<s>[INST] Prove that $[\mathbb{Q}(\sqrt{ 3+4i}+\sqrt{ 3-4i}):\mathbb{Q} ] =1 $ let $\sqrt{ 3+  4i}$   denote the square root of the complex number $3 + 4i$ that lies in the first quadrant and let    $\sqrt{ 3-  4i}$   denote the square root of $3 - 4i$ that lies in the fourth quadrant. Prove that $[\mathbb{Q}(\sqrt{ 3+4i}+\sqrt{ 3-4i}):\mathbb{Q} ] =1    $
My attempt:         $\sqrt{ 3+  4i}=\sqrt{ 3+  \sqrt{(-16)}}$
$\sqrt{ 3-  4i}=\sqrt{ 3-  \sqrt{(-16)}}$
take       $x=\sqrt{ 3+  \sqrt{(-16)}}+\sqrt{ 3-  \sqrt{(-16)}}$
taking square on both sides,we get
$x^2=3+  \sqrt{(-16)}+3-  \sqrt{(-16)}+2\sqrt{ 3+  \sqrt{(-16)}}\sqrt{ 3-  \sqrt{(-16)}}$
$x^2=6+10=16$ this implies $x=4$
degree of $4$ is $1$
Therefore $ [\mathbb{Q(4):\mathbb{Q}}]=1$
Is its correct ?
 [/INST] It is enough to verify that $\sqrt{3+4i}=2+i$ and $\sqrt{3-4i}=2-i$ from which
$$\sqrt{ 3+4i}+\sqrt{ 3-4i}=4\in\mathbb Q$$
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