Find the simplest form for T = (√(1+(√(−3)))) +(√(1−(√(−3))))


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Question Number 142829 by liberty last updated on 06/Jun/21

$F i n d t h e s i m p l e s t f o r m f o r$ $T = \sqrt{1 + \sqrt{- 3}} + \sqrt{1 - \sqrt{- 3}}$
	Answered by EDWIN88 last updated on 06/Jun/21
	$Let (√a) +(√(−b)) = (√(1+(√(−3)))) with a>0 ,b>0 ⇒a−b+2i(√(ab)) = 1+i(√3) we get { ((a−b=1)),((2(√(ab)) =(√3))) :} squaring { ((a^2 −2ab+b^2 =1...(1))),((4ab=3...(2))) :} addition of (1) and (2) we have a^2 +2ab+b^2 = 4 ⇒a+b = 2 now we solve { ((a−b=1)),((a+b=2)) :} ⇒ { ((a=(3/2))),((b=(1/2))) :} hence (√(1+(√(−3)))) = (√(3/2)) +(√(−(1/2))) in a similar manner we obtain (√(1−(√(−3)))) = (√(3/2))−(√(−(1/2))) Therefore T= (√(3/2))+(√(−(1/2)))+(√(3/2))−(√(−(1/2))) T = 2(√(3/2)) =(√6) □$
	$Let \sqrt{a} + \sqrt{- b} = \sqrt{1 + \sqrt{- 3}} with a > 0, b > 0$ $\Rightarrow a - b + 2 i \sqrt{ab} = 1 + i \sqrt{3} we get {\begin{cases} a - b = 1 \\ 2 \sqrt{ab} = \sqrt{3} \end{cases}$ $squaring {\begin{cases} a^{2} - 2 ab + b^{2} = 1 . . . (1) \\ 4 ab = 3 . . . (2) \end{cases}$ $addition of (1) and (2) we have$ $a^{2} + 2 ab + b^{2} = 4 \Rightarrow a + b = 2$ $now we solve {\begin{cases} a - b = 1 \\ a + b = 2 \end{cases} \Rightarrow {\begin{cases} a = \frac{3}{2} \\ b = \frac{1}{2} \end{cases}$ $hence \sqrt{1 + \sqrt{- 3}} = \sqrt{\frac{3}{2}} + \sqrt{- \frac{1}{2}}$ $in a similar manner we obtain$ $\sqrt{1 - \sqrt{- 3}} = \sqrt{\frac{3}{2}} - \sqrt{- \frac{1}{2}}$ $Therefore T = \sqrt{\frac{3}{2}} + \sqrt{- \frac{1}{2}} + \sqrt{\frac{3}{2}} - \sqrt{- \frac{1}{2}}$ $T = 2 \sqrt{\frac{3}{2}} = \sqrt{6} ◻$
	Commented by liberty last updated on 06/Jun/21

	$t h a n k y o u$
	Answered by mr W last updated on 06/Jun/21

	$1 + \sqrt{- 3} = 1 + \sqrt{3} i = 2 e^{\frac{π i}{3}}$ $\Rightarrow \sqrt{1 + \sqrt{- 3}} = \sqrt{2} e^{\frac{π i}{6}}$ $s i m i l a r l y$ $\Rightarrow \sqrt{1 - \sqrt{- 3}} = \sqrt{2} e^{- \frac{π i}{6}}$ $\sqrt{1 + \sqrt{- 3}} + \sqrt{1 - \sqrt{- 3}} = \sqrt{2} (e^{\frac{π i}{6}} + e^{- \frac{π i}{6}})$ $= \sqrt{2} \times 2 \times \cos \frac{π}{6} = \sqrt{2} \times 2 \times \frac{\sqrt{3}}{2} = \sqrt{6}$
	Commented by liberty last updated on 06/Jun/21

	$t h a n k y o u$
	Answered by mathmax by abdo last updated on 06/Jun/21

	$T = \sqrt{1 + \sqrt{- 3} + \sqrt{1 - \sqrt{- 3}}}$ $we have 1 + \sqrt{- 3} = 1 + i \sqrt{3} = 2 (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = {2 e}^{\frac{i π}{3}} \Rightarrow$ $\sqrt{1 + \sqrt{- 3}} = \sqrt{2} e^{\frac{i π}{6}}$ $also 1 - \sqrt{- 3} = 1 - i \sqrt{3} = {2 e}^{- \frac{i π}{3}} \Rightarrow \sqrt{1 - \sqrt{- 3}} = \sqrt{2} e^{- \frac{i π}{6}} \Rightarrow$ $T = \sqrt{2} (e^{\frac{i π}{6}} + e^{- \frac{i π}{6}}) = \sqrt{2} (2 \cos (\frac{π}{6})) = 2 \sqrt{2} . \frac{\sqrt{3}}{2} = \sqrt{6}$
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