Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$

Question

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

Answer 1

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 $$ \sqrt{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right)= 2 \sqrt{2} \cos(\pi/6)= 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}= \sqrt{6}. $$

Answer 2

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.

Answer 3

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

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

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

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

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 $\sqrt 6$.

Answer 4

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

Answer 5

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

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

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

Answer 6

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}