Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$
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9Then be so kind and provide a correct one... – 2012-09-27
6 Answers
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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0How 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
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+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
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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1@AgustíRoig Quite so, but $\sqrt{1+i\sqrt 3}$ does not *have* a positive square root. – 2012-09-27
$\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?
$1 + i \sqrt 3 = 2 \exp \left( \dfrac {\pi}{3}i + 2 \pi n i \right) \quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} = \sqrt 2 \exp \left( \dfrac {\pi}{6}i + \pi n i \right) \quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} = \pm \left( \dfrac{\sqrt 6}{2} + \dfrac{\sqrt 2}{2} i \right)$
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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0I 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
$\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=\sqrt{6}$
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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0Why is the square root of the conjugate equal to the conjugate of the square root (line 3)? – 2016-05-23
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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2Why is $\sqrt{z^2}=z$ (line 7 of the set of displayed equations)? – 2016-05-23