Find-the-simplest-form-for-T-1-3-1-3-

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 \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 \dots (1) \\ 4 ab = 3 \dots (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}