$$$$x=3$$$$y=4^{\dfrac{1}{3}-\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$$$$y=4^{\dfrac{\dfrac{1}{3}}{1+\dfrac{1}{3}}}$$$$y=4^{\dfrac{1}{4}}$$$$y=2^{\dfrac{1}{2}}$$$$y=\sqrt{2}$$$$z=\dfrac{1}{(1+i)^{1}}+\dfrac{1}{(1+i)^{2}}+\dfrac{1}{(1+i)^{3}}...\infty$$Since it is a G.P with a common ratio of $$\dfrac{1}{(1+i)}$$ we get the sum as$$=\dfrac{\dfrac{1}{1+i}}{1-\dfrac{1}{1+i}}$$$$=\dfrac{1}{i}$$$$=-i$$.Hence $$x+yz =3-\sqrt{2}i$$$$\tan\theta=\dfrac{-\sqrt{2}}{3}$$$$\theta=\tan^{-1}(\dfrac{-\sqrt{2}}{3})$$$$=-\tan^{-1}(\dfrac{\sqrt{2}}{3})$$
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