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Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!

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I upvoted this because I think any question that can attract four wrong answers so quickly has something interesting going on. – MJD Sep 27 '12 at 14:19
Then be so kind and provide a correct one... – draks ... Sep 27 '12 at 14:29
up vote -13 down vote accepted

$\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ $/^2$

$(\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3})^2=(\sqrt 6)^2$

$1+i\sqrt 3+1-i\sqrt 3+2\sqrt{(1+i\sqrt 3)(1-i\sqrt 3)}=6$

$2\sqrt{(1+i\sqrt 3)(1-i\sqrt 3)}=4$$/:2$

$\sqrt{(1+i\sqrt 3)(1-i\sqrt 3)}=2$$/^2$

$1-i^2\sqrt 3^2=2^2$



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That's not a proof; squaring isn't reversible. – joriki Sep 27 '12 at 14:12
How do you know $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}\neq-\sqrt 6$ – Arthur Sep 27 '12 at 14:12
The exact same argument "proves" $\sqrt{1 + i\sqrt{3}} + \sqrt{1 - i\sqrt{3}} = -\sqrt{6}$. Oh noes, $\sqrt{6} = -\sqrt{6}$!!! – Zarrax Sep 27 '12 at 16:39

Changing into polar form we have $ 1 + i \sqrt{3} = 2 e^{i\pi/3}$ and $1 - i \sqrt{3} = 2e^{-i\pi/3}$ so the left hand side is $$ \sqrt{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right)= 2 \sqrt{2} \cos(\pi/6)= 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}= \sqrt{6}. $$

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How did you decide to choose $e^{i\pi/6}$ rather than $e^{7i\pi/6}$? – MJD Sep 27 '12 at 14:17
@MJD I used the principal branch of the square root function, if we chose to work with the other then the result is the same but the order of the terms to be added are simply switched. – Ragib Zaman Sep 27 '12 at 14:23

Use $\sqrt{1\pm i\sqrt 3}=\sqrt{2}e^{\pm i\pi/6}$ (EDIT we are picking the principal branch here) to get $$ \sqrt{2}\left( e^{i\pi/6}+e^{-i\pi/6}\right)=2\sqrt{2}\cos(\pi /6)=2\sqrt{2}\frac{\sqrt{3}}2=\sqrt{6} $$

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How did you decide to choose $e^{i\pi/6}$ rather than $e^{7i\pi/6}$? – MJD Sep 27 '12 at 14:17
Where's is the difference to Ragib's answer? – draks ... Sep 27 '12 at 14:22
I've chosen it, since obviously $\frac12+i\frac{\sqrt{3}}2=\cos(\pi/3)+i\sin(\pi/3)$ and then I halved the angle. – draks ... Sep 27 '12 at 14:27
@MJD: good question! Indeed why not use $\exp(7\pi i/6)$ for one and $\exp(-\pi i/6)$ for the other? – André Nicolas Sep 27 '12 at 14:29
@AndréNicolas Well if we chose different branches for each square root, then the result simply doesn't hold true. Is one completely wrong if we do the standard thing and choose the principal branch unless otherwise specified, rather than mentioning all the different ways things could go for every possible branch cut every time we apply a multi-valued function? – Ragib Zaman Sep 27 '12 at 14:34

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