Explanation
sin6π5+i(1+cos6π5)=2sin3π5cos3π5+2icos23π5=−2cos3π5(−sin3π5−icos3π5) =−2cos3π5(cos9π10+isin9π10)So argument is 9π10
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=3y=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}}...\inftySince 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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