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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.2012-09-27
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    Then be so kind and provide a correct one...2012-09-27

6 Answers 6

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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 $e^{7i\pi/6}$?2012-09-27
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    @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.2012-09-27
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    Oh if only we could override accepted answers that are not in fact correct!2016-05-23
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    How do we know that $\sqrt{e^{i\pi/3}}=e^{i\pi/6}$ and $\sqrt{e^{-i\pi/3}}=e^{-i\pi/6}$?2016-05-23
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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 $e^{7i\pi/6}$?2012-09-27
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    Where's is the difference to Ragib's answer?2012-09-27
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    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.2012-09-27
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    @MJD: good question! Indeed why not use $\exp(7\pi i/6)$ for one and $\exp(-\pi i/6)$ for the other?2012-09-27
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    @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?2012-09-27
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    @RagibZaman: Of course not wrong, and certainly not completely wrong. However, in an answer a brief mention that one is picking the principal branch may be useful. We are all familiar with proofs of $1=0$ though plausible manipulation of square roots of complex numbers.2012-09-27
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    @AndréNicolas I did not mean to sound rude to you, in case that is how it came off. I was just a bit peeved that I was downvoted (and my answer was said to be "wrong") because I didn't mention that I am using the principal branch. I thought the polite thing to do would have been to leave a comment saying that it may be helpful to clarify that point.2012-09-27
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    @RagibZaman: In case you are wondering whether I downvoted, the answer is no. For one thing, I have not downvoted any answer for $18$ months, maybe more. The comment to MJD was partly for fun: it is natural to choose the *same* branch.2012-09-27
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    @AgustíRoig Quite so, but $\sqrt{1+i\sqrt 3}$ does not *have* a positive square root.2012-09-27
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$\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?

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

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

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

Similarly

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

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)2016-05-23
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    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.2016-05-23
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    @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.2016-05-23
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    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} \ $ .2016-05-23
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    @RecklessReckoner - It's interesting to me that we think of $i$ as positive and $-i$ as negative.2016-05-23
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    @StevenGregory Many of us refer to $i$ as the **principal** square root of $-1$, or as **a** square root of $-1$, not as the square root of $-1$. And I did not meet many people thinking of $i$ and $-i$ as positive or negative. Sorry.2016-05-23
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    @Did - I don't either. Which is why I made a joke with RecklessReckoner . I have however told hundreds of college and high school students that it is technically wrong to call $i$ __the__ square root of negative one. I have even heard student making fun of me by saying "__principal__ squre root."2016-05-23
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    I think the matter of considering $ \ i \ $ as "positive" comes from the fact that $ \ 0 \ + \ 1 \cdot i \ $ lies on the positive $ \ Im(z) \ $ axis. It becomes rather a subtlety to beginning students that a sense can be given to ordering complex numbers with zero imaginary part or zero real part, but not elsewhere on the complex plane. And saying $ \ i \ $ is "the square root of $ \ -1 \ $ with positive imaginary part" sounds cumbersome. So many popular math books as well as textbooks repeat " $ \ i \ = \ \sqrt{-1} \ $ " that it becomes quite a job to correct people on the point.2016-05-23
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Let $x=\sqrt{1+i\sqrt3}+\sqrt{1-i\sqrt3}$. Then

$$x^2=(1+i\sqrt3)+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}+(1-i\sqrt3)=2+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}$$

so

$$(x^2-2)^2=4(1+i\sqrt3)(1-i\sqrt3)=4(1+3)=16$$

so

$$x^4-4x^2-12=(x^2-6)(x^2+2)=0$$

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...)2016-05-23
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$\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=\sqrt{6}$

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 $a=\frac{\sqrt{6}}{2}$

Therefore

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

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    Please format using MathJax. Your current answer is very hard to read.2016-05-23
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    Done. Sorry, It's my first contribution to math.stackexchange.com2016-05-23
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    Thanks! That's much better! :)2016-05-23
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    Why is the square root of the conjugate equal to the conjugate of the square root (line 3)?2016-05-23
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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)?2016-05-23