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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

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
Oh if only we could override accepted answers that are not in fact correct! – Oscar Lanzi May 23 at 1:04
How do we know that $\sqrt{e^{i\pi/3}}=e^{i\pi/6}$ and $\sqrt{e^{-i\pi/3}}=e^{-i\pi/6}$? – Did May 23 at 14:46

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
@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
@AgustíRoig Quite so, but $\sqrt{1+i\sqrt 3}$ does not have a positive square root. – MJD Sep 27 '12 at 18:17

$\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?

$1 + i \sqrt 3 = 2 \exp \left( \dfrac {\pi}{3}i + 2 \pi n i \right) \quad \{ n \in \mathbb Z \}$

$\sqrt{1 + i \sqrt 3} = \sqrt 2 \exp \left( \dfrac {\pi}{6}i + \pi n i \right) \quad \{ n \in \mathbb Z \}$

$\sqrt{1 + i \sqrt 3} = \pm \left( \dfrac{\sqrt 6}{2} + \dfrac{\sqrt 2}{2} i \right)$


$\sqrt{1 - i \sqrt 3} = \pm \left( \dfrac{\sqrt 6}{2} - \dfrac{\sqrt 2}{2} i \right)$

So there are four possible values of $\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3}$

One of then is $\sqrt 6$.

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I prefer this answer because it does not attempt to make the right-hand side of the equation equal $ \ \sqrt{6} $ . Of course the sum of the left-hand side is equal to that, but the sum of the square-root(s) of a complex number and the square-root(s) of its conjugate must yield two real and two imaginary values. (The question of "the square-root" of a complex number has arisen many times on this site.) It is not clear Leibniz would have known about DeMoivre's "formula": he derived it in 1707, the "calculus priority" dispute broke out the next year, and (continued) – RecklessReckoner May 23 at 2:04
Leibniz was living "under a cloud" for the remaining few years of his life, while a rift between English (DeMoivre being there) and Continental mathematicians developed. The careful handling of complex values was still emerging, so it seems reasonable that Leibniz would have found the one (positive) value from the sum of the two (positive) square-root values and concluded this was complete. – RecklessReckoner May 23 at 2:07
@RecklessReckoner - Even now we refer to $i$ as the square root of $-1$ and in the next breath say that every non zero complex number has two square roots. – Steven Gregory May 23 at 5:35
Certainly we should say "a square root..." or "the 'positive' square root...". I think the "positive" one is intended when we write $ \ i \ = \ \sqrt{-1} \ $ , but there is a tendency to state that and fail to mention "the other one", since we only have one for, say, $ \ \sqrt{6} \ $ . – RecklessReckoner May 23 at 5:43
@RecklessReckoner - It's interesting to me that we think of $i$ as positive and $-i$ as negative. – Steven Gregory May 23 at 11:01

Let $x=\sqrt{1+i\sqrt3}+\sqrt{1-i\sqrt3}$. Then






Thus $x\in\{\sqrt6,-\sqrt6,i\sqrt2,-i\sqrt2\}$, so it remains to determine which of these four roots is meant.

The answer depends on which convention you decide to use when computing the square root of a complex number, $\sqrt{a+ib}$ with $b\not=0$. There are two common conventions: require $\sqrt{a+ib}$ with $b\not=0$ to have positive real part, or require it to have positve imaginary part. (In terms of polar coordinates, this amounts to saying $\sqrt{re^{i\theta}}=\sqrt re^{i\theta/2}$, but with $-\pi\lt\theta\le\pi$ for the first convention and $0\le\theta\lt2\pi$ for the second.)

If we assume the first convention, then $x$, being the sum of two square roots, must have positive real part, hence must equal $\sqrt6$. If we assume the second convention, then $x$ must have positive imaginary part, hence must equal $i\sqrt2$. Since the desired answer is $\sqrt6$, we see that the problem is tacitly assuming the first convention.

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+1, if only because this avoids the $\sqrt{z}\sqrt{w}=\sqrt{zw}$ fallacy. (I know, Barry, no big deal in real life, but on this site, well worth an upvote...) – Did May 23 at 14:44

I'm pretty new to proofs, but it seems like you could just evaluate $\sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}}$ and if it equals $\sqrt{6}$, then it's proof enough (though this is admittedly a convoluted step-by-step that I derived from WolframAlpha):

\begin{aligned} \sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}} &= \sqrt{\frac{3}{2} + i\sqrt{3} -\frac{1}{2}} + \sqrt{\frac{3}{2} - i\sqrt{3} - \frac{1}{2}} \\ &= \sqrt{\frac{9}{6} + \frac{6i\sqrt{3}}{6} -\frac{3}{6}} + \sqrt{\frac{9}{6}-\frac{6i\sqrt{3}}{6}-\frac{3}{6}} \\ &= \sqrt{\frac{9+6i\sqrt{3}-3}{6}} + \sqrt{\frac{9-6i\sqrt{3}-3}{6}} \\ &= \sqrt{\frac{9+6i\sqrt{3}+(i\sqrt{3})^2}{6}} + \sqrt{\frac{9-6i\sqrt{3}+(i\sqrt{3})^2}{6}} \\ &= \sqrt{\frac{(3+i\sqrt{3})^2}{6}}+\sqrt{\frac{(3-i\sqrt{3})^2}{6}} \\ &= \frac{\sqrt{(3+i\sqrt{3})^2}}{\sqrt{6}}+\frac{\sqrt{(3-i\sqrt{3})^2}}{\sqrt{6}} \\ &= \frac{3+i\sqrt{3}}{\sqrt{6}}+\frac{3-i\sqrt{3}}{\sqrt{6}} \\ &= \frac{\sqrt{6}(3+i\sqrt{3})}{6}+\frac{\sqrt{6}(3-i\sqrt{3})}{6} \\ &= \frac{3\sqrt{6}+3i\sqrt{2}}{6}+\frac{3\sqrt{6}-3i\sqrt{2}}{6} \\ &= \frac{3\sqrt{6}+3i\sqrt{2}+3\sqrt{6}-3i\sqrt{2}}{6} \\ &= \frac{6\sqrt{6}}{6} = \sqrt{6} \end{aligned}

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Why is $\sqrt{z^2}=z$ (line 7 of the set of displayed equations)? – Did May 23 at 14:38


Is an addition of complex conjugates of the form

$(a+b\sqrt{-1}) + (a-b\sqrt{-1})$

So then

$1+\sqrt{-3}=(a+b\sqrt{-1})^2$ $=a^2+2ab\sqrt{-1}-b^2$

If we equal the real and complex parts we get:

$1=a^2-b^2$ and $\sqrt{-3}=2ab\sqrt{-1}$

Solving for a in the complex equation gives: $a=\frac{\sqrt{3}}{2b}$

Plugging in $a=\frac{\sqrt{3}}{2b}$ into $1=a^2-b^2$ and multiplying both sides by $b^2$ yields:

$b^4+b^2-\frac{3}{4}$ which is a quadratic for $b^2$

The quadratic equations then gives $b^2=\frac{-1\pm2}{2}$

Using only the positive root means $b=\frac{\sqrt{2}}{2}$

Plug in $b=\frac{\sqrt{2}}{2}$ into $1=a^2-b^2$ to get $a=\frac{\sqrt{6}}{2}$



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Please format using MathJax. Your current answer is very hard to read. – M47145 May 23 at 0:19
Done. Sorry, It's my first contribution to – Carlos May 23 at 1:33
Thanks! That's much better! :) – M47145 May 23 at 1:37
Why is the square root of the conjugate equal to the conjugate of the square root (line 3)? – Did May 23 at 14:42

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