Find : arg( (((2(√3)+2i)^8 )/((1−i)^6 )) + (((1+i)^6 )/((2(√3)−2i)^8 ))) ?


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Question Number 49331 by rahul 19 last updated on 05/Dec/18

$F i n d :$ $a r g (\frac{{(2 \sqrt{3} + 2 i)}^{8}}{{(1 - i)}^{6}} + \frac{{(1 + i)}^{6}}{{(2 \sqrt{3} - 2 i)}^{8}}) ?$
Commented by maxmathsup by imad last updated on 05/Dec/18

$l e t u = 2 \sqrt{3} + 2 i a n d v = 1 + i \Rightarrow$ $A = \frac{{(2 \sqrt{3} + 2 i)}^{8}}{{(1 - i)}^{6}} + \frac{{(1 + i)}^{6}}{{(2 \sqrt{3} - 2 i)}^{8}} = \frac{{(u . \bar{u})}^{8} + {(v . \bar{v})}^{6}}{\overset{-^{6}}{v} . \overset{-^{8}}{u}}$ $w e h a v e ∣ u ∣ = 2 ∣ \sqrt{3} + i ∣ = 2.2 = 4 \Rightarrow {(u . \bar{u})}^{8} = {(4^{2})}^{8} = 4^{16}$ $v = 1 + i \Rightarrow∣ v ∣ = \sqrt{2} \Rightarrow {(v . \bar{v})}^{6} = 2^{6} \Rightarrow A = \frac{4^{16} + 2^{6}}{\overset{-^{8}}{u} . \overset{-^{6}}{v}} \Rightarrow$ $a r g (A) = - a r g (\overset{-^{8}}{u} . \overset{-^{6}}{v}) \equiv - 8 a r g (\bar{u}) - 6 a r g (\bar{v}) [2 π] b u t$ $v = \sqrt{2} e^{\frac{i π}{4}} \Rightarrow \bar{v} = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow a r g (\bar{v}) \equiv - \frac{π}{4} [2 π]$ $a l s o w e h a v e u = 4 (\frac{\sqrt{3}}{2} + \frac{i}{2}) = 4 e^{\frac{i π}{6}} \Rightarrow \bar{u} = 4 e^{- \frac{i π}{6}} \Rightarrow a r g (\bar{u}) \equiv - \frac{π}{6} [2 π] \Rightarrow$ $a r g (A) \equiv \frac{8 π}{6} + \frac{6 π}{4} [2 π] \Rightarrow a r g (A) \equiv \frac{4 π}{3} + \frac{3 π}{2} [2 π] \Rightarrow a r g (A) \equiv \frac{17 π}{6} [2 π] b u t$ $\frac{17 π}{6} = 2 π + \frac{5 π}{6} \equiv \frac{5 π}{6} \Rightarrow a r g (A) \equiv \frac{5 π}{6} [2 π] .$
Commented by rahul 19 last updated on 05/Dec/18

$I^{'} m g e t t i n g \frac{5 π}{3}$ $A n s . i s \frac{5 π}{6}$ $K i n d l y c h e c k . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
	$a=(2(√3) +2i) a=4(((2(√3))/4)+(2/4)i)=4(cos(π/6)+isin(π/6))=4e^((iπ)/6) b=(1+i)=(√2) ((1/(√2))+i(1/(√2)))=(√2) e^((iπ)/4) {(((4e^((iπ)/6) )^8 )/(((√2) e^((−iπ)/4) )^6 ))+((((√2) e^((iπ)/4) )^6 )/((4e^((−iπ)/6) )^8 ))} ={(2^(16) /2^3 )×e^(((i4π)/3)+((i3π)/2)) +(2^3 /2^(16) )×e^(((i3π)/2)+((i4π)/3)) } =e^((i17π)/6) ×(2^(13) +2^(−13) ) now [cos(((17π)/6))+isin((17π)/6)]×(2^(13) +2^(−13) ) =[cos(2π+((5π)/6))+isin(2π+((5π)/6))]×(2^(13) +2^(−13) ) =[cos(((5π)/6))+isin(((5π)/6))]×(2^(13) +2^(−13) ) so tanθ=tan(((5π)/6)) θ=((5π)/6)$
	$a = (2 \sqrt{3} + 2 i)$ $a = 4 (\frac{2 \sqrt{3}}{4} + \frac{2}{4} i) = 4 (c o s \frac{π}{6} + i s i n \frac{π}{6}) = 4 e^{\frac{i π}{6}}$ $b = (1 + i) = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = \sqrt{2} e^{\frac{i π}{4}}$ ${\frac{{(4 e^{\frac{i π}{6}})}^{8}}{{(\sqrt{2} e^{\frac{- i π}{4}})}^{6}} + \frac{{(\sqrt{2} e^{\frac{i π}{4}})}^{6}}{{(4 e^{\frac{- i π}{6}})}^{8}}}$ $= {\frac{2^{16}}{2^{3}} \times e^{\frac{i 4 π}{3} + \frac{i 3 π}{2}} + \frac{2^{3}}{2^{16}} \times e^{\frac{i 3 π}{2} + \frac{i 4 π}{3}}}$ $= e^{\frac{i 17 π}{6}} \times (2^{13} + 2^{- 13})$ $n o w [c o s (\frac{17 π}{6}) + i s i n \frac{17 π}{6}] \times (2^{13} + 2^{- 13})$ $= [c o s (2 π + \frac{5 π}{6}) + i s i n (2 π + \frac{5 π}{6})] \times (2^{13} + 2^{- 13})$ $= [c o s (\frac{5 π}{6}) + i s i n (\frac{5 π}{6})] \times (2^{13} + 2^{- 13})$ $s o t a n θ = t a n (\frac{5 π}{6}) θ = \frac{5 π}{6}$
	Commented by maxmathsup by imad last updated on 05/Dec/18

	$c o r r e c t a n s w e r t h a n k s s i r T a n m a y .$
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